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Hypothesis Testing

Hypothesis testing is a tool for making statistical inferences about the population data. It is an analysis tool that tests assumptions and determines how likely something is within a given standard of accuracy. Hypothesis testing provides a way to verify whether the results of an experiment are valid.

A null hypothesis and an alternative hypothesis are set up before performing the hypothesis testing. This helps to arrive at a conclusion regarding the sample obtained from the population. In this article, we will learn more about hypothesis testing, its types, steps to perform the testing, and associated examples.

What is Hypothesis Testing in Statistics?

Hypothesis testing uses sample data from the population to draw useful conclusions regarding the population probability distribution . It tests an assumption made about the data using different types of hypothesis testing methodologies. The hypothesis testing results in either rejecting or not rejecting the null hypothesis.

Hypothesis Testing Definition

Hypothesis testing can be defined as a statistical tool that is used to identify if the results of an experiment are meaningful or not. It involves setting up a null hypothesis and an alternative hypothesis. These two hypotheses will always be mutually exclusive. This means that if the null hypothesis is true then the alternative hypothesis is false and vice versa. An example of hypothesis testing is setting up a test to check if a new medicine works on a disease in a more efficient manner.

Null Hypothesis

The null hypothesis is a concise mathematical statement that is used to indicate that there is no difference between two possibilities. In other words, there is no difference between certain characteristics of data. This hypothesis assumes that the outcomes of an experiment are based on chance alone. It is denoted as \(H_{0}\). Hypothesis testing is used to conclude if the null hypothesis can be rejected or not. Suppose an experiment is conducted to check if girls are shorter than boys at the age of 5. The null hypothesis will say that they are the same height.

Alternative Hypothesis

The alternative hypothesis is an alternative to the null hypothesis. It is used to show that the observations of an experiment are due to some real effect. It indicates that there is a statistical significance between two possible outcomes and can be denoted as \(H_{1}\) or \(H_{a}\). For the above-mentioned example, the alternative hypothesis would be that girls are shorter than boys at the age of 5.

Hypothesis Testing P Value

In hypothesis testing, the p value is used to indicate whether the results obtained after conducting a test are statistically significant or not. It also indicates the probability of making an error in rejecting or not rejecting the null hypothesis.This value is always a number between 0 and 1. The p value is compared to an alpha level, \(\alpha\) or significance level. The alpha level can be defined as the acceptable risk of incorrectly rejecting the null hypothesis. The alpha level is usually chosen between 1% to 5%.

Hypothesis Testing Critical region

All sets of values that lead to rejecting the null hypothesis lie in the critical region. Furthermore, the value that separates the critical region from the non-critical region is known as the critical value.

Hypothesis Testing Formula

Depending upon the type of data available and the size, different types of hypothesis testing are used to determine whether the null hypothesis can be rejected or not. The hypothesis testing formula for some important test statistics are given below:

• z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). \(\overline{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation and n is the size of the sample.
• t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\). s is the sample standard deviation.
• \(\chi ^{2} = \sum \frac{(O_{i}-E_{i})^{2}}{E_{i}}\). \(O_{i}\) is the observed value and \(E_{i}\) is the expected value.

We will learn more about these test statistics in the upcoming section.

Types of Hypothesis Testing

Selecting the correct test for performing hypothesis testing can be confusing. These tests are used to determine a test statistic on the basis of which the null hypothesis can either be rejected or not rejected. Some of the important tests used for hypothesis testing are given below.

Hypothesis Testing Z Test

A z test is a way of hypothesis testing that is used for a large sample size (n ≥ 30). It is used to determine whether there is a difference between the population mean and the sample mean when the population standard deviation is known. It can also be used to compare the mean of two samples. It is used to compute the z test statistic. The formulas are given as follows:

• One sample: z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).
• Two samples: z = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\).

Hypothesis Testing t Test

The t test is another method of hypothesis testing that is used for a small sample size (n < 30). It is also used to compare the sample mean and population mean. However, the population standard deviation is not known. Instead, the sample standard deviation is known. The mean of two samples can also be compared using the t test.

• One sample: t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\).
• Two samples: t = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}\).

Hypothesis Testing Chi Square

The Chi square test is a hypothesis testing method that is used to check whether the variables in a population are independent or not. It is used when the test statistic is chi-squared distributed.

One Tailed Hypothesis Testing

One tailed hypothesis testing is done when the rejection region is only in one direction. It can also be known as directional hypothesis testing because the effects can be tested in one direction only. This type of testing is further classified into the right tailed test and left tailed test.

Right Tailed Hypothesis Testing

The right tail test is also known as the upper tail test. This test is used to check whether the population parameter is greater than some value. The null and alternative hypotheses for this test are given as follows:

\(H_{0}\): The population parameter is ≤ some value

\(H_{1}\): The population parameter is > some value.

If the test statistic has a greater value than the critical value then the null hypothesis is rejected

Left Tailed Hypothesis Testing

The left tail test is also known as the lower tail test. It is used to check whether the population parameter is less than some value. The hypotheses for this hypothesis testing can be written as follows:

\(H_{0}\): The population parameter is ≥ some value

\(H_{1}\): The population parameter is < some value.

The null hypothesis is rejected if the test statistic has a value lesser than the critical value.

Two Tailed Hypothesis Testing

In this hypothesis testing method, the critical region lies on both sides of the sampling distribution. It is also known as a non - directional hypothesis testing method. The two-tailed test is used when it needs to be determined if the population parameter is assumed to be different than some value. The hypotheses can be set up as follows:

\(H_{0}\): the population parameter = some value

\(H_{1}\): the population parameter ≠ some value

The null hypothesis is rejected if the test statistic has a value that is not equal to the critical value.

Hypothesis Testing Steps

Hypothesis testing can be easily performed in five simple steps. The most important step is to correctly set up the hypotheses and identify the right method for hypothesis testing. The basic steps to perform hypothesis testing are as follows:

• Step 1: Set up the null hypothesis by correctly identifying whether it is the left-tailed, right-tailed, or two-tailed hypothesis testing.
• Step 2: Set up the alternative hypothesis.
• Step 3: Choose the correct significance level, \(\alpha\), and find the critical value.
• Step 4: Calculate the correct test statistic (z, t or \(\chi\)) and p-value.
• Step 5: Compare the test statistic with the critical value or compare the p-value with \(\alpha\) to arrive at a conclusion. In other words, decide if the null hypothesis is to be rejected or not.

Hypothesis Testing Example

The best way to solve a problem on hypothesis testing is by applying the 5 steps mentioned in the previous section. Suppose a researcher claims that the mean average weight of men is greater than 100kgs with a standard deviation of 15kgs. 30 men are chosen with an average weight of 112.5 Kgs. Using hypothesis testing, check if there is enough evidence to support the researcher's claim. The confidence interval is given as 95%.

Step 1: This is an example of a right-tailed test. Set up the null hypothesis as \(H_{0}\): \(\mu\) = 100.

Step 2: The alternative hypothesis is given by \(H_{1}\): \(\mu\) > 100.

Step 3: As this is a one-tailed test, \(\alpha\) = 100% - 95% = 5%. This can be used to determine the critical value.

1 - \(\alpha\) = 1 - 0.05 = 0.95

0.95 gives the required area under the curve. Now using a normal distribution table, the area 0.95 is at z = 1.645. A similar process can be followed for a t-test. The only additional requirement is to calculate the degrees of freedom given by n - 1.

Step 4: Calculate the z test statistic. This is because the sample size is 30. Furthermore, the sample and population means are known along with the standard deviation.

z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).

\(\mu\) = 100, \(\overline{x}\) = 112.5, n = 30, \(\sigma\) = 15

z = \(\frac{112.5-100}{\frac{15}{\sqrt{30}}}\) = 4.56

Step 5: Conclusion. As 4.56 > 1.645 thus, the null hypothesis can be rejected.

Hypothesis Testing and Confidence Intervals

Confidence intervals form an important part of hypothesis testing. This is because the alpha level can be determined from a given confidence interval. Suppose a confidence interval is given as 95%. Subtract the confidence interval from 100%. This gives 100 - 95 = 5% or 0.05. This is the alpha value of a one-tailed hypothesis testing. To obtain the alpha value for a two-tailed hypothesis testing, divide this value by 2. This gives 0.05 / 2 = 0.025.

Related Articles:

• Probability and Statistics
• Data Handling

Important Notes on Hypothesis Testing

• Hypothesis testing is a technique that is used to verify whether the results of an experiment are statistically significant.
• It involves the setting up of a null hypothesis and an alternate hypothesis.
• There are three types of tests that can be conducted under hypothesis testing - z test, t test, and chi square test.
• Hypothesis testing can be classified as right tail, left tail, and two tail tests.

