The NCAA is interested in estimating the difference in mean number of daily training hours for men and women athletes on college campuses. It wants 95 percent confidence and will select a sample of 10 men and 10 women for the study. The variances are assumed equal and the populations normally distributed. The sample results are:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Data given

[tex] \bar X_M = 2.7[/tex] represent the sample mean for men

[tex] \bar X_F = 2.4[/tex] represent the sample mean for women

[tex] s_M = 0.3[/tex] represent the sample deviation for men

[tex] s_F = 0.4[/tex] represent the sample deviation for women

[tex] n_M = 10[/tex] sample size of male

[tex] n_F =10[/tex] sample size of women

The confidence interval is given by:

[tex] (\bar X_M -\bar X_F) \pm t_{\alpha/2} S_p \sqrt{\frac{1}{n_M}+\frac{1}{n_F}}[/tex] (1)

The polled variance can be calculated with this formula:

[tex]s^2_p = \frac{9*0.3^2 +9*0.4^2}{10+10-2}=0.125[/tex]

[tex] s_p =\sqrt{0.125}=0.354[/tex]

For a confidence of 95% the value for the significance is [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 = 0.025[/tex], the degrees of freedom are given by:

[tex] df = n_M + n_F -2= 10+10-2=18[/tex]

And the critical value can be calculated with the following formula in excel: "=T.INV(1-0.025,18)" and we got [tex] t_{\alpha/2}=2.1[/tex]

Now we can replace into the confidence interval:

[tex] (2.7 -2.4) - 2.1*0.354\sqrt{\frac{1}{10}+\frac{1}{10}}=-0.032[/tex]

[tex] (2.7 -2.4)+ 2.1*0.354\sqrt{\frac{1}{10}+\frac{1}{10}}=0.632[/tex]