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Chapter 9: Q. 70. (page 607)

One-sided test Suppose you want to perform a test of H0: 渭=5 versus Ha: 渭<5

at the 伪=0.05 significance level. A random sample of size n=20 from the population of interest yields x炉=4.7 and sx=0.74 . Assume that the conditions for carrying out the test are met.

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

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

Short Answer

Expert verified

Part a) The sample mean 4.7is less than 5

Part b)

t=-1.8130.25<P<0.05OrP=0.04283

Step by step solution

01

Part a) Step 1: Given information

H0:渭=5H1:渭<5伪=0.05n=20x炉=4.7s=0.74

02

Part a) Step 2: Calculation

The sample mean of 4.7is less than 5,which agrees with the alternative hypothesis that the mean is less than 5,and thus the sample result provides some evidence for the alternative hypothesis.

03

Part b) Step 1: Given information

H0:渭=5H1:渭<5伪=0.05n=20x炉=4.7s=0.74

04

Part b) Step 2: Calculation

We know,

t=x炉-渭0sln

The test statistic is

t=x炉-渭0sln=4.7-50.74/20=-1.813

If the null hypothesis is true, the P-value is the probability of getting the test statistic's value or a value that is more extreme.

df=n-1=20-1=19.0.25<P<0.05

Command for Ti83/84-calculator: (-1E99,-1.813,19)which will return a P-value of 0.04283. It could replace -1E99 by any other very small negative number.

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Most popular questions from this chapter

18%Members of the city council want to know if a majority of city residents supports a 1%increase in the sales tax to fund road repairs. To investigate, they survey a random sample of 300city residents and use the results to test the following hypotheses:

H0:p=0.50

Ha:p>0.50

where pis the proportion of all city residents who support a 1%increase in the sales tax to fund road repairs.

In the sample, p^=158/300=0.527, The resulting P-value is 0.18. What is the correct interpretation of this P-value?

a. Only 18% of the city residents support the tax increase.

b. There is an 18%chance that the majority of residents supports the tax increase.

c. Assuming that 50%of residents support the tax increase, there is an 18%probability that the sample proportion would be 0.527or greater by chance alone.

d. Assuming that more than 50%of residents support the tax increase, there is an 18%probability that the sample proportion would be 0.527or greater by chance alone.

e. Assuming that 50%of residents support the tax increase, there is an 18% chance that the null hypothesis is true by chance alone.

Opening a restaurant You are thinking about opening a restaurant and are

searching for a good location. From 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 location. Based on the mean income of this sample, you will perform a test of

H0:渭=\)85,000

Ha:渭>$85,000

where 渭is the true mean income in the population of people who live near the restaurant. Describe a Type I error and a Type II error in this setting, and give a possible consequence of each.

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).

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.

Flu vaccine A drug company has developed a new vaccine for preventing the flu. The company claims that fewer than 5% of adults who use its vaccine will get the flu. To test the claim, researchers give the vaccine to a random sample of 1000 adults.

a. State appropriate hypotheses for testing the company鈥檚 claim. Be sure to define your parameter.

b. Describe a Type I error and a Type II error in this setting, and give the consequences Page Number: 615 of each.

c. Would you recommend a significance level of 0.01, 0.05, or 0.10 for this test? Justify your choice.

d. The power of the test to detect the fact that only 3% of adults who use this vaccine would develop flu using 伪=0.05 is 0.9437. Interpret this value.

e. Explain two ways that you could increase the power of the test from part (d).
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