Examples on Hypothesis Testing

• Example 1: The average weight of a dumbbell in a gym is 90lbs. However, a physical trainer believes that the average weight might be higher. A random sample of 5 dumbbells with an average weight of 110lbs and a standard deviation of 18lbs. Using hypothesis testing check if the physical trainer's claim can be supported for a 95% confidence level. Solution: As the sample size is lesser than 30, the t-test is used. \(H_{0}\): \(\mu\) = 90, \(H_{1}\): \(\mu\) > 90 \(\overline{x}\) = 110, \(\mu\) = 90, n = 5, s = 18. \(\alpha\) = 0.05 Using the t-distribution table, the critical value is 2.132 t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\) t = 2.484 As 2.484 > 2.132, the null hypothesis is rejected. Answer: The average weight of the dumbbells may be greater than 90lbs
• Example 2: The average score on a test is 80 with a standard deviation of 10. With a new teaching curriculum introduced it is believed that this score will change. On random testing, the score of 38 students, the mean was found to be 88. With a 0.05 significance level, is there any evidence to support this claim? Solution: This is an example of two-tail hypothesis testing. The z test will be used. \(H_{0}\): \(\mu\) = 80, \(H_{1}\): \(\mu\) ≠ 80 \(\overline{x}\) = 88, \(\mu\) = 80, n = 36, \(\sigma\) = 10. \(\alpha\) = 0.05 / 2 = 0.025 The critical value using the normal distribution table is 1.96 z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\) z = \(\frac{88-80}{\frac{10}{\sqrt{36}}}\) = 4.8 As 4.8 > 1.96, the null hypothesis is rejected. Answer: There is a difference in the scores after the new curriculum was introduced.
• Example 3: The average score of a class is 90. However, a teacher believes that the average score might be lower. The scores of 6 students were randomly measured. The mean was 82 with a standard deviation of 18. With a 0.05 significance level use hypothesis testing to check if this claim is true. Solution: The t test will be used. \(H_{0}\): \(\mu\) = 90, \(H_{1}\): \(\mu\) < 90 \(\overline{x}\) = 110, \(\mu\) = 90, n = 6, s = 18 The critical value from the t table is -2.015 t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\) t = \(\frac{82-90}{\frac{18}{\sqrt{6}}}\) t = -1.088 As -1.088 > -2.015, we fail to reject the null hypothesis. Answer: There is not enough evidence to support the claim.

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FAQs on Hypothesis Testing

What is hypothesis testing.

Hypothesis testing in statistics is a tool that is used to make inferences about the population data. It is also used to check if the results of an experiment are valid.

What is the z Test in Hypothesis Testing?

The z test in hypothesis testing is used to find the z test statistic for normally distributed data . The z test is used when the standard deviation of the population is known and the sample size is greater than or equal to 30.

What is the t Test in Hypothesis Testing?

The t test in hypothesis testing is used when the data follows a student t distribution . It is used when the sample size is less than 30 and standard deviation of the population is not known.

What is the formula for z test in Hypothesis Testing?

The formula for a one sample z test in hypothesis testing is z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\) and for two samples is z = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\).

What is the p Value in Hypothesis Testing?

The p value helps to determine if the test results are statistically significant or not. In hypothesis testing, the null hypothesis can either be rejected or not rejected based on the comparison between the p value and the alpha level.

What is One Tail Hypothesis Testing?

When the rejection region is only on one side of the distribution curve then it is known as one tail hypothesis testing. The right tail test and the left tail test are two types of directional hypothesis testing.

What is the Alpha Level in Two Tail Hypothesis Testing?

To get the alpha level in a two tail hypothesis testing divide \(\alpha\) by 2. This is done as there are two rejection regions in the curve.

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S.3 hypothesis testing.

In reviewing hypothesis tests, we start first with the general idea. Then, we keep returning to the basic procedures of hypothesis testing, each time adding a little more detail.

The general idea of hypothesis testing involves:

• Making an initial assumption.
• Collecting evidence (data).
• Based on the available evidence (data), deciding whether to reject or not reject the initial assumption.

Every hypothesis test — regardless of the population parameter involved — requires the above three steps.

Example S.3.1

Is normal body temperature really 98.6 degrees f section  .

Consider the population of many, many adults. A researcher hypothesized that the average adult body temperature is lower than the often-advertised 98.6 degrees F. That is, the researcher wants an answer to the question: "Is the average adult body temperature 98.6 degrees? Or is it lower?" To answer his research question, the researcher starts by assuming that the average adult body temperature was 98.6 degrees F.

Then, the researcher went out and tried to find evidence that refutes his initial assumption. In doing so, he selects a random sample of 130 adults. The average body temperature of the 130 sampled adults is 98.25 degrees.

Then, the researcher uses the data he collected to make a decision about his initial assumption. It is either likely or unlikely that the researcher would collect the evidence he did given his initial assumption that the average adult body temperature is 98.6 degrees:

• If it is likely , then the researcher does not reject his initial assumption that the average adult body temperature is 98.6 degrees. There is not enough evidence to do otherwise.
• either the researcher's initial assumption is correct and he experienced a very unusual event;
• or the researcher's initial assumption is incorrect.

In statistics, we generally don't make claims that require us to believe that a very unusual event happened. That is, in the practice of statistics, if the evidence (data) we collected is unlikely in light of the initial assumption, then we reject our initial assumption.

Example S.3.2

Criminal trial analogy section  .

One place where you can consistently see the general idea of hypothesis testing in action is in criminal trials held in the United States. Our criminal justice system assumes "the defendant is innocent until proven guilty." That is, our initial assumption is that the defendant is innocent.

In the practice of statistics, we make our initial assumption when we state our two competing hypotheses -- the null hypothesis ( H 0 ) and the alternative hypothesis ( H A ). Here, our hypotheses are:

• H 0 : Defendant is not guilty (innocent)
• H A : Defendant is guilty

In statistics, we always assume the null hypothesis is true . That is, the null hypothesis is always our initial assumption.

The prosecution team then collects evidence — such as finger prints, blood spots, hair samples, carpet fibers, shoe prints, ransom notes, and handwriting samples — with the hopes of finding "sufficient evidence" to make the assumption of innocence refutable.

In statistics, the data are the evidence.

The jury then makes a decision based on the available evidence:

• If the jury finds sufficient evidence — beyond a reasonable doubt — to make the assumption of innocence refutable, the jury rejects the null hypothesis and deems the defendant guilty. We behave as if the defendant is guilty.
• If there is insufficient evidence, then the jury does not reject the null hypothesis . We behave as if the defendant is innocent.

In statistics, we always make one of two decisions. We either "reject the null hypothesis" or we "fail to reject the null hypothesis."

Errors in Hypothesis Testing Section  

Did you notice the use of the phrase "behave as if" in the previous discussion? We "behave as if" the defendant is guilty; we do not "prove" that the defendant is guilty. And, we "behave as if" the defendant is innocent; we do not "prove" that the defendant is innocent.

This is a very important distinction! We make our decision based on evidence not on 100% guaranteed proof. Again:

• If we reject the null hypothesis, we do not prove that the alternative hypothesis is true.
• If we do not reject the null hypothesis, we do not prove that the null hypothesis is true.

We merely state that there is enough evidence to behave one way or the other. This is always true in statistics! Because of this, whatever the decision, there is always a chance that we made an error .

Let's review the two types of errors that can be made in criminal trials:

Table S.3.2 shows how this corresponds to the two types of errors in hypothesis testing.

Note that, in statistics, we call the two types of errors by two different  names -- one is called a "Type I error," and the other is called  a "Type II error." Here are the formal definitions of the two types of errors:

There is always a chance of making one of these errors. But, a good scientific study will minimize the chance of doing so!

Making the Decision Section  

Recall that it is either likely or unlikely that we would observe the evidence we did given our initial assumption. If it is likely , we do not reject the null hypothesis. If it is unlikely , then we reject the null hypothesis in favor of the alternative hypothesis. Effectively, then, making the decision reduces to determining "likely" or "unlikely."

In statistics, there are two ways to determine whether the evidence is likely or unlikely given the initial assumption:

• We could take the " critical value approach " (favored in many of the older textbooks).
• Or, we could take the " P -value approach " (what is used most often in research, journal articles, and statistical software).

In the next two sections, we review the procedures behind each of these two approaches. To make our review concrete, let's imagine that μ is the average grade point average of all American students who major in mathematics. We first review the critical value approach for conducting each of the following three hypothesis tests about the population mean $\mu$:

In Practice

• We would want to conduct the first hypothesis test if we were interested in concluding that the average grade point average of the group is more than 3.
• We would want to conduct the second hypothesis test if we were interested in concluding that the average grade point average of the group is less than 3.
• And, we would want to conduct the third hypothesis test if we were only interested in concluding that the average grade point average of the group differs from 3 (without caring whether it is more or less than 3).

Upon completing the review of the critical value approach, we review the P -value approach for conducting each of the above three hypothesis tests about the population mean \(\mu\). The procedures that we review here for both approaches easily extend to hypothesis tests about any other population parameter.

Statistics Made Easy

4 Examples of Hypothesis Testing in Real Life

In statistics, hypothesis tests are used to test whether or not some hypothesis about a population parameter is true.

To perform a hypothesis test in the real world, researchers will obtain a random sample from the population and perform a hypothesis test on the sample data, using a null and alternative hypothesis:

• Null Hypothesis (H 0 ): The sample data occurs purely from chance.
• Alternative Hypothesis (H A ): The sample data is influenced by some non-random cause.

If the p-value of the hypothesis test is less than some significance level (e.g. α = .05), then we can reject the null hypothesis and conclude that we have sufficient evidence to say that the alternative hypothesis is true.

The following examples provide several situations where hypothesis tests are used in the real world.

Example 1: Biology

Hypothesis tests are often used in biology to determine whether some new treatment, fertilizer, pesticide, chemical, etc. causes increased growth, stamina, immunity, etc. in plants or animals.

