Subjects in a study by Dabonneville et al. (A-9) included a sample of 40 men who claimed to engage in a variety of sports activities (multisport). The mean body mass index (BMI) for these men was 22.41 with a standard deviation of 1.27. A sample of 24 male rugby players had a mean BMI of 27.75 with a standard deviation of 2.64. Is there sufficient evidence for one to claim that, in general, rugby players have a higher BMI than the multisport men? Let a ¼ :01.

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Answer:

The responses can be defined as follows:

Step-by-step explanation:

Given:

[tex]H_o : \mu_1 = \mu_2\\\\H_a : \mu_1 < \mu_2[/tex]

Testing the hypothesis:

[tex]Z=\frac{( \bar{x_1}-\bar{x_2})}{\sqrt{\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2}}}[/tex]

   [tex]=\frac{(22.41-27.75)}{\sqrt{(\frac{1.27^2}{40}+\frac{2.64^2}{24})}}\\\\=-9.29\\\\[/tex]

[tex]a=0.01[/tex]

critical value [tex]Z(0.01)=-2.33[/tex] (using the standard normal table)

[tex]Z=-9.29<-2.33[/tex] so, we reject [tex]H_o[/tex].

Using the t-distribution, it is found that since the test statistic is greater than the critical value for the right-tailed test, there is sufficient evidence for one to claim that, in general, rugby players have a higher BMI than the multisport men.

At the null hypothesis, it is tested that rugby players do not have a higher BMI than the multi-sport men, that is, the subtraction of the means is not greater than 0, hence:

[tex]H_0: \mu_R - \mu_M \leq 0[/tex]

At the alternative hypothesis, it is tested if they have a higher BMI, that is:

[tex]H_1: \mu_R - \mu_M > 0[/tex]

We have the standard deviation for the samples, hence, the t-distribution is used.

The standard errors are given by:

[tex]s_M = \frac{1.27}{\sqrt{40}} = 0.2008[/tex]

[tex]s_R = \frac{2.64}{\sqrt{24}} = 0.5389[/tex]

For the distribution of differences, the mean and the standard error are given by:

[tex]\overline{x} = \mu_R - \mu_M = 27.75 - 22.41 = 5.34[/tex]

[tex]s = \sqrt{s_M^2 + s_R^2} = \sqrt{0.2008^2 + 0.5389^2} = 0.5751[/tex]

The test statistic is given by:

[tex]t = \frac{\overline{x} - \mu}{s}[/tex]

In which [tex]\mu = 0[/tex] is the value tested at the null hypothesis.

Hence:

[tex]t = \frac{\overline{x} - \mu}{s}[/tex]

[tex]t = \frac{5.34 - 0}{0.5751}[/tex]

[tex]t = 9.29[/tex]

Using a t-distribution calculator, for a right-tailed test, as we are testing if the mean is greater than a value, with a significance level of 0.05 and 40 + 24 - 2 = 62 df, the critical value is of [tex]t^{\ast} = 1.67[/tex]

Since the test statistic is greater than the critical value for the right-tailed test, there is sufficient evidence for one to claim that, in general, rugby players have a higher BMI than the multisport men.

You can learn more about the use of the t-distribution to test an hypothesis at https://brainly.com/question/16313918