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Chapter 9: Q. 10. (page 564)

Attitudes Refer to Exercise 4. In the study of older students’ attitudes, the sample mean SSHA score was 125.7 and 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 the possibility of1.01%of getting a random sample with a sample mean of125.7or more.

Step by step solution

01

Part a) Step 1: Given information

From the previous exercise,

H0:μ=115Ha:μ>115

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

02

Part a) Step 2: The objective is to explain the mean for the null hypothesis to be true in this setting

If the null hypothesis H0:μ=115is correct, the true mean score for students who are at least 30years old is115

03

Part b) Step 1: Given information

n=45x¯=125.7s=29.8P=0.0101=1.01%

04

Part b) Step 2: The objective is to explain the p value

Result of the previous exercise:

H0:μ=115Ha:μ>115

If the population means is115then there is a 1.01%chance of getting a random sample with a sample means of125.7 or higher.

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Most popular questions from this chapter

Experiments on learning in animals sometimes measure how long it takes mice to find their way through a maze. The mean time is 18 seconds for one particular maze. A researcher thinks that a loud noise will cause the mice to complete the maze faster. She measures how long each of 10 mice takes with a loud noise as stimulus. The appropriate hypotheses for the significance test are

a. H0:μ=18;Ha:μ≠18

b. H0:μ=18;Ha:μ>18

c. H0:μ<18;Ha:μ=18

d. H0:μ=18;Ha:μ<18

e. H0:x¯=18;Ha:x¯<18

Fast connection? How long does it take for a chunk of information to travel

from one server to another and back on the Internet? According to the site

internettrafficreport.com, the average response time is 200 milliseconds (about one-fifth of a second). Researchers wonder if this claim is true, so they collect data on response times (in milliseconds) for a random sample of 14 servers in Europe. A graph of the data reveals no strong skewness or outliers.

a. State an appropriate pair of hypotheses for a significance test in this setting. Be sure to define the parameter of interest.

b. Check conditions for performing the test in part (a).

c. The 95% confidence interval for the mean response time is 158.22 to 189.64

milliseconds. Based on this interval, what conclusion would you make for a test of the hypotheses in part (a) at the 5% significance level?

d. Do we have convincing evidence that the mean response time of servers in the United States is different from 200 milliseconds? Justify your answer.

How much juice? Refer to Exercises 3 and 11 .

a. What conclusion would you make at the α=0.10α=0.10level?

b. Would your conclusion from part (a) change if a 5 \% significance level was used instead? Explain your reasoning.

Do you Tweet? The Pew Internet and American Life Project asked a random sample of U.S. adults, “Do you ever … use Twitter or another service to share updates about yourself or to see updates about others?” According to Pew, the resulting 95% confidence interval is (0.123, 0.177).11 Based on the confidence interval, is there convincing evidence that the true proportion of U.S. adults who would say they use Twitter or another service to share updates differs from 0.17? Explain your reasoning.

Proposition XA political organization wants to determine if there is convincing evidence that a majority of registered voters in a large city favor Proposition X. In an SRS of 1000registered voters, 482favor the proposition. Explain why it isn’t necessary to carry out a significance test in this setting.
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