For example, suppose a biologist believes that a certain fertilizer will cause plants to grow more during a one-month period than they normally do, which is currently 20 inches. To test this, she applies the fertilizer to each of the plants in her laboratory for one month.

She then performs a hypothesis test using the following hypotheses:

• H 0 : μ = 20 inches (the fertilizer will have no effect on the mean plant growth)
• H A : μ > 20 inches (the fertilizer will cause mean plant growth to increase)

If the p-value of the test is less than some significance level (e.g. α = .05), then she can reject the null hypothesis and conclude that the fertilizer leads to increased plant growth.

Example 2: Clinical Trials

Hypothesis tests are often used in clinical trials to determine whether some new treatment, drug, procedure, etc. causes improved outcomes in patients.

For example, suppose a doctor believes that a new drug is able to reduce blood pressure in obese patients. To test this, he may measure the blood pressure of 40 patients before and after using the new drug for one month.

He then performs a hypothesis test using the following hypotheses:

• H 0 : μ after = μ before (the mean blood pressure is the same before and after using the drug)
• H A : μ after < μ before (the mean blood pressure is less after using the drug)

If the p-value of the test is less than some significance level (e.g. α = .05), then he can reject the null hypothesis and conclude that the new drug leads to reduced blood pressure.

Example 3: Advertising Spend

Hypothesis tests are often used in business to determine whether or not some new advertising campaign, marketing technique, etc. causes increased sales.

For example, suppose a company believes that spending more money on digital advertising leads to increased sales. To test this, the company may increase money spent on digital advertising during a two-month period and collect data to see if overall sales have increased.

They may perform a hypothesis test using the following hypotheses:

• H 0 : μ after = μ before (the mean sales is the same before and after spending more on advertising)
• H A : μ after > μ before (the mean sales increased after spending more on advertising)

If the p-value of the test is less than some significance level (e.g. α = .05), then the company can reject the null hypothesis and conclude that increased digital advertising leads to increased sales.

Example 4: Manufacturing

Hypothesis tests are also used often in manufacturing plants to determine if some new process, technique, method, etc. causes a change in the number of defective products produced.

For example, suppose a certain manufacturing plant wants to test whether or not some new method changes the number of defective widgets produced per month, which is currently 250. To test this, they may measure the mean number of defective widgets produced before and after using the new method for one month.

They can then perform a hypothesis test using the following hypotheses:

• H 0 : μ after = μ before (the mean number of defective widgets is the same before and after using the new method)
• H A : μ after ≠ μ before (the mean number of defective widgets produced is different before and after using the new method)

If the p-value of the test is less than some significance level (e.g. α = .05), then the plant can reject the null hypothesis and conclude that the new method leads to a change in the number of defective widgets produced per month.

Additional Resources

Introduction to Hypothesis Testing Introduction to the One Sample t-test Introduction to the Two Sample t-test Introduction to the Paired Samples t-test

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Calcworkshop

Hypothesis Testing w/ 21 Step-by-Step Examples!

// Last Updated: October 9, 2020 - Watch Video //

In statistical testing, also referred to as hypothesis testing, our goal is to show the credibility of a claim regarding the population.

Jenn, Founder Calcworkshop ® , 15+ Years Experience (Licensed & Certified Teacher)

What Is Hypothesis Testing

Now it would be unreasonable to assume that we can test the entire population to determine the feasibility of every claim one might have.

Thus, we need a way to conclude an assumption is true or false by taking an appropriate sample and calculating a relevant statistic.

And knowing that we must expect that there will be some variation between the sample statistic that is calculated and the true population parameter, leads us to the understanding of statistical inferences (hypotheses).

Hypothesis Testing Steps

First, we must identify the parameter of interest.

Remember that a parameter always points to the population so that it will be either a population mean, population proportion, population slope, or some other population parameter.

Types of Hypothesis Tests

Then we will write a declaration of our significance test, which will include a null hypothesis statement and an alternative hypothesis.

The null hypothesis is the expected value of the population parameter, similar to the status quo, whereas the alternative hypothesis is a statement of negation of the null hypothesis as discussed by Penn State .

Next, we will calculate the desired test statistic from our random sample. This test statistic is a numerical quantity that measures the difference between the observed value and the expected value, divided by the standard error, which is the sample standard deviation.

Then we will compare this test statistic with a specified level of significance (alpha), just like we did with confidence intervals.

If the probability of yielding the sample statistic is as extreme or more extreme is smaller than our significance level, then we declare the sample statistic to be significant and reject the null hypothesis in favor of the alternative. In other words, if the probability is inside our shaded critical region then it is considered more extreme; thus, rejecting the hypothesis. But if it is outside the critical region, we will fail to reject our claim in favor of the alternative.

Null and Alternative Hypothesis

Additionally, we will also learn how to determine whether our study calls for a one or two-tailed test.

Type 1 And Type 2 Errors

Now, with all inferences and tests of significance, there is always room for error. A Type I error occurs if we reject the null hypothesis, when in actuality, the null hypothesis is true. Similarly, if we fail to reject the null hypothesis when, in reality, the null hypothesis is false, this is considered a Type II error .

Type 1 Vs. Type 2 Error

Imagine you are in a court of law, where a defendant is presumed innocent until proven guilty. What possible errors could a jury make regarding the outcome of the trial?

First, let’s state the following:

• The Null Hypothesis: The defendant is innocent.
• The Alternative Hypothesis: the defendant is guilty.

Now, a Type I Error would happen if the jury rejects the null hypothesis as false when, in reality, the null hypothesis is true. In other words, the jury finds the defendant guilty of a crime they didn’t commit.

And a Type II Error is when a jury accepts the null hypothesis as true when, in reality, the null hypothesis is false. Meaning, the defendant is found innocent of a crime they did commit.

Let’s look at an example where we put all of these ideas together.

Worked Example

Imagine we have a textile manufacturer investigating a new yarn, which claims it has a thread elongation of 12 kilograms with a standard deviation of 0.5 kilograms.

Using a random sample of 4 specimens, the manufacturer wishes to test the claim that the mean thread elongation is less than 12 kilograms.

Write a hypothesis statement for this scenario and using a normal distribution, find the Type 1 error if the sample mean is less than 11.5 kilograms.

Type 1 Error — Example

As we can see, from the example above, the likelihood of a type I error, where the manufacturer rejects the null hypothesis when the null hypothesis is actually true, is approximately 0.023 or 2.3% likely.

Together, we will look at these two types of error and how they affect decision-making and begin to explore the notion of a probability value and how it helps us determine the validity or falsity of our claim.

Hypothesis Testing – Lesson & Examples (Video)

1 hr 17 min

• Introduction to Video: Statistical Hypotheses
• 00:00:38 – Overview of Hypothesis Testing and determining a correctly stated hypothesis testing problem (Examples #1-7)
• Exclusive Content for Members Only
• 00:14:34 – State the Null Hypothesis and the Alternative Hypothesis for each scenario (Examples #8-12)
• 00:25:46 – Hypothesis Testing Steps and Overview of Type I and Type II errors (Examples #13-14)
• 00:40:32 – Describe a Type 1 and Type 2 error (Examples #15-16)
• 00:46:32 – Overview of p-value and Tails of the Hypothesis Test
• 00:55:55 – Find the probability of a Type I and Type II error (Example #17)
• 01:06:08 – Identify null hypothesis, alternative hypothesis, and state whether the scenario is a one-tail or two-tailed test (Examples #18-21)
• Practice Problems with Step-by-Step Solutions
• Chapter Tests with Video Solutions

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9.4 Full Hypothesis Test Examples

Tests on means, example 9.8.

Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds . His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims . For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal.

Set up the Hypothesis Test:

Since the problem is about a mean, this is a test of a single population mean .

H 0 : μ = 16.43   H a : μ < 16.43

For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "<" tells you this is left-tailed.

Determine the distribution needed:

Random variable: X ¯ X ¯ = the mean time to swim the 25-yard freestyle.

Distribution for the test: X ¯ X ¯ is normal (population standard deviation is known: σ = 0.8)

X ¯ ~ N ( μ , σ X n ) X ¯ ~ N ( μ , σ X n ) Therefore, X ¯ ~ N ( 16.43 , 0.8 15 ) X ¯ ~ N ( 16.43 , 0.8 15 )

μ = 16.43 comes from H 0 and not the data. σ = 0.8, and n = 15.

Calculate the p -value using the normal distribution for a mean:

p -value = P ( x ¯ x ¯ < 16) = 0.0187 where the sample mean in the problem is given as 16.

p -value = 0.0187 (This is called the actual level of significance .) The p -value is the area to the left of the sample mean is given as 16.

μ = 16.43 comes from H 0 . Our assumption is μ = 16.43.

Interpretation of the p -value: If H 0 is true , there is a 0.0187 probability (1.87%)that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.

Compare α and the p -value:

α = 0.05 p -value = 0.0187 α > p -value

Make a decision: Since α > α > p -value, reject H 0 .

This indicates that you reject the null hypothesis that the mean time to swim the 25-yard freestyle is at least 16.43 seconds.

Conclusion: At the 5% significance level, there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds. Thus, based on the sample data, we conclude that Jeffrey swims faster using the new goggles.

The Type I and Type II errors for this problem are as follows: The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in at least 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)

The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)

The mean throwing distance of a football for Marco, a high school quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset α = 0.05. Assume the throw distances for footballs are normal.

First, determine what type of test this is, set up the hypothesis test, find the p -value, sketch the graph, and state your conclusion.

