Chapter 10.2, Problem 55E

(a)

To determine

To Explain: the suitable test of this belief and the conclusion.

Answer to Problem 55E

Yes

Explanation of Solution

Given:

  $\begin{array}{l}{\overline{x}}_1=4.182\\ {\overline{x}}_2=3.010\\ s_1=0.479\\ s_2=0.959\\ n_1=10\\ n_2=8\end{array}$

Formula used:

  $t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$

Calculation:

Test hypothesis:

  $\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_1:{\mu }_1>{\mu }_2\end{array}$

The test statistic:

  $\begin{array}{l}t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{4.182-3.010}{\sqrt{\frac{{0.479}^2}{10}+\frac{{0.959}^2}{8}}}\\ =3.156\end{array}$

Degrees of freedom

  $df=\min \left(n_1-1,n_2-1\right)=\min \left(10-1,8-1\right)=7$

The P-value is

  $0.005<P<0.01$

In the output it is observed that the exact P-value is 0.0104/2 = 0.0052, because the test is one-sided and the P-value in output is two-sided).

The null hypothesis is rejected if the P-value is less than or equal to the significance level:

  $P<0.05\text{Reject }{H}_0$

There is strong proof to help the claim that the knee velocity would be greater for skilled rowers.

(b)

To determine

To construct: and interpret a 90 percent confidence interval for the mean difference for the provided information.

Answer to Problem 55E

0.4683, 1.8757

Explanation of Solution

Given:

  $\begin{array}{l}{\overline{x}}_1=4.182\\ {\overline{x}}_2=3.010\\ s_1=0.479\\ s_2=0.959\\ n_1=10\\ n_2=8\end{array}$

Formula used:

For the confidence interval

  $\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Calculation:

Determine the degrees of freedom:

  $df=\min \left(n_1-1,n_2-1\right)=\min \left(10-1,8-1\right)=7$

Finding $t^{\ast }$ with $df=7$ and c = 90%

  $t^{\ast }=1.895$

Confidence interval for ${\mu }_1={\mu }_2$ are:

  $\begin{array}{l}\left({\overline{x}}_1-{\overline{x}}_2\right)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =\left(4.182-3.010\right)-1.895\times \sqrt{\frac{{0.479}^2}{10}+\frac{{0.959}^2}{8}}\\ =0.4683\end{array}$

  $\begin{array}{l}\left({\overline{x}}_1-{\overline{x}}_2\right)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =\left(4.182-3.010\right)+1.895\times \sqrt{\frac{{0.479}^2}{10}+\frac{{0.959}^2}{8}}\\ =1.8757\end{array}$

90 percent believe that the mean difference is between 0.4683 and 1.8757.

(c)

To determine

To Explain: the two results with the comparison with the clarification.

Explanation of Solution

Technology uses a higher degree of freedom than there would use it when using table B, which would result in technology providing a smaller t-value $t^{\ast }$ for the technology The confidence interval using technology is narrower than the confidence interval using the table B, because technology has a lower $t^{\ast }$ value.