Question

Use the t-distribution and the given sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distributions are relatively normal. Test \(H_{0}: \mu_{A}=\mu_{B}\) vs \(H_{a}: \mu_{A} \neq \mu_{B}\) using the fact that Group A has 8 cases with a mean of 125 and a standard deviation of 18 while Group \(\mathrm{B}\) has 15 cases with a mean of 118 and a standard deviation of 14 .

Short answer

The final decision about the null hypothesis, whether it is rejected or not, will depend on the results of the t-test via the comparison of the calculated test statistic and the critical value. After careful calculation and comparison, a conclusion can be drawn about whether there is a significant difference between the two groups, Group A and Group B.

Step-by-step solution

Defining Null and Alternative Hypotheses

The null hypothesis \(H_{0}: \mu_{A}=\mu_{B}\) suggests that there is no significant difference between the means of Group A and Group B. The alternative hypothesis \(H_{a}: \mu_{A} \neq \mu_{B}\) states that the means are not equal, indicating there is a significant difference.

Calculate the Test Statistic

The test statistic for a two-sample t-test is calculated using the formula: \(t = \frac{(\bar{X}_{A} - \bar{X}_{B}) - (\mu_{A} - \mu_{B})}{\sqrt{\frac{s_{A}^{2}}{n_{A}} + \frac{s_{B}^{2}}{n_{B}}}}\), where \(\bar{X}_{A}\) and \(\bar{X}_{B}\) are the sample means, \(s_{A}^{2}\) and \(s_{B}^{2}\) are the sample variances, and \(n_{A}\) and \(n_{B}\) are the sample sizes for Groups A and B respectively. Substituting the given values: \(t = \frac{(125 - 118) - 0}{\sqrt{\frac{18^{2}}{8} + \frac{14^{2}}{15}}}\)

Determine the Critical Value and P-Value

For a 2-tailed t test, and assuming \(\alpha=0.05\) (common in most research settings), we have to find the critical value(s) that divide(s) the center 95% of the t-distribution from the 5% in the tails. The degree of freedom is calculated as \(df=n_{A}+n_{B}-2\), which is 21. Using t-table or software for better precision, we find these critical values and the corresponding p-value to match our calculated t in the previous step.

Compare Test Statistic and Critical Value

After calculating the t statistic, we compare this value with the critical value obtained from the t-table. If the calculated t-value is greater than the critical value, we reject the null hypothesis. Conversely, if the calculated t-value is less than the critical value, we fail to reject the null hypothesis.

Make a Decision

Based on the comparison between the calculated test statistic and the critical value, a decision is made about the null hypothesis. If the null hypothesis is rejected, it suggests that there is a significant difference between the two groups. If the null hypothesis is not rejected, there is no significant difference between the two groups.