Example 9.9

Jasmine has just begun her new job on the sales force of a very competitive company. In a sample of 16 sales calls it was found that she closed the contract for an average value of 108 dollars with a standard deviation of 12 dollars. Test at 5% significance that the population mean is at least 100 dollars against the alternative that it is less than 100 dollars. Company policy requires that new members of the sales force must exceed an average of $100 per contract during the trial employment period. Can we conclude that Jasmine has met this requirement at the significance level of 95%?

• H 0 : µ ≤ 100 H a : µ > 100 The null and alternative hypothesis are for the parameter µ because the number of dollars of the contracts is a continuous random variable. Also, this is a one-tailed test because the company has only an interested if the number of dollars per contact is below a particular number not "too high" a number. This can be thought of as making a claim that the requirement is being met and thus the claim is in the alternative hypothesis.
• Test statistic: t c = x ¯ − µ 0 s n = 108 − 100 ( 12 16 ) = 2.67 t c = x ¯ − µ 0 s n = 108 − 100 ( 12 16 ) = 2.67
• Critical value: t a = 1.753 t a = 1.753 with n-1 degrees of freedom= 15

The test statistic is a Student's t because the sample size is below 30; therefore, we cannot use the normal distribution. Comparing the calculated value of the test statistic and the critical value of t t ( t a ) ( t a ) at a 5% significance level, we see that the calculated value is in the tail of the distribution. Thus, we conclude that 108 dollars per contract is significantly larger than the hypothesized value of 100 and thus we cannot accept the null hypothesis. There is evidence that supports Jasmine's performance meets company standards.

It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, state your conclusion, and identify the Type I errors.

Example 9.10

A manufacturer of salad dressings uses machines to dispense liquid ingredients into bottles that move along a filling line. The machine that dispenses salad dressings is working properly when 8 ounces are dispensed. Suppose that the average amount dispensed in a particular sample of 35 bottles is 7.91 ounces with a variance of 0.03 ounces squared, s 2 s 2 . Is there evidence that the machine should be stopped and production wait for repairs? The lost production from a shutdown is potentially so great that management feels that the level of significance in the analysis should be 99%.

Again we will follow the steps in our analysis of this problem.

STEP 1 : Set the Null and Alternative Hypothesis. The random variable is the quantity of fluid placed in the bottles. This is a continuous random variable and the parameter we are interested in is the mean. Our hypothesis therefore is about the mean. In this case we are concerned that the machine is not filling properly. From what we are told it does not matter if the machine is over-filling or under-filling, both seem to be an equally bad error. This tells us that this is a two-tailed test: if the machine is malfunctioning it will be shutdown regardless if it is from over-filling or under-filling. The null and alternative hypotheses are thus:

STEP 2 : Decide the level of significance and draw the graph showing the critical value.

This problem has already set the level of significance at 99%. The decision seems an appropriate one and shows the thought process when setting the significance level. Management wants to be very certain, as certain as probability will allow, that they are not shutting down a machine that is not in need of repair. To draw the distribution and the critical value, we need to know which distribution to use. Because this is a continuous random variable and we are interested in the mean, and the sample size is greater than 30, the appropriate distribution is the normal distribution and the relevant critical value is 2.575 from the normal table or the t-table at 0.005 column and infinite degrees of freedom. We draw the graph and mark these points.

STEP 3 : Calculate sample parameters and the test statistic. The sample parameters are provided, the sample mean is 7.91 and the sample variance is .03 and the sample size is 35. We need to note that the sample variance was provided not the sample standard deviation, which is what we need for the formula. Remembering that the standard deviation is simply the square root of the variance, we therefore know the sample standard deviation, s, is 0.173. With this information we calculate the test statistic as -3.07, and mark it on the graph.

STEP 4 : Compare test statistic and the critical values Now we compare the test statistic and the critical value by placing the test statistic on the graph. We see that the test statistic is in the tail, decidedly greater than the critical value of 2.575. We note that even the very small difference between the hypothesized value and the sample value is still a large number of standard deviations. The sample mean is only 0.08 ounces different from the required level of 8 ounces, but it is 3 plus standard deviations away and thus we cannot accept the null hypothesis.

STEP 5 : Reach a Conclusion

Three standard deviations of a test statistic will guarantee that the test will fail. The probability that anything is within three standard deviations is almost zero. Actually it is 0.0026 on the normal distribution, which is certainly almost zero in a practical sense. Our formal conclusion would be “ At a 99% level of significance we cannot accept the hypothesis that the sample mean came from a distribution with a mean of 8 ounces” Or less formally, and getting to the point, “At a 99% level of significance we conclude that the machine is under filling the bottles and is in need of repair”.

Try It 9.10

A company records the mean time of employees working in a day. The mean comes out to be 475 minutes, with a standard deviation of 45 minutes. A manager recorded times of 20 employees. The times of working were (frequencies are in parentheses) 460(3); 465(2); 470(3); 475(1); 480(6); 485(3); 490(2).

Conduct a hypothesis test using a 2.5% level of significance to determine if the mean time is more than 475 .

Hypothesis Test for Proportions

Just as there were confidence intervals for proportions, or more formally, the population parameter p of the binomial distribution, there is the ability to test hypotheses concerning p .

The population parameter for the binomial is p . The estimated value (point estimate) for p is p′ where p′ = x/n , x is the number of successes in the sample and n is the sample size.

When you perform a hypothesis test of a population proportion p , you take a simple random sample from the population. The conditions for a binomial distribution must be met, which are: there are a certain number n of independent trials meaning random sampling, the outcomes of any trial are binary, success or failure, and each trial has the same probability of a success p . The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np′ and nq′ must both be greater than five ( np′ > 5 and nq′ > 5). In this case the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = np μ = np and σ = npq σ = npq . Remember that q = 1 – p q = 1 – p . There is no distribution that can correct for this small sample bias and thus if these conditions are not met we simply cannot test the hypothesis with the data available at that time. We met this condition when we first were estimating confidence intervals for p .

Again, we begin with the standardizing formula modified because this is the distribution of a binomial.

Substituting p 0 p 0 , the hypothesized value of p , we have:

This is the test statistic for testing hypothesized values of p , where the null and alternative hypotheses take one of the following forms:

The decision rule stated above applies here also: if the calculated value of Z c shows that the sample proportion is "too many" standard deviations from the hypothesized proportion, the null hypothesis cannot be accepted. The decision as to what is "too many" is pre-determined by the analyst depending on the level of significance required in the test.

Example 9.11

The mortgage department of a large bank is interested in the nature of loans of first-time borrowers. This information will be used to tailor their marketing strategy. They believe that 50% of first-time borrowers take out smaller loans than other borrowers. They perform a hypothesis test to determine if the percentage is the same or different from 50% . They sample 100 first-time borrowers and find 53 of these loans are smaller that the other borrowers. For the hypothesis test, they choose a 5% level of significance.

STEP 1 : Set the null and alternative hypothesis.

H 0 : p = 0.50   H a : p ≠ 0.50

The words "is the same or different from" tell you this is a two-tailed test. The Type I and Type II errors are as follows: The Type I error is to conclude that the proportion of borrowers is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true). The Type II error is there is not enough evidence to conclude that the proportion of first time borrowers differs from 50% when, in fact, the proportion does differ from 50%. (You fail to reject the null hypothesis when the null hypothesis is false.)

STEP 2 : Decide the level of significance and draw the graph showing the critical value

The level of significance has been set by the problem at the 5% level. Because this is two-tailed test one-half of the alpha value will be in the upper tail and one-half in the lower tail as shown on the graph. The critical value for the normal distribution at the 95% level of confidence is 1.96. This can easily be found on the student’s t-table at the very bottom at infinite degrees of freedom remembering that at infinity the t-distribution is the normal distribution. Of course the value can also be found on the normal table but you have go looking for one-half of 95 (0.475) inside the body of the table and then read out to the sides and top for the number of standard deviations.

STEP 3 : Calculate the sample parameters and critical value of the test statistic.

The test statistic is a normal distribution, Z, for testing proportions and is:

For this case, the sample of 100 found 53 of these loans were smaller than those of other borrowers. The sample proportion, p′ = 53/100= 0.53 The test question, therefore, is : “Is 0.53 significantly different from .50?” Putting these values into the formula for the test statistic we find that 0.53 is only 0.60 standard deviations away from .50. This is barely off of the mean of the standard normal distribution of zero. There is virtually no difference from the sample proportion and the hypothesized proportion in terms of standard deviations.

STEP 4 : Compare the test statistic and the critical value.

The calculated value is well within the critical values of ± 1.96 standard deviations and thus we cannot reject the null hypothesis. To reject the null hypothesis we need significant evident of difference between the hypothesized value and the sample value. In this case the sample value is very nearly the same as the hypothesized value measured in terms of standard deviations.

STEP 5 : Reach a conclusion

The formal conclusion would be “At a 5% level of significance we cannot reject the null hypothesis that 50% of first-time borrowers take out smaller loans than other borrowers.” Notice the length to which the conclusion goes to include all of the conditions that are attached to the conclusion. Statisticians, for all the criticism they receive, are careful to be very specific even when this seems trivial. Statisticians cannot say more than they know, and the data constrain the conclusion to be within the metes and bounds of the data.

Try It 9.11

A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. The teacher performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.

Example 9.12

Suppose a consumer group suspects that the proportion of households that have three or more cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three or more cell phones.

