Following is a sample of five matched pairs.
$\begin{array}{llllll}
\hline \text { Sample 1 } & 18 & 26 & 20 & 20 & 20 \
\text { Sample 2 } & 17 & 20 & 18 & 19 & 18 \
\hline
\end{array}$

Let $\mu _ { 1 }$ and $\mu _ { 2 }$ represent the population means and let $\mu _ { \mathrm { d } } = \mu _ { 1 } - \mu _ { 2 }$. A test will be made of the hypotheses $H _ { 0 } : \mu _ { \mathrm { d } } = 0$ versus $H _ { 1 } : \mu _ { \mathrm { d } } 0$. Can you reject $H _ { 0 }$ at the $\alpha = 0.05$ level of significance?

Question

Following is a sample of five matched pairs. 
$\begin{array}{llllll}
\hline \text { Sample 1 } & 18 & 26 & 20 & 20 & 20 \
\text { Sample 2 } & 17 & 20 & 18 & 19 & 18 \
\hline
\end{array}$

Let $\mu _ { 1 }$ and $\mu _ { 2 }$ represent the population means and let $\mu _ { \mathrm { d } } = \mu _ { 1 } - \mu _ { 2 }$. A test will be made of the hypotheses $H _ { 0 } : \mu _ { \mathrm { d } } = 0$ versus $H _ { 1 } : \mu _ { \mathrm { d } } > 0$. Can you reject $H _ { 0 }$ at the $\alpha = 0.05$ level of significance?

Quizplus · Accepted Answer

The sample mean difference is 1.2, and the sample standard deviation of the differences is 1.414. The test statistic is 1.2/1.414 = 0.85. The critical value for a one-tailed test with $\alpha = 0.05$ and 4 degrees of freedom is 2.132. Since the test statistic is less than the critical value, we cannot reject $H _ { 0 }$ at the $\alpha = 0.05$ level of significance.

Following Is a Sample of Five Matched Pairs Sample 1 1826202020 Sample 2 1720181918\begin{array}{llllll}\hline \text { Sample 1 } & 18 & 26 & 20 & 20 & 20 \\\text { Sample 2 } & 17 & 20 & 18 & 19 & 18 \\\hline\end{array} Sample 1 Sample 2 ​1817​2620​2018​2019​2018​​

Following Is a Sample of Five Matched Pairs $\begin{array}{llllll}\hline \text { Sample 1 } & 18 & 26 & 20 & 20 & 20 \\\text { Sample 2 } & 17 & 20 & 18 & 19 & 18 \\\hline\end{array}$