Question

Use a t-distribution and the given matched pair 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 distribution of the differences is relatively normal. Assume that differences are computed using \(d=x_{1}-x_{2}\). Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}<\mu_{2}\) using the paired data in the following table: $$ \begin{array}{lllllllll} \hline \text { Treatment } 1 & 16 & 12 & 18 & 21 & 15 & 11 & 14 & 22 \\ \text { Treatment } 2 & 18 & 20 & 25 & 21 & 19 & 8 & 15 & 20 \\ \hline \end{array} $$

Short answer

The null hypothesis \(H_{0}: \mu_{1}=\mu_{2}\) is rejected in favor of the alternative hypothesis \(H_{a}: \mu_{1}<\mu_{2}\), with t-statistic \(-2.0366\) and p-value less than \(0.05\). It's suggested that the mean of treatment 1 is less than treatment 2.

Step-by-step solution

Calculate differences

Calculate the differences between each pair of treatments 1 and 2 by using the formula \(d=x_{1}-x_{2}\). So, differences would be \(-2, -8, -7, 0, -4, 3, -1, 2\).

Compute mean of differences

Calculate the mean of these differences using the formula \(\bar{d} = \frac{1}{n} \sum_{i=1}^{n} d_{i}\). With the given differences and \(n=8\), the mean of differences, \(\bar{d}\), is \(-2.125\).

Calculate standard deviation of differences

Calculate the standard deviation of the differences using the formula \(s_{d} = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (d_{i}-\bar{d})^{2}}\). The standard deviation, \(s_{d}\), of these differences turns out to be approximately \(3.3066\).

Compute t-statistic

Calculate the t-statistic using the formula \(t = \frac{\bar{d}}{s_{d}/\sqrt{n}}\). The t-score is \(-2.0366\).

Determine p-value

For a one-tailed test with \(n-1=7\) degrees of freedom, the critical t-value for a significance level of \(0.05\) is approximately \(1.895\). The computed t-statistic is less than this, hence, the p-value is less than \(0.05\).

Make a decision

Because the p-value is less than \(0.05\), we reject the null hypothesis in favor of the alternative hypothesis, suggesting that there is a significant difference and the mean of treatment 1 is less than treatment 2.