Here is an abbreviate version of the system to solve hypothesis tests applied to a test on a proportions.

Try It 9.12

Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the p -value, state your conclusion, and identify the Type I and Type II errors.

Example 9.13

The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.

1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98; 1.02; .95; .95 Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05.

Let’s follow a four-step process to answer this statistical question.

• H 0 : μ ≤ 1
• H a : μ > 1
• Plan : We are testing a sample mean without a known population standard deviation with less than 30 observations. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal.
• Do the calculations and draw the graph .
• State the Conclusions : We cannot accept the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.

Try It 9.13

The boiling point of a specific liquid is measured for 15 samples, and the boiling points are obtained as follows:

205; 206; 206; 202; 199; 194; 197; 198; 198; 201; 201; 202; 207; 211; 205

Is there convincing evidence that the average boiling point is greater than 200? Use a significance level of 0.1. Assume the population is normal.

Example 9.14

In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.

• H 0 : p ≤ 0.00034
• H a : p > 0.00034

If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.

• We will be testing a sample proportion with x = 172 and n = 420,019. The sample is sufficiently large because we have np' = 420,019(0.00034) = 142.8, nq' = 420,019(0.99966) = 419,876.2, two independent outcomes, and a fixed probability of success p' = 0.00034. Thus we will be able to generalize our results to the population.

Try It 9.14

In a study of 390,000 moisturizer users, 138 of the subjects developed skin diseases. Test the claim that moisturizer users developed skin diseases at a greater rate than that for non-moisturizer users (the rate of skin diseases for non-moisturizer users is 0.041%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.

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8.4: Hypothesis Test Examples for Proportions

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• In a hypothesis test problem, you may see words such as "the level of significance is 1%." The "1%" is the preconceived or preset \(\alpha\).
• The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data.
• If no level of significance is given, a common standard to use is \(\alpha = 0.05\).
• When you calculate the \(p\)-value and draw the picture, the \(p\)-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.
• The alternative hypothesis, \(H_{a}\), tells you if the test is left, right, or two-tailed. It is the key to conducting the appropriate test.
• \(H_{a}\) never has a symbol that contains an equal sign.
• Thinking about the meaning of the \(p\)-value: A data analyst (and anyone else) should have more confidence that he made the correct decision to reject the null hypothesis with a smaller \(p\)-value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large p -value such as 0.4, as opposed to a \(p\)-value of 0.056 (\(\alpha = 0.05\) is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules.

Full Hypothesis Test Examples

Example \(\PageIndex{7}\)

Joon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is the same or different from 50% . Joon samples 100 first-time brides and 53 reply that they are younger than their grooms. For the hypothesis test, she uses a 1% level of significance.

Set up the hypothesis test:

The 1% level of significance means that α = 0.01. This is a test of a single population proportion .

\(H_{0}: p = 0.50\)  \(H_{a}: p \neq 0.50\)

The words "is the same or different from" tell you this is a two-tailed test.

Calculate the distribution needed:

Random variable: \(P′ =\) the percent of of first-time brides who are younger than their grooms.

Distribution for the test: The problem contains no mention of a mean. The information is given in terms of percentages. Use the distribution for P′ , the estimated proportion.

\[P' - N\left(p, \sqrt{\frac{p-q}{n}}\right)\nonumber \]

\[P' - N\left(0.5, \sqrt{\frac{0.5-0.5}{100}}\right)\nonumber \]

where \(p = 0.50, q = 1−p = 0.50\), and \(n = 100\)

Calculate the p -value using the normal distribution for proportions:

\[p\text{-value} = P(p′ < 0.47 or p′ > 0.53) = 0.5485\nonumber \]

where \[x = 53, p' = \frac{x}{n} = \frac{53}{100} = 0.53\nonumber \].

Interpretation of the \(p\text{-value})\: If the null hypothesis is true, there is 0.5485 probability (54.85%) that the sample (estimated) proportion \(p'\) is 0.53 or more OR 0.47 or less (see the graph in Figure).

\(\mu = p = 0.50\) comes from \(H_{0}\), the null hypothesis.

\(p′ = 0.53\). Since the curve is symmetrical and the test is two-tailed, the \(p′\) for the left tail is equal to \(0.50 – 0.03 = 0.47\) where \(\mu = p = 0.50\). (0.03 is the difference between 0.53 and 0.50.)

Compare \(\alpha\) and the \(p\text{-value}\):

Since \(\alpha = 0.01\) and \(p\text{-value} = 0.5485\). \(\alpha < p\text{-value}\).

Make a decision: Since \(\alpha < p\text{-value}\), you cannot reject \(H_{0}\).

Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50%.

The \(p\text{-value}\) can easily be calculated.

Press STAT and arrow over to TESTS . Press 5:1-PropZTest . Enter .5 for \(p_{0}\), 53 for \(x\) and 100 for \(n\). Arrow down to Prop and arrow to not equals \(p_{0}\). Press ENTER . Arrow down to Calculate and press ENTER . The calculator calculates the \(p\text{-value}\) (\(p = 0.5485\)) and the test statistic (\(z\)-score). Prop not equals .5 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with \(\(z\) = 0.6\) (test statistic) and \(p = 0.5485\) (\(p\text{-value}\)). Make sure when you use Draw that no other equations are highlighted in \(Y =\) and the plots are turned off.

The Type I and Type II errors are as follows:

The Type I error is to conclude that the proportion of first-time brides who are younger than their grooms is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true).

The Type II error is there is not enough evidence to conclude that the proportion of first time brides who are younger than their grooms differs from 50% when, in fact, the proportion does differ from 50%. (Do not reject the null hypothesis when the null hypothesis is false.)

Exercise \(\PageIndex{7}\)

A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.

First, determine what type of test this is, set up the hypothesis test, find the \(p\text{-value}\), sketch the graph, and state your conclusion.

Since the problem is about percentages, this is a test of single population proportions.

• \(H_{0} : p = 0.85\)
• \(H_{a}: p \neq 0.85\)
• \(p = 0.7554\)

Because \(p > \alpha\), we fail to reject the null hypothesis. There is not sufficient evidence to suggest that the proportion of students that want to go to the zoo is not 85%.

Example \(\PageIndex{8}\)

Suppose a consumer group suspects that the proportion of households that have three cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three cell phones.

Set up the Hypothesis Test:

\(H_{0}: p = 0.30, H_{a}: p \neq 0.30\)

Determine the distribution needed:

The random variable is \(P′ =\) proportion of households that have three cell phones.

The distribution for the hypothesis test is \(P' - N\left(0.30, \sqrt{\frac{(0.30 \cdot 0.70)}{150}}\right)\)

Exercise 9.6.8.2

a. The value that helps determine the \(p\text{-value}\) is \(p′\). Calculate \(p′\).

a. \(p' = \frac{x}{n}\) where \(x\) is the number of successes and \(n\) is the total number in the sample.

\(x = 43, n = 150\)

\(p′ = 43150\)

Exercise 9.6.8.3

b. What is a success for this problem?

b. A success is having three cell phones in a household.

Exercise 9.6.8.4

c. What is the level of significance?

c. The level of significance is the preset \(\alpha\). Since \(\alpha\) is not given, assume that \(\alpha = 0.05\).

Exercise 9.6.8.5

d. Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately.

Calculate the \(p\text{-value}\).

d. \(p\text{-value} = 0.7216\)

Exercise 9.6.8.6

e. Make a decision. _____________(Reject/Do not reject) \(H_{0}\) because____________.

e. Assuming that \(\alpha = 0.05, \alpha < p\text{-value}\). The decision is do not reject \(H_{0}\) because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30%.

Exercise \(\PageIndex{8}\)

Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the p -value, state your conclusion, and identify the Type I and Type II errors.

• \(H_{0}: p = 0.92\)
• \(H_{a}: p < 0.92\)
• \(p\text{-value} = 0.0046\)

Because \(p < 0.05\), we reject the null hypothesis. There is sufficient evidence to conclude that fewer than 92% of American adults own cell phones.

• Type I Error: To conclude that fewer than 92% of American adults own cell phones when, in fact, 92% of American adults do own cell phones (reject the null hypothesis when the null hypothesis is true).
• Type II Error: To conclude that 92% of American adults own cell phones when, in fact, fewer than 92% of American adults own cell phones (do not reject the null hypothesis when the null hypothesis is false).

The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the null and alternate hypotheses use the parameter \(p\). The distribution for the test is normal. The estimated proportion \(p′\) is the proportion of fleas killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived \(\alpha = 0.01\), for comparison, and a 95% confidence interval computation. The poem is clever and humorous, so please enjoy it!

Example \(\PageIndex{9}\)

My dog has so many fleas,

They do not come off with ease. As for shampoo, I have tried many types Even one called Bubble Hype, Which only killed 25% of the fleas, Unfortunately I was not pleased.

I've used all kinds of soap, Until I had given up hope Until one day I saw An ad that put me in awe.

A shampoo used for dogs Called GOOD ENOUGH to Clean a Hog Guaranteed to kill more fleas.

I gave Fido a bath And after doing the math His number of fleas Started dropping by 3's! Before his shampoo I counted 42.

At the end of his bath, I redid the math And the new shampoo had killed 17 fleas. So now I was pleased.

Now it is time for you to have some fun With the level of significance being .01, You must help me figure out

Use the new shampoo or go without?

