Answer to Question #258112 in Statistics and Probability for Barbs

Question

A researcher wishes to see if there is a difference in the cholesterol levels of two groups of men. A random sample of 30 men between the ages of 25 and 40 is selected and tested. The average level is 223 with standard deviation of 6. A second sample of 25 men between the ages of 41 and 56 is selected and tested. The average of this group is 229 with standard deviation of 11. Use α = 0.01, to test the difference in the cholesterol levels between the two groups. What is the critical value and the computed test statistic?

Answer 1

$n_1=30 \\ \bar{x_1} = 223 \\ s_1 = 6 \\ n_2 = 25 \\ \bar{x_2} = 229 \\ s_2 = 11 \\ H_0: \mu_2 -\mu_1 = 0 \\ H_1: \mu_2 -\mu_1 ≠ 0$

Test-statistic

$t = \frac{(\bar{x_2} - \bar{x_1}) -(\mu_2 -\mu_1)}{s_p \sqrt{(\frac{1}{n_1}) + (\frac{1}{n_2})}} \\ s^2_p = \frac{(n_1-1)s^2_1 +(n_2-1)s^2_2}{n_1+n_2-2} \\ s^2_p = \frac{(29 \times 36) + (24 \times 121)}{30+25-2} \\ = \frac{1044+2904}{53} = 74.49 \\ s_p = 8.63 \\ t = \frac{(229 - 223) -0}{8.63 \sqrt{(\frac{1}{30}) + (\frac{1}{25})}} \\ t = \frac{6}{2.182} = 2.567 \\ df = n_1+n_2-2 = 53 \\ α=0.01 \\ t_{crit} = 2.3988$

Reject the null hypothesis if $|t| ≥ t_{crit}$

$t = 2.567 > t_{crit} = 2.3988$

So, we reject the null hypothesis at a 0.01 level of significance.

We can conclude that there is NO significant difference in the cholesterol levels between the two groups.