Answer to Question #137763 in Statistics and Probability for Swarshita Ramkorun

Question

The mean height of 50 Male students who showed above average participation in college athletics was 68.2 inches with a sd of 2.5 inches, while 50 students who showed no interest in such participation had a mean height of 67.5 inches with a standard deviation of 2.8 inches. Test the significance that male students who participate in college athletics are taller than the other male students
Hypothesis
Ho : Mu 1 <= Mu 2
H1 : Mu 1 > Mu 2

Answer 1

Unpaired t-test for equality of means, unequal variances.

$H_0:\mu_1\le\mu_2$

$H_1:\mu_1>\mu_2$

$t = \frac{\overline{x}_{1} - \overline{x}_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}}$

$t = \frac{68.2 - 67.5}{\sqrt{\frac{2.5^{2}}{50} + \frac{2.8^{2}}{50}}}=1.3186$

One tailed test. Cv=$t_{{\alpha}df}$

Let$\alpha=0.05$

$df = \frac{ \left ( \frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}} \right ) ^{2} }{ \frac{1}{n_{1}-1} \left ( \frac{s_{1}^2}{n_{1}} \right ) ^{2} + \frac{1}{n_{2}-1} \left ( \frac{s_{2}^2}{n_{2}} \right ) ^{2}}$

$df = \frac{ \left ( \frac{2.5^2}{50} + \frac{2.8^2}{50} \right ) ^{2} }{ \frac{1}{50-1} \left ( \frac{2.5^2}{50} \right ) ^{2} + \frac{1}{50-1} \left ( \frac{2.8^2}{50} \right ) ^{2}}$

$Df=96.8=97$

$Cv=t_{0.05,97}=1.661$

Since the test statistic value 1.3186 is less than the critical value 1.661, we fail to reject the null hypothesis and conclude that there is no sufficient evidence to support the claim that make students who participated in the college athletics are taller than the other male students.