\(H_{0}: p \leq 0.25\)   \(H_{a}: p > 0.25\)

In words, CLEARLY state what your random variable \(\bar{X}\) or \(P′\) represents.

\(P′ =\) The proportion of fleas that are killed by the new shampoo

State the distribution to use for the test.

\[N\left(0.25, \sqrt{\frac{(0.25){1-0.25}}{42}}\right)\nonumber \]

Test Statistic: \(z = 2.3163\)

Calculate the \(p\text{-value}\) using the normal distribution for proportions:

\[p\text{-value} = 0.0103\nonumber \]

In one to two complete sentences, explain what the p -value means for this problem.

If the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated) proportion is 0.4048 \(\left(\frac{17}{42}\right)\) or more.

Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the \(p\text{-value}\).

Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences.

Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of fleas that are killed by the new shampoo is more than 25%.

Construct a 95% confidence interval for the true mean or proportion. Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval.

Confidence Interval: (0.26,0.55) We are 95% confident that the true population proportion p of fleas that are killed by the new shampoo is between 26% and 55%.

This test result is not very definitive since the \(p\text{-value}\) is very close to alpha. In reality, one would probably do more tests by giving the dog another bath after the fleas have had a chance to return.

Example \(\PageIndex{11}\)

In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.

We will follow the four-step process.

• \(H_{0}: p \leq 0.00034\)
• \(H_{a}: p > 0.00034\)

If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.

• We will be testing a sample proportion with \(x = 172\) and \(n = 420,019\). The sample is sufficiently large because we have \(np = 420,019(0.00034) = 142.8\), \(nq = 420,019(0.99966) = 419,876.2\), two independent outcomes, and a fixed probability of success \(p = 0.00034\). Thus we will be able to generalize our results to the population.

Figure 9.6.11.

Figure 9.6.12.

• Since the \(p\text{-value} = 0.0073\) is greater than our alpha value \(= 0.005\), we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher brain cancer rates for the cell phone users.

Example \(\PageIndex{12}\)

According to the US Census there are approximately 268,608,618 residents aged 12 and older. Statistics from the Rape, Abuse, and Incest National Network indicate that, on average, 207,754 rapes occur each year (male and female) for persons aged 12 and older. This translates into a percentage of sexual assaults of 0.078%. In Daviess County, KY, there were reported 11 rapes for a population of 37,937. Conduct an appropriate hypothesis test to determine if there is a statistically significant difference between the local sexual assault percentage and the national sexual assault percentage. Use a significance level of 0.01.

We will follow the four-step plan.

• We need to test whether the proportion of sexual assaults in Daviess County, KY is significantly different from the national average.
• \(H_{0}: p = 0.00078\)
• \(H_{a}: p \neq 0.00078\)

Figure 9.6.13.

Figure 9.6.14.

• Since the \(p\text{-value}\), \(p = 0.00063\), is less than the alpha level of 0.01, the sample data indicates that we should reject the null hypothesis. In conclusion, the sample data support the claim that the proportion of sexual assaults in Daviess County, Kentucky is different from the national average proportion.

The hypothesis test itself has an established process. This can be summarized as follows:

• Determine \(H_{0}\) and \(H_{a}\). Remember, they are contradictory.
• Determine the random variable.
• Determine the distribution for the test.
• Draw a graph, calculate the test statistic, and use the test statistic to calculate the \(p\text{-value}\). (A z -score and a t -score are examples of test statistics.)
• Compare the preconceived α with the p -value, make a decision (reject or do not reject H 0 ), and write a clear conclusion using English sentences.

Notice that in performing the hypothesis test, you use \(\alpha\) and not \(\beta\). \(\beta\) is needed to help determine the sample size of the data that is used in calculating the \(p\text{-value}\). Remember that the quantity \(1 – \beta\) is called the Power of the Test . A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same.If the power is low, the null hypothesis might not be rejected when it should be.

• Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.
• Data from Bloomberg Businessweek . Available online at http://www.businessweek.com/news/2011- 09-15/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html.
• Data from energy.gov. Available online at http://energy.gov (accessed June 27. 2013).
• Data from Gallup®. Available online at www.gallup.com (accessed June 27, 2013).
• Data from Growing by Degrees by Allen and Seaman.
• Data from La Leche League International. Available online at www.lalecheleague.org/Law/BAFeb01.html.
• Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013).
• Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013).
• Data from the Bureau of Labor Statistics. Available online at http://www.bls.gov/oes/current/oes291111.htm .
• Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013)
• Data from the U.S. Census Bureau, available online at quickfacts.census.gov/qfd/states/00000.html (accessed June 27, 2013).
• Data from the United States Census Bureau. Available online at www.census.gov/hhes/socdemo/language/.
• Data from Toastmasters International. Available online at http://toastmasters.org/artisan/deta...eID=429&Page=1 .
• Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013).
• Federal Bureau of Investigations. “Uniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.” Available online at http://www.disastercenter.com/kentucky/crime/3868.htm (accessed June 27, 2013).
• “Foothill-De Anza Community College District.” De Anza College, Winter 2006. Available online at research.fhda.edu/factbook/DA...t_da_2006w.pdf.
• Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed June 27, 2013).
• Rape, Abuse & Incest National Network. “How often does sexual assault occur?” RAINN, 2009. Available online at www.rainn.org/get-information...sexual-assault (accessed June 27, 2013).

Contributors and Attributions

Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] .

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• Fundamental Analysis

Hypothesis to Be Tested: Definition and 4 Steps for Testing with Example

What Is Hypothesis Testing?

Hypothesis testing, sometimes called significance testing, is an act in statistics whereby an analyst tests an assumption regarding a population parameter. The methodology employed by the analyst depends on the nature of the data used and the reason for the analysis.

Hypothesis testing is used to assess the plausibility of a hypothesis by using sample data. Such data may come from a larger population, or from a data-generating process. The word "population" will be used for both of these cases in the following descriptions.

Key Takeaways

• Hypothesis testing is used to assess the plausibility of a hypothesis by using sample data.
• The test provides evidence concerning the plausibility of the hypothesis, given the data.
• Statistical analysts test a hypothesis by measuring and examining a random sample of the population being analyzed.
• The four steps of hypothesis testing include stating the hypotheses, formulating an analysis plan, analyzing the sample data, and analyzing the result.

How Hypothesis Testing Works

In hypothesis testing, an  analyst  tests a statistical sample, with the goal of providing evidence on the plausibility of the null hypothesis.

Statistical analysts test a hypothesis by measuring and examining a random sample of the population being analyzed. All analysts use a random population sample to test two different hypotheses: the null hypothesis and the alternative hypothesis.

The null hypothesis is usually a hypothesis of equality between population parameters; e.g., a null hypothesis may state that the population mean return is equal to zero. The alternative hypothesis is effectively the opposite of a null hypothesis (e.g., the population mean return is not equal to zero). Thus, they are mutually exclusive , and only one can be true. However, one of the two hypotheses will always be true.

The null hypothesis is a statement about a population parameter, such as the population mean, that is assumed to be true.

4 Steps of Hypothesis Testing

All hypotheses are tested using a four-step process:

• The first step is for the analyst to state the hypotheses.
• The second step is to formulate an analysis plan, which outlines how the data will be evaluated.
• The third step is to carry out the plan and analyze the sample data.
• The final step is to analyze the results and either reject the null hypothesis, or state that the null hypothesis is plausible, given the data.

Real-World Example of Hypothesis Testing

If, for example, a person wants to test that a penny has exactly a 50% chance of landing on heads, the null hypothesis would be that 50% is correct, and the alternative hypothesis would be that 50% is not correct.

Mathematically, the null hypothesis would be represented as Ho: P = 0.5. The alternative hypothesis would be denoted as "Ha" and be identical to the null hypothesis, except with the equal sign struck-through, meaning that it does not equal 50%.

A random sample of 100 coin flips is taken, and the null hypothesis is then tested. If it is found that the 100 coin flips were distributed as 40 heads and 60 tails, the analyst would assume that a penny does not have a 50% chance of landing on heads and would reject the null hypothesis and accept the alternative hypothesis.

If, on the other hand, there were 48 heads and 52 tails, then it is plausible that the coin could be fair and still produce such a result. In cases such as this where the null hypothesis is "accepted," the analyst states that the difference between the expected results (50 heads and 50 tails) and the observed results (48 heads and 52 tails) is "explainable by chance alone."

Some staticians attribute the first hypothesis tests to satirical writer John Arbuthnot in 1710, who studied male and female births in England after observing that in nearly every year, male births exceeded female births by a slight proportion. Arbuthnot calculated that the probability of this happening by chance was small, and therefore it was due to “divine providence.”

What is Hypothesis Testing?

Hypothesis testing refers to a process used by analysts to assess the plausibility of a hypothesis by using sample data. In hypothesis testing, statisticians formulate two hypotheses: the null hypothesis and the alternative hypothesis. A null hypothesis determines there is no difference between two groups or conditions, while the alternative hypothesis determines that there is a difference. Researchers evaluate the statistical significance of the test based on the probability that the null hypothesis is true.

What are the Four Key Steps Involved in Hypothesis Testing?

Hypothesis testing begins with an analyst stating two hypotheses, with only one that can be right. The analyst then formulates an analysis plan, which outlines how the data will be evaluated. Next, they move to the testing phase and analyze the sample data. Finally, the analyst analyzes the results and either rejects the null hypothesis or states that the null hypothesis is plausible, given the data.

