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91Ó°ÊÓ

In the study of older students’ attitudes, the sample mean SSHA score was 125.7and the sample standard deviation was 29.8. A significance test yields a P-value of 0.0101.

a. Explain what it would mean for the null hypothesis to be true in this setting.

b. Interpret the P-value.

Short Answer

Expert verified

Part a. The correct mean score for the students who are at least 30years of age is 115.

Part b. The P-value is if the population mean is equal to115, then there is possibility of 1.01%of getting a random sample with a sample mean of 125.7or more.

Step by step solution

01

Part a. Step 1. Given information

From the previous exercise

H0:μ=115Ha:μ>115

02

Part a. Step 2. Explanation

μis the population mean score of students who are at least 30years of age.

If the null hypothesis H0:μ=115is true, then this means that the true mean score for students who are at least 30years of age is115.

03

Part b. Step 1. Given information

n=45x¯=125.7s=29.8P=0.0101=1.01%

Result previous exercise:

H0:μ=115Ha:μ>115

04

Part b. Step 2. Explanation

The P-value is if the population mean is equal to 115, then there is possibility of 1.01%of getting a random sample with a sample mean of125.7 or more.

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Most popular questions from this chapter

A company that manufactures classroom chairs for high school students

claims that the mean breaking strength of the chairs is 300 pounds. One of the chairs collapsed beneath a 220-pound student last week. You suspect that the manufacturer is exaggerating the breaking strength of the chairs, so you would like to perform a test of H0:μ=300Ha:μ<300where μ is the true mean breaking strength of this company’s classroom chairs.

a. The power of the test to detect that μ=294 based on a random sample of 30

chairs and a significance level of α=0.05 is 0.71. Interpret this value.

b. Find the probability of a Type I error and the probability of a Type II error for the test in part (a).

c. Describe two ways to increase the power of the test in part (a).

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effective for this variety of grass?

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