Answer to Question #143533 in Statistics and Probability for danica

Question

If the following information test for the significant difference between the
two means of the samples
Particulars Sample 1 Sample 2
Mean 20.86 18.94
S.D 2.98 2.66
Size 10 8

Answer 1

The following null and alternative hypotheses need to be tested:

$Ho: \mu_1 ​ = \mu_2\\ Ha: \mu_1 ​ ≠ \mu_2$

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Based on the information provided, the significance level isα=0.05, and the degrees of freedom are df = 16

In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal(since variances given are approximately equal)

Hence, it is found that the critical value for this two-tailed test is $t_c = 2.12$

, for α=0.05 and df = 16

The t-statistic is computed as follows:

$T = \frac{\bar{Y_{1}} - \bar{Y_{2}}} {s_{p}\sqrt{1/N_{1} + 1/N_{2}}}$ where $s_{p}^{2} = \frac{(N_{1}-1){s^{2}_{1}} + (N_{2}-1){s^{2}_{2}}} {N_{1} + N_{2} - 2}$

substituting the values , $s_{p}^{2} = \frac{(10-1){2.98^2} + (8-1){2.66^2}} {10 +8 - 2}$ = 8.0908

Hence T= $\frac{20.86 - 18.94} {2.8444\sqrt{\frac{1}{10} + \frac{1}{8}}}$ = 1.23

Since it is observed that $|t| = 1.23 \le t_c = 2.12$ , it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

Using ti84, the p-value is p=0.2364≥0.05, it is concluded that the null hypothesis is not rejected. It is concluded that the null hypothesis Ho is not rejected and  is not statistically significant. Therefore, there is not enough evidence to claim that the population mean μ1 is different than $\mu_2$ at the 0.05 significance level.