Question

Fifty senior students in statistic got an average grade of 85 with standard deviation of 10.2, while a group of 60 senior students have an average grade of 80 with standard deviation of 8.90 can the difference in the mean grade be attributed to chance, using 0.05 level of significance?

Answer 1

$H_0: \mu_1 = \mu_2 \\ H_1: \mu_1 ≠ \mu_2$

Test statistic :

To test the hypothesis the most appropriate test is two independent samples t-test assuming equal population variances. The test statistic is given as follows :

$t= \frac{\bar{x_1} -\bar{x_2}}{\sqrt{S^2_{Pooled}(\frac{1}{n_1} + \frac{1}{n_2}) }}$

Where, $\bar{x_1}, \bar{x_2}$ are sample means, n₁, n₂ are sample sizes and

$S^2_{Pooled} = \frac{(n_1-1)s^2_1+(n_2-1)s^2_2}{n_1+n_2 -2} \\ \bar{x_1} = 85 \\ \bar{x_2} = 80 \\ s_1 = 10.2 \\ s_2= 8.9 \\ n_1 = 50 \\ n_2 = 60$

$S^2_{Pooled} = \frac{(50-1)(10.2)^2+(60-1)(8.9)^2}{50+60 -2} = 90.4755 \\ t = \frac{85-80}{\sqrt{90.4755 \times (\frac{1}{50} + \frac{1}{60})}}= 2.7452$

The value of the test statistic is 2.7452.

P-value :

Our test is two-tailed test. The p-value for two-tailed t-test is given as follows :

P-value = 2.P(T > | t |)

P-value = 2.P(T > 2.7452)

P-value = 0.0071

Decision: Since the p-value is less than the significance level of 0.05, therefore, we shall reject the null hypothesis at a 0.05 significance level.

Conclusion: At 0.05 significance level there is sufficient evidence to conclude that the difference in the mean grade is not attributed to chance only.