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Chapter 9: Q6 (page 424)

Interpreting Displays.

In Exercises 5 and 6, use the results from the given displays.

Treating Carpal Tunnel Syndrome Carpal tunnel syndrome is a common wrist complaintresulting from a compressed nerve, and it is often the result of extended use of repetitive wristmovements, such as those associated with the use of a keyboard. In a randomized controlledtrial, 73 patients were treated with surgery and 67 were found to have successful treatments.Among 83 patients treated with splints, 60 were found to have successful treatments (based ondata from 鈥淪plinting vs Surgery in the Treatment of Carpal Tunnel Syndrome,鈥� by Gerritsenet al., Journal of the American Medical Association, Vol. 288, No. 10). Use the accompanyingStatCrunch display with a 0.01 significance level to test the claim that the success rate is better with surgery.

Short Answer

Expert verified

There is sufficient evidence to support the claim that the success rate for surgery is better than the success rate of splints in the treatment at a 0.01 level of significance.

Step by step solution

01

Given information

The output is known

02

Describe the hypothesis to be tested

Let p1be the population proportion of success rate in splinting and p2be population proportion of success rate in surgery.

Mathematically, the test hypothesis is,

H0:p1=p2H1:p1<p2

The test is one-tailed.

03

State the result

The decision rule:

If p value is less than the level of significance then reject the null hypothesis atlevel of significance.

Here, the p-value is 0.0009 which is less than the level of significance.

Therefore, reject null hypothesis at 0.01significance level.

04

Interpret the result

Therefore, there is sufficient evidence to support the claim that the success rate for surgery is better than the success rate of splinting.

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Most popular questions from this chapter

In Exercises 5鈥�20, assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with 鈥淭able鈥� answers based on Table A-3 with df equal to the smaller of\({n_1} - 1\)and\({n_2} - 1\).) Car and Taxi Ages When the author visited Dublin, Ireland (home of Guinness Brewery employee William Gosset, who first developed the t distribution), he recorded the ages of randomly selected passenger cars and randomly selected taxis. The ages can be found from the license plates. (There is no end to the fun of traveling with the author.) The ages (in years) are listed below. We might expect that taxis would be newer, so test the claim that the mean age of cars is greater than the mean age of taxis.

Car

Ages

4

0

8

11

14

3

4

4

3

5

8

3

3

7

4

6

6

1

8

2

15

11

4

1

1

8

Taxi Ages

8

8

0

3

8

4

3

3

6

11

7

7

6

9

5

10

8

4

3

4

Are Flights Cheaper When Scheduled Earlier? Listed below are the costs (in dollars) of flights from New York (JFK) to Los Angeles (LAX). Use a 0.01 significance level to test the claim that flights scheduled one day in advance cost more than flights scheduled 30 days in advance. What strategy appears to be effective in saving money when flying?

Delta

Jet Blue

American

Virgin

Alaska

United

1 day in advance

501

634

633

646

633

642

30 days in advance

148

149

156

156

252

313

Testing Claims About Proportions. In Exercises 7鈥�22, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim.

Denomination Effect A trial was conducted with 75 women in China given a 100-yuan bill, while another 75 women in China were given 100 yuan in the form of smaller bills (a 50-yuan bill plus two 20-yuan bills plus two 5-yuan bills). Among those given the single bill, 60 spent some or all of the money. Among those given the smaller bills, 68 spent some or all of the money (based on data from 鈥淭he Denomination Effect,鈥� by Raghubir and Srivastava, Journal of Consumer Research, Vol. 36). We want to use a 0.05 significance level to test the claim that when given a single large bill, a smaller proportion of women in China spend some or all of the money when compared to the proportion of women in China given the same amount in smaller bills.

a. Test the claim using a hypothesis test.

b. Test the claim by constructing an appropriate confidence interval.

c. If the significance level is changed to 0.01, does the conclusion change?

Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with 鈥淭able鈥� answers based on Table A-3 with df equal to the smaller of\({n_1} - 1\)and\({n_2} - 1\))

Regular Coke and Diet Coke Data Set 26 鈥淐ola Weights and Volumes鈥� in Appendix B includesweights (lb) of the contents of cans of Diet Coke (n= 36,\(\overline x \)= 0.78479 lb, s= 0.00439 lb) and of the contents of cans of regular Coke (n= 36,\(\overline x \)= 0.81682 lb, s= 0.00751 lb).

a. Use a 0.05 significance level to test the claim that the contents of cans of Diet Coke have weights with a mean that is less than the mean for regular Coke.

b. Construct the confidence interval appropriate for the hypothesis test in part (a).

c. Can you explain why cans of Diet Coke would weigh less than cans of regular Coke?

Degrees of Freedom

For Example 1 on page 431, we used df\( = \)smaller of\({n_1} - 1\)and\({n_2} - 1\), we got\(df = 11\), and the corresponding critical values are\(t = \pm 2.201.\)If we calculate df using Formula 9-1, we get\(df = 19.063\), and the corresponding critical values are\( \pm 2.093\). How is using the critical values of\(t = \pm 2.201\)more 鈥渃onservative鈥� than using the critical values of\( \pm 2.093\).
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