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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 “Splinting vs Surgery in the Treatment of Carpal Tunnel Syndrome,” 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\({p_1}\)be the population proportion of success rate in splinting and\({p_2}\)be population proportion of success rate in surgery.

Mathematically, the test hypothesis is,

\(\begin{array}{l}{H_0}:{p_1} = {p_2}{\rm{ }}\\{H_1}:{p_1} < {p_2}\end{array}\)

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 at\({\rm{\alpha }}\)level of significance.

Here, the p-value is 0.0009 which is less than the level of significance\(\alpha = 0.01\).

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

Determining Sample Size The sample size needed to estimate the difference between two population proportions to within a margin of error E with a confidence level of 1 - a can be found by using the following expression:

\({\bf{E = }}{{\bf{z}}_{\frac{{\bf{\alpha }}}{{\bf{2}}}}}\sqrt {\frac{{{{\bf{p}}_{\bf{1}}}{{\bf{q}}_{\bf{1}}}}}{{{{\bf{n}}_{\bf{1}}}}}{\bf{ + }}\frac{{{{\bf{p}}_{\bf{2}}}{{\bf{q}}_{\bf{2}}}}}{{{{\bf{n}}_{\bf{2}}}}}} \)

Replace \({{\bf{n}}_{\bf{1}}}\;{\bf{and}}\;{{\bf{n}}_{\bf{2}}}\) by n in the preceding formula (assuming that both samples have the same size) and replace each of \({{\bf{p}}_{\bf{1}}}{\bf{,}}{{\bf{q}}_{\bf{1}}}{\bf{,}}{{\bf{p}}_{\bf{2}}}\;{\bf{and}}\;{{\bf{q}}_{\bf{2}}}\)by 0.5 (because their values are not known). Solving for n results in this expression:

\({\bf{n = }}\frac{{{\bf{z}}_{\frac{{\bf{\alpha }}}{{\bf{2}}}}^{\bf{2}}}}{{{\bf{2}}{{\bf{E}}^{\bf{2}}}}}\)

Use this expression to find the size of each sample if you want to estimate the difference between the proportions of men and women who own smartphones. Assume that you want 95% confidence that your error is no more than 0.03.

In Exercises 5–20, 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 “Table” answers based on Table A-3 with df equal to the smaller of\({n_1} - 1\)and\({n_2} - 1\).)

BMI We know that the mean weight of men is greater than the mean weight of women, and the mean height of men is greater than the mean height of women. A person’s body mass index (BMI) is computed by dividing weight (kg) by the square of height (m). Given below are the BMI statistics for random samples of females and males taken from Data Set 1 “Body Data” in Appendix B.

a. Use a 0.05 significance level to test the claim that females and males have the same mean BMI.

b. Construct the confidence interval that is appropriate for testing the claim in part (a).

c. Do females and males appear to have the same mean BMI?

Female BMI: n = 70, \(\bar x\) = 29.10, s = 7.39

Male BMI: n = 80, \(\bar x\) = 28.38, s = 5.37

In Exercises 5–20, assume that the two samples are independent random samples selected from normally distributed populations. 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 “Table” answers based on Table A-3 with df equal to the smaller of n1−1 and n2−1).

Are male and female professors rated differently? According to Data Set 17 “Course Evaluations” Appendix B, given below are student evaluation scores of female professors and male professors. The test claims that female and male professors have the same mean evaluation ratings. Does there appear to be a difference?

Females

4.4

3.4

4.8

2.9

4.4

4.9

3.5

3.7

3.4

4.8

Males

4.0

3.6

4.1

4.1

3.5

4.6

4.0

4.3

4.5

4.3

Testing Claims About Proportions. In Exercises 7–22, 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.

Accuracy of Fast Food Drive-Through Orders In a study of Burger King drive-through orders, it was found that 264 orders were accurate and 54 were not accurate. For McDonald’s, 329 orders were found to be accurate while 33 orders were not accurate (based on data from QSR magazine). Use a 0.05 significance level to test the claim that Burger King and McDonald’s have the same accuracy rates.

a. Test the claim using a hypothesis test.

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

c. Relative to accuracy of orders, does either restaurant chain appear to be better?

A sample size that will ensure a margin of error of at most the one specified.

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