Hypothesis Testing Using a P-Value In Exercises 33–38,
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Question

Hypothesis Testing Using a P-Value In Exercises 33–38,

a. identify the claim and state and .

b. find the standardized test statistic z.

c. find the corresponding P-value.

d. decide whether to reject or fail to reject the null hypothesis.

e. interpret the decision in the context of the original claim.

MCAT Scores A random sample of 100 medical school applicants at a university has a mean total score of 505 on the MCAT. According to a report, the mean total score for the school’s applicants is more than 503. Assume the population standard deviation is 10.6. At alpha=0.01, is there enough evidence to support the report’s claim?

Accepted Answer

Hello, everyone. Let's take a look at this question together. A random sample of 64 engineering students at a university has a mean GRE quantitative score of 162. The department claims that the mean GRE quantitative score for its applicants is greater than 160. Assume the population standard deviation is 8.2. At an alpha of 0.01, is there sufficient evidence to support the department's claim? Use a P value. So in order to solve this question, we have to recall how we can determine whether there is sufficient evidence to support the claim of the department. That the mean GRE quantitative score for its applicants is greater than 160. And so to solve this question, the first thing is that we should note that the sample size is N equals 64. Which this is important since the population standard deviation is known, the sample is random and our N is greater than 30, so we know that we can use a P value for a Z test. And so the next step in solving this problem is to state the null and alternative hypotheses, which we know from the information provided that the department claims that the mean GRE quantitative score for its applicants is greater than 160. Which this claim can be expressed as mu is greater than 160. And since it is a statement of strict inequality, it is the alternative hypothesis, so our alternative hypothesis is m is greater than 160, and we know that the complement of the alternative hypothesis is the null hypothesis, so the null hypothesis is mu is less than or equal to 160. Therefore, the null and alternative hypotheses are the null hypothesis is mu is less than 160, and the alternative hypothesis, which is the claim, is mu is greater than 160, and we know that the given level of significance, which is represented as alpha, is equal to 0.01. And then the next step. is to find the standardized test statistic, which is Z, and is given by the following formula where Z is equal to the sample mean subtracted by the population mean divided by the population standard deviation, which is divided by the square root of our sample size N. and using the values that we are provided from the question, we can. Substitute them into our equation so that Z is equal to 162, subtracted by 160 divided by 8.2, which is divided by the square root of 64. And then simplifying our equation, we know Z is equal to 1.95, 12, 19512. And using that value we can find the area. That corresponds to Z by using the table for standard normal distribution, which we should note that the table values represent the cumulative area to the left of the Z score, and since our value for Z, which is 1.95, 12, 19512, is between 1.95 and 1.96, we can interpolate. And using the table for standard normal distribution, we can see the values for our Z between 1.95 and 1.96 are highlighted in the table, and we will let the area corresponding to our Z score represent A. And then by interpolation, we get the following equation where we have a minus 0.9744 divided by 0.975. subtracted by 0.9744 is equal to 80 divided by 41 minus 1.95 divided by 1.96 subtracted by 1.95, which this equation can be rearranged so that we are solving for A. And when we simplify this equation, we know that A is equal to 0.9744731707. And then the next step is to find the P value, and since our alternative hypothesis contains the greater than inequality symbol, the hypothesis test is a right-tailed test, which we know that for a right-tailed test, he is equal to the area in the right tail. And since the area obtained from the previous step is a cumulative area to the left. Of the Z score, we know that P is equal to 1 minus A, which substituting the value for A P is equal to 1 minus 0.9744731707, which when simplified, we know that P is approximately 0.0255. And now we have to make our decision. And since P is greater than alpha, Which is 0.0255 is greater than 0.01, we fail to reject the null hypothesis. And then interpreting our decision, since the claim is the alternative hypothesis, and we fail to reject the null hypothesis, the interpretation for this decision is that there is insufficient evidence to support the claim that the means. GRE quantitative score for the department's applicants is greater than 160. Therefore, looking at our answer choices, we can identify that answer choice D is the correct answer. As we know, there is insufficient evidence to support the claim that the mean GRE quantitative score for the department's applicants is greater than 160. And all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and goodbye.

Key Concepts

Hypothesis Testing

P-Value

Standardized Test Statistic (z)

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