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Chapter 9: Q 86. (page 608)

Upscale restaurant You are thinking about opening a restaurant and are searching for a good location. From the research you have done, you know that the mean income of those living near the restaurant must be over \(85,000to support the type of upscale restaurant you wish to open. You decide to take a simple random sample of 50people living near one potential site. Based on the mean income of this sample, you will perform a test at the

伪=0.05 significance level of H0:渭=\)85,000versus Ha:渭>\(85,000, where 渭 is the true mean income in the population of people who live near the restaurant. The power of the test to detect that 渭=\)86,000is 0.64 Interpret this value.

Short Answer

Expert verified

There is 64 percent probability that finds the convincing proof to help the alternative hypothesis渭>$85000

Step by step solution

01

Given information

H0:渭=$85000H1:渭>$85000渭A=alternativemean=$86000

伪=significancelevel=0.05

Power =0.64=64%

02

Concept

When the alternative hypothesis is true, the power is the probability of rejecting the null hypothesis.

03

Explanation

If the genuine mean income in the population of persons who live near the restaurant is $86000 there is a 64 percent chance that the alternative hypothesis 渭>$85000 will be supported.

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Most popular questions from this chapter

Restaurant power problems Refer to Exercises 86 and 88

a. Explain one disadvantage of using 伪=0.10 instead of 伪=0.05 when

performing the test.

b. Explain one disadvantage of taking a random sample of 50 people instead of 30 people.

Candy! A machine is supposed to fill bags with an average of 19.2 ounces of candy. The manager of the candy factory wants to be sure that the machine does not consistently underfill or overfill the bags. So the manager plans to conduct a significance test at the 伪=0.10significance level of

H0:渭=19.2Ha:渭notequalto19.2

where 渭=the true mean amount of candy (in ounces) that the machine put in all bags filled that day. The manager takes a random sample of 75 bags of candy produced that day and weighs each bag. Check if the conditions for performing the test are met.

Walking to school Refer to Exercise 36.

a. Explain why the sample result gives some evidence for the alternative hypothesis.

b. Calculate the standardized test statistic and P-value.

c. What conclusion would you make?

Jump around Student researchers Haley, Jeff, and Nathan saw an article on the Internet claiming that the average vertical jump for teens was 15 inches. They wondered if the average vertical jump of students at their school differed from 15 inches, so they obtained a list of student names and selected a random sample of 20 students. After contacting these students several times, they finally convinced them to allow their vertical jumps to be measured. Here are the data (in inches):

Do these data provide convincing evidence at the 伪=0.10 level that the average vertical jump of students at this school differs from 15 inches?

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:渭=300Ha:渭<300where 渭 is the true mean breaking strength of this company鈥檚 classroom chairs.

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

chairs and a significance level of 伪=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).
See all solutions

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