What are the Benefits of Hypothesis Testing?

Hypothesis testing helps assess the accuracy of new ideas or theories by testing them against data. This allows researchers to determine whether the evidence supports their hypothesis, helping to avoid false claims and conclusions. Hypothesis testing also provides a framework for decision-making based on data rather than personal opinions or biases. By relying on statistical analysis, hypothesis testing helps to reduce the effects of chance and confounding variables, providing a robust framework for making informed conclusions.

What are the Limitations of Hypothesis Testing?

Hypothesis testing relies exclusively on data and doesn’t provide a comprehensive understanding of the subject being studied. Additionally, the accuracy of the results depends on the quality of the available data and the statistical methods used. Inaccurate data or inappropriate hypothesis formulation may lead to incorrect conclusions or failed tests. Hypothesis testing can also lead to errors, such as analysts either accepting or rejecting a null hypothesis when they shouldn’t have. These errors may result in false conclusions or missed opportunities to identify significant patterns or relationships in the data.

The Bottom Line

Hypothesis testing refers to a statistical process that helps researchers and/or analysts determine the reliability of a study. By using a well-formulated hypothesis and set of statistical tests, individuals or businesses can make inferences about the population that they are studying and draw conclusions based on the data presented. There are different types of hypothesis testing, each with their own set of rules and procedures. However, all hypothesis testing methods have the same four step process, which includes stating the hypotheses, formulating an analysis plan, analyzing the sample data, and analyzing the result. Hypothesis testing plays a vital part of the scientific process, helping to test assumptions and make better data-based decisions.

Sage. " Introduction to Hypothesis Testing. " Page 4.

Elder Research. " Who Invented the Null Hypothesis? "

Formplus. " Hypothesis Testing: Definition, Uses, Limitations and Examples. "

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Lecture 14: Hypothesis Test for One Variance

STAT 205: Introduction to Mathematical Statistics

University of British Columbia Okanagan

March 17, 2024

Introduction

We have covered three hypothesis tests for a single sample:

• Hypothesis test for the mean \(\mu\) with \(\sigma\) known ( \(Z\) - test)
• Hypothesis tests for the proportion \(p\) ( \(Z\) - test)
• Hypothesis test for the mean \(\mu\) with \(\sigma\) unknown ( \(t\) -test)

Today we consider hypothesis tests involve the population variance \(\sigma^2\)

Assumptions: \(X_1, X_2, \dots, X_n\) are i.i.d + assumptions in the rhombuses.

In Lecture 7 we saw how to construct a confidence interval for \(\sigma^2\) based on the sampling distribution derived in Lecture 8 .

For random samples from normal populations , we know:

\[ \dfrac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1} \]

where \(S^2 = \frac{\sum_{i = 1}^n (X_i - \bar{X})}{n-1}\) is the sample variance and \(\chi^2_{n-1}\) is the Chi-squared distribution with \(n-1\) degrees of freedom.

We may which to test if there is evidence to suggest that population variance differs for some hypothesized value \(\sigma_0^2\) .

As before, we start with a null hypothesis ( \(H_0\) ) that the population variance equals a specified value ( \(\sigma^2 = \sigma_0^2\) )

We test this against the alternative hypothesis \(H_A\) which can either be one-sided ( \(\sigma^2 < \sigma_0^2\) or \(\sigma^2 > \sigma_0^2\) ) or two-sided ( \(\sigma^2 \neq \sigma_0^2\) ).

Test Statistic

Recall that our test statistic is calculated assuming the null hypothesis is true . Hence, if we are testing \(H_0: \sigma^2 = \sigma_0^2\) , the test statistic we use is : \[ \chi^2 = \dfrac{(n-1)S^2}{\sigma_0^2} \] where \(\chi^2 \sim \chi^2_{n-1}\) .

Chi-square distrbituion

Assumptions

For the following inference procedures to be valid we require:

• A simple random sample from the population
• A normally distributed population (very important, even for large sample sizes)

It is important to note that if the population is not approximately normally distributed, chi-squared distribution may not accurately represent the sampling distribution of the test statistic.

Critical Region (upper-tailed)

The rejection region associated with an upper-tailed test for the population variance. Note that the critical value will depend on the chosen significance level ( \(\alpha\) ) and the d.f.

Critical Region (lower-tailed)

Critical Region (two-tailed)

Similarly we can find \(p\) -values from Chi-squared tables or R

\(p\) -value for lower-tailed: \[\Pr(\chi^2 < \chi^2_{\text{obs}})\] \(p\) -value for upper-tailed: \[\Pr(\chi^2 > \chi^2_{\text{obs}})\] \(p\) -value for two-tailed:

\[2\cdot \min \{ \Pr(\chi^2 < \chi^2_{\text{obs}}), \Pr(\chi^2 > \chi^2_{\text{obs}})\}\]

Exercise 1: Beyond Burger Fat

Beyond Burgers claim to have 18g grams of fat. A random sample of 6 burgers had a mean of 19.45 and a variance of 0.85 grams \(^2\) . Suppose that the quality assurance team at the company will on accept at most a \(\sigma\) of 0.5. Use the 0.05 level of significance to test the null hypotehsis \(\sigma = 0.5\) against the appropriate alternative.

Distribution of Test Statistic

Under the null hypothesis, the test statistic follows \(\chi^2 = (n-1)S^2/0.5^2\) a chi-square distribution with df = 5

Critical value

The critical value can be found by determining what value on the chi-square curve with 5 df yield a 5 percent probability in the upper tail (since we are doing an upper-tailed test). In R: qchisq(alpha, df=n-1, lower.tail = FALSE) . Verify using \(\chi^2\) table.

Observed Test Statistic

Compute the observed test statistic which we denote by \(\chi^2_{\text{obs}}\)

Since the observed test statistic falls in the rejection region, i.e.  \(\chi^2_{\text{obs}} > \chi^2_{\alpha}\) , we rejection the null hypothesis in favour of the alternative.

P-value in R

Alternatively we could compute the p-value which in this case is 0.013. Since this is smaller than the alpha-level of 0.05, we reject the null hypothesis in favour of the alternative. Verify using \(\chi^2\) table.

P-value from tables

Using the chi-square distribution table we can see that our observed test statistic falls between two values. We can use the neigbouring values to approximate our p-value.

Approximate P-value

It is clear from the visualization that \[\begin{align} \Pr(\chi^2_{5} > \chi^2_{0.025}) > \Pr(\chi^2_{5} > \chi^2_{\text{obs}})\\ \Pr(\chi^2_{5} > \chi^2_{\text{obs}}) < \Pr(\chi^2_{5} > \chi^2_{0.01}) \\ \end{align}\]

The \(p\) -value, \(\Pr(\chi^2_{5} > 14.45)\) can then be expressed as: \[\begin{align} 0.01 < p\text{-value } < 0.025 \end{align}\]

• the \(p\) -value (0.013) is less than \(\alpha\) = 0.05 OR
• the the observed test statistic ( \(\chi^2_{\text{obs}}\) = 14.45) is larger than the critical value \(\chi^2_{\alpha}\)

we reject the null hypothesis in favour of the alternative. More specifically, there is very strong evidence to suggest that the population variance \(\sigma^2\) is greater than \(0.5^2\) .

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Hypothesis testing involves formulating assumptions about population parameters based on sample statistics and rigorously evaluating these assumptions against empirical evidence. This article sheds light on the significance of hypothesis testing and the critical steps involved in the process.

What is Hypothesis Testing?

Hypothesis testing is a statistical method that is used to make a statistical decision using experimental data. Hypothesis testing is basically an assumption that we make about a population parameter. It evaluates two mutually exclusive statements about a population to determine which statement is best supported by the sample data. 

Example: You say an average height in the class is 30 or a boy is taller than a girl. All of these is an assumption that we are assuming, and we need some statistical way to prove these. We need some mathematical conclusion whatever we are assuming is true.

Defining Hypotheses

Key Terms of Hypothesis Testing

• P-value: The P value , or calculated probability, is the probability of finding the observed/extreme results when the null hypothesis(H0) of a study-given problem is true. If your P-value is less than the chosen significance level then you reject the null hypothesis i.e. accept that your sample claims to support the alternative hypothesis.
• Test Statistic: The test statistic is a numerical value calculated from sample data during a hypothesis test, used to determine whether to reject the null hypothesis. It is compared to a critical value or p-value to make decisions about the statistical significance of the observed results.
• Critical value : The critical value in statistics is a threshold or cutoff point used to determine whether to reject the null hypothesis in a hypothesis test.
• Degrees of freedom: Degrees of freedom are associated with the variability or freedom one has in estimating a parameter. The degrees of freedom are related to the sample size and determine the shape.

Why do we use Hypothesis Testing?

Hypothesis testing is an important procedure in statistics. Hypothesis testing evaluates two mutually exclusive population statements to determine which statement is most supported by sample data. When we say that the findings are statistically significant, thanks to hypothesis testing. 

One-Tailed and Two-Tailed Test

One tailed test focuses on one direction, either greater than or less than a specified value. We use a one-tailed test when there is a clear directional expectation based on prior knowledge or theory. The critical region is located on only one side of the distribution curve. If the sample falls into this critical region, the null hypothesis is rejected in favor of the alternative hypothesis.

One-Tailed Test

There are two types of one-tailed test:

Two-Tailed Test

A two-tailed test considers both directions, greater than and less than a specified value.We use a two-tailed test when there is no specific directional expectation, and want to detect any significant difference.

