Answer to Question #172602 in Statistics and Probability for zar

Question

1.      Two examples are randomly selected from two groups of students who have been taught using different teaching methods. An examination is given and the results are shown below:

Group 1                               Group 2

N1 = 8                                  n2 = 10

Mean of Grp 1 =85         Mean of Grp 2 =87

Variance of Grp 1 =46    Variance of Grp 2 = 36.

               Using a .05 level of significance, can we conclude that the two different teaching methods are equally effective?

Answer 1

$H_0 : μ_1 = μ_2 \\ H_1 : μ_1 ≠ μ_2 \\ n_1 = 8 \\ \bar{X_1} = 85 \\ s^2_1 = 46 \\ n_2 = 10 \\ \bar{X_2} = 87 \\ s^2_2=36$

Because both samples are small (< 30), we use the t test statistic. Before implementing the formula, we first check whether the assumption of equality of population variances is reasonable. The ratio of the sample variances, $\frac{s_1^2}{s_2^2} = \frac{46}{36} = 1.27$ , which falls between 0.5 and 2, suggesting that the assumption of equality of population variances is reasonable. The appropriate test statistic is:

$t = \frac{\bar{X_1} – \bar{X_2}}{S_p \sqrt{ \frac{1}{n_1} + \frac{1}{n_2}}}$

This is a two-tailed test, using a t statistic and a 5% level of significance. The appropriate critical value can be found in the t Table. In order to determine the critical value of t we need degrees of freedom, df, defined as df=n₁+n₂-2 = 8+10-2=16. The critical regions for a two-tailed test with df=16 and α=0.05 are t ≥ 1.745 and t ≤ -1.745 and the decision rule is: Reject H₀ if t ≥ 1.745 or t ≤ -1.745.

We now substitute the sample data into the formula for the test statistic. Before substituting, we will first compute Sp, the pooled estimate of the common standard deviation.

$S_p = \sqrt{ \frac{(n_1-1)s^2_1+(n_2-1)s^2_2}{n_1+n_2-2} } \\ S_p = \sqrt{ \frac{(8-1)46+(10-1)36}{8+10-2} } \\ S_p = 6.35 \\ t = \frac{85 – 87}{6.35 \sqrt{ \frac{1}{8} + \frac{1}{10}}} = -0.66$

Since -0.66 > -1.745, we cannot reject the null hypothesis.

Using a 0.05 level of significance, we can conclude that the two different teaching methods are equally effective.