A paired difference experiment produced the following results:

${\mathbf{\text{n}}}_{\mathbf{\text{d}}}\mathbf{\text{=38,}}{\overline{\mathbf{\text{x}}}}_{\mathbf{\text{1}}}\mathbf{\text{=92,}}{\overline{\mathbf{\text{x}}}}_{\mathbf{\text{2}}}\mathbf{\text{=95.5,}}\overline{\mathbf{\text{d}}}{{\mathbf{\text{=-3.5,s}}}_{\mathbf{\text{d}}}}^{\mathbf{\text{2}}}\mathbf{\text{=21}}$

a. Determine the values $\mathit{z}$for which the null hypothesis ${\mathbf{\text{μ}}}_{\mathbf{\text{1}}}\mathbf{-}{\mathbf{\text{μ}}}_{\mathbf{\text{2}}}\mathbf{\text{=0}}$would be rejected in favor of the alternative hypothesis ${\mathbf{\text{μ}}}_{\mathbf{\text{1}}}\mathbf{-}{\mathbf{\text{μ}}}_{\mathbf{\text{2}}}\mathbf{\text{<0}}$ Use .role="math" localid="1652704322912" $\mathbf{\text{α=.10}}$

b. Conduct the paired difference test described in part a. Draw the appropriate conclusions.

c. What assumptions are necessary so that the paired difference test will be valid?

d. Find a$\mathbf{90}\mathbf{\%}$ confidence interval for the mean difference ${\mathbf{\mu }}_{\mathbf{d}}$.

e. Which of the two inferential procedures, the confidence interval of part d or the test of the hypothesis of part b, provides more information about the differences between the population means?

Null Hypothesis:$H_0:{\mu }_1-{\mu }_2=0$

Alternate Hypothesis:$H_a:{\mu }_1-{\mu }_2<0$

The critical value for a left-tailed test$\alpha =0.1$ is$-1.28$ .

If $z_{data}<-1.28$then, reject the null hypothesis.

$\begin{array}{l}z=\frac{\overline{d}-0}{\sqrt{\left(\frac{{s_d}^2}{n_d}\right)}}\\ =\frac{-3.5-0}{\sqrt{\left(\frac{21}{38}\right)}}\\ =\frac{-3.5}{0.7434}\\ =-4.71\end{array}$

Therefore, the value of paired difference test is $-4.71$.

The null hypothesis will be rejected, and it can be concluded that $H_a:{\mu }_1-{\mu }_2<0$.

1. By Central Limit Theorem, the sampling distribution$\overline{d}$ follows a normal distribution.
2. The sample differences are randomly selected from the population differences.
3. The sample size of the no. of pairs is large, which is more than $30$.

Confidence interval $=90\%$

Level of Significance $\left(\alpha \right)=0.10$

Confidence Interval $=\overline{d}\pm z_{\alpha /2}\left(\sqrt{\frac{{s_d}^2}{n_d}}\right)$

$\begin{array}{l}=-3.5\pm z_{0.05}\left(\sqrt{\frac{21}{38}}\right)\\ =-3.5\pm 1.645\left(0.7434\right)\\ =-3.5\pm 1.223\\ =(-4.723,-2.277)\end{array}$

Therefore,$90\%$the confidence interval for the mean difference ${\mu }_d$is $(-4.723,-2.277)$.

The confidence intervals tell us about the difference between population means of an interval for all possible values. So, it gives more information than the z-test.