What are Type 1 and Type 2 errors in Hypothesis Testing?

In hypothesis testing, Type I and Type II errors are two possible errors that researchers can make when drawing conclusions about a population based on a sample of data. These errors are associated with the decisions made regarding the null hypothesis and the alternative hypothesis.

How does Hypothesis Testing work?

Step 1: define null and alternative hypothesis.

We first identify the problem about which we want to make an assumption keeping in mind that our assumption should be contradictory to one another, assuming Normally distributed data.

Step 2 – Choose significance level

Step 3 – Collect and Analyze data.

Gather relevant data through observation or experimentation. Analyze the data using appropriate statistical methods to obtain a test statistic.

Step 4-Calculate Test Statistic

The data for the tests are evaluated in this step we look for various scores based on the characteristics of data. The choice of the test statistic depends on the type of hypothesis test being conducted.

There are various hypothesis tests, each appropriate for various goal to calculate our test. This could be a Z-test , Chi-square , T-test , and so on.

• Z-test : If population means and standard deviations are known. Z-statistic is commonly used.
• t-test : If population standard deviations are unknown. and sample size is small than t-test statistic is more appropriate.
• Chi-square test : Chi-square test is used for categorical data or for testing independence in contingency tables
• F-test : F-test is often used in analysis of variance (ANOVA) to compare variances or test the equality of means across multiple groups.

We have a smaller dataset, So, T-test is more appropriate to test our hypothesis.

T-statistic is a measure of the difference between the means of two groups relative to the variability within each group. It is calculated as the difference between the sample means divided by the standard error of the difference. It is also known as the t-value or t-score.

Step 5 – Comparing Test Statistic:

In this stage, we decide where we should accept the null hypothesis or reject the null hypothesis. There are two ways to decide where we should accept or reject the null hypothesis.

Method A: Using Crtical values

Comparing the test statistic and tabulated critical value we have,

• If Test Statistic>Critical Value: Reject the null hypothesis.
• If Test Statistic≤Critical Value: Fail to reject the null hypothesis.

Note: Critical values are predetermined threshold values that are used to make a decision in hypothesis testing. To determine critical values for hypothesis testing, we typically refer to a statistical distribution table , such as the normal distribution or t-distribution tables based on.

Method B: Using P-values

We can also come to an conclusion using the p-value,

Note : The p-value is the probability of obtaining a test statistic as extreme as, or more extreme than, the one observed in the sample, assuming the null hypothesis is true. To determine p-value for hypothesis testing, we typically refer to a statistical distribution table , such as the normal distribution or t-distribution tables based on.

Step 7- Interpret the Results

At last, we can conclude our experiment using method A or B.

Calculating test statistic

To validate our hypothesis about a population parameter we use statistical functions . We use the z-score, p-value, and level of significance(alpha) to make evidence for our hypothesis for normally distributed data .

1. Z-statistics:

When population means and standard deviations are known.

• μ represents the population mean, 
• σ is the standard deviation
• and n is the size of the sample.

2. T-Statistics

T test is used when n<30,

t-statistic calculation is given by:

• t = t-score,
• x̄ = sample mean
• μ = population mean,
• s = standard deviation of the sample,
• n = sample size

3. Chi-Square Test

Chi-Square Test for Independence categorical Data (Non-normally distributed) using:

• i,j are the rows and columns index respectively.

Real life Hypothesis Testing example

Let’s examine hypothesis testing using two real life situations,

Case A: D oes a New Drug Affect Blood Pressure?

Imagine a pharmaceutical company has developed a new drug that they believe can effectively lower blood pressure in patients with hypertension. Before bringing the drug to market, they need to conduct a study to assess its impact on blood pressure.

• Before Treatment: 120, 122, 118, 130, 125, 128, 115, 121, 123, 119
• After Treatment: 115, 120, 112, 128, 122, 125, 110, 117, 119, 114

Step 1 : Define the Hypothesis

• Null Hypothesis : (H 0 )The new drug has no effect on blood pressure.
• Alternate Hypothesis : (H 1 )The new drug has an effect on blood pressure.

Step 2: Define the Significance level

Let’s consider the Significance level at 0.05, indicating rejection of the null hypothesis.

If the evidence suggests less than a 5% chance of observing the results due to random variation.

Step 3 : Compute the test statistic

Using paired T-test analyze the data to obtain a test statistic and a p-value.

The test statistic (e.g., T-statistic) is calculated based on the differences between blood pressure measurements before and after treatment.

t = m/(s/√n)

• m  = mean of the difference i.e X after, X before
• s  = standard deviation of the difference (d) i.e d i ​= X after, i ​− X before,
• n  = sample size,

then, m= -3.9, s= 1.8 and n= 10

we, calculate the , T-statistic = -9 based on the formula for paired t test

Step 4: Find the p-value

The calculated t-statistic is -9 and degrees of freedom df = 9, you can find the p-value using statistical software or a t-distribution table.

thus, p-value = 8.538051223166285e-06

Step 5: Result

• If the p-value is less than or equal to 0.05, the researchers reject the null hypothesis.
• If the p-value is greater than 0.05, they fail to reject the null hypothesis.

Conclusion: Since the p-value (8.538051223166285e-06) is less than the significance level (0.05), the researchers reject the null hypothesis. There is statistically significant evidence that the average blood pressure before and after treatment with the new drug is different.

Python Implementation of Hypothesis Testing

Let’s create hypothesis testing with python, where we are testing whether a new drug affects blood pressure. For this example, we will use a paired T-test. We’ll use the scipy.stats library for the T-test.

Scipy is a mathematical library in Python that is mostly used for mathematical equations and computations.

We will implement our first real life problem via python,

In the above example, given the T-statistic of approximately -9 and an extremely small p-value, the results indicate a strong case to reject the null hypothesis at a significance level of 0.05. 

• The results suggest that the new drug, treatment, or intervention has a significant effect on lowering blood pressure.
• The negative T-statistic indicates that the mean blood pressure after treatment is significantly lower than the assumed population mean before treatment.

Case B : Cholesterol level in a population

Data: A sample of 25 individuals is taken, and their cholesterol levels are measured.

Cholesterol Levels (mg/dL): 205, 198, 210, 190, 215, 205, 200, 192, 198, 205, 198, 202, 208, 200, 205, 198, 205, 210, 192, 205, 198, 205, 210, 192, 205.

Populations Mean = 200

Population Standard Deviation (σ): 5 mg/dL(given for this problem)

Step 1: Define the Hypothesis

• Null Hypothesis (H 0 ): The average cholesterol level in a population is 200 mg/dL.
• Alternate Hypothesis (H 1 ): The average cholesterol level in a population is different from 200 mg/dL.

As the direction of deviation is not given , we assume a two-tailed test, and based on a normal distribution table, the critical values for a significance level of 0.05 (two-tailed) can be calculated through the z-table and are approximately -1.96 and 1.96.

Step 4: Result

Since the absolute value of the test statistic (2.04) is greater than the critical value (1.96), we reject the null hypothesis. And conclude that, there is statistically significant evidence that the average cholesterol level in the population is different from 200 mg/dL

Limitations of Hypothesis Testing

• Although a useful technique, hypothesis testing does not offer a comprehensive grasp of the topic being studied. Without fully reflecting the intricacy or whole context of the phenomena, it concentrates on certain hypotheses and statistical significance.
• The accuracy of hypothesis testing results is contingent on the quality of available data and the appropriateness of statistical methods used. Inaccurate data or poorly formulated hypotheses can lead to incorrect conclusions.
• Relying solely on hypothesis testing may cause analysts to overlook significant patterns or relationships in the data that are not captured by the specific hypotheses being tested. This limitation underscores the importance of complimenting hypothesis testing with other analytical approaches.

Hypothesis testing stands as a cornerstone in statistical analysis, enabling data scientists to navigate uncertainties and draw credible inferences from sample data. By systematically defining null and alternative hypotheses, choosing significance levels, and leveraging statistical tests, researchers can assess the validity of their assumptions. The article also elucidates the critical distinction between Type I and Type II errors, providing a comprehensive understanding of the nuanced decision-making process inherent in hypothesis testing. The real-life example of testing a new drug’s effect on blood pressure using a paired T-test showcases the practical application of these principles, underscoring the importance of statistical rigor in data-driven decision-making.

Frequently Asked Questions (FAQs)

1. what are the 3 types of hypothesis test.

There are three types of hypothesis tests: right-tailed, left-tailed, and two-tailed. Right-tailed tests assess if a parameter is greater, left-tailed if lesser. Two-tailed tests check for non-directional differences, greater or lesser.

2.What are the 4 components of hypothesis testing?

Null Hypothesis ( ): No effect or difference exists. Alternative Hypothesis ( ): An effect or difference exists. Significance Level ( ): Risk of rejecting null hypothesis when it’s true (Type I error). Test Statistic: Numerical value representing observed evidence against null hypothesis.

3.What is hypothesis testing in ML?

Statistical method to evaluate the performance and validity of machine learning models. Tests specific hypotheses about model behavior, like whether features influence predictions or if a model generalizes well to unseen data.

4.What is the difference between Pytest and hypothesis in Python?

Pytest purposes general testing framework for Python code while Hypothesis is a Property-based testing framework for Python, focusing on generating test cases based on specified properties of the code.

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