9. Grip and Leg Strength. In the paper, "Sex Differences in Static Strength and Fatigability in Three Different Muscle Groups "(Re- search Quarterly for Erercise and Sport, Vol. 61(3), pp. 238-242), J. Misner et al. published results of a study on grip and leg strength of males and females. The following data, in newtons, is based on their measurements of right-leg strength.

Preliminary data analyses indicate that you can reasonably presume leg strength is normally distributed for both males and females and that the standard deviations of leg strength are approximately equal. At the 5% significance level, do the data provide sufficient evidence to conclude that mean right-leg strength of males exceeds that of females? (Note: x1¯=2127, s1=513, =1843, x2¯and s2=446

Short Answer

Expert verified

Therefore, the data do not provide sufficient evidence to conclude that the mean right -leg strength of males exceeds that of female.

Step by step solution

01

Step 1:Given information

For the given data sets of males and females we are given the following information.

Males:

x¯1=2127

s1=513

Females:

x¯2=1843

s2=446


02

Step 2:Explaination

From the given data sets we can easily identify the sample sizes of two samples as follows.

n1=13

n2=14

The significance level is5%.

Here we need to compare two independent small samples for the difference of the means. So we have to consider the student t-test.

Let μ1and μ2denote, respectively the mean right leg strength of males exceeds that of females. To perform the hypothesis test the null and alternative hypothesis are as follows.

H0:μ1=μ2(Mean right leg strength of males not exceeds that of females)

Ha:μ1>μ2(Mean right leg strength of males exceeds that of females)

03

Step 3:Observation

Use MINITAB software to perform the t-test as follows.

1) Choose Stat > basic Statistics>2-sample t..

2) Select samples in different columns.

3) Click in first text box and click specifyMales.

4) Click in second text box and click specifyFemales.

5) Uncheck the Assume equal variances check box

6) Click theOptions.... Button

7) Click in the Confidence level text box and type95

8) Click in the test differencetext box and type 0

9) Click arrow button at the right of the Alternativedrop-down list box and select greater than

10) Click Oktwice

The following is the Minitab out put for the hypothesis test.

04

Step 4:Result

From the above out put, the P-value for the hypothesis test is 0.068 Because the P-value is greater than the specified significance level of 0.05, we fail to rejectH0

Conclusion:

Therefore, the data do not provide sufficient evidence to conclude that the mean right -leg strength of males exceeds that of female.

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Most popular questions from this chapter

Recess and Wasted Food. Refer to Exercise 10.50 and find a 98% confidence interval for the difference between the mean amount of food wasted for lunches before recess and that for lunches after recess.

In each of exercise 10.13-10.18, we have presented a confidence interval for the difference,μ1-μ2, between two population means. interpret each confidence interval

90%CI from-10to-5

Two-Tailed Hypothesis Tests and CIs. As we mentioned on page 413, the following relationship holds between hypothesis tests and confidence intervals: For a two-tailed hypothesis test at the significance level α, the null hypothesis H0:μ1=μ2 will be rejected in favor of the alternative hypothesis H2:μ1μ2 if and only if the (1-α)-level confidence interval for μ1-μ2 does not contain 0. In each case, illustrate the preceding relationship by comparing the results of the hypothesis test and confidence interval in the specified exercises.

a. Exercises 10.81 and 10.87

b. Excrcises 10.86 and 10.92

In each of exercise 10.13-10.18, we have presented a confidence interval for the difference,μ1-μ2, between two population means. interpret each confidence interval.

95% CI is from15to20

Two-Tailed Hypothesis Tests and CIs. As we mentioned on page 413, the following relationship holds between hypothesis tests and confidence intervals: For a two-tailed hypothesis test at the significance level α, the null hypothesis H0:μ1=μ2will be rejected in favor of the alternative hypothesis Ha:μ1μ2if and only if the ( 1-α)-level confidence interval for μ1-μ2does not contain 0. In each case, illustrate the preceding relationship by comparing the reults of the hypothesis test and confidence interval in the specified xercises.

a. Exercises 10.48 and 10.54.

b. Exercises 10.49 and 10.55.

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