91Ó°ÊÓ

The data is given as

The given data is written as

${\overline{x}}_1=4.182$

${\overline{x}}_2=3.010$

$s_1=0.479$

$s_2=0.959$

$n_1=10$

$n_2=8$

Find the hypothesis

$H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2$

$H_a:{\mathrm{μ}}_1>{\mathrm{μ}}_2$

Determine the test statistic

localid="1650519272086" $$\begin{gathered} t=\frac{{\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ =\frac{4.182-3.010}{\sqrt{\frac{0.{479}^2}{10}+\frac{0.{959}^2}{8}}} \\ ≈3.156 \end{gathered}$$

Determine the degrees of freedom

localid="1650519294480" $$\begin{gathered} df=\min \left(n_1-1,n_2-1\right) \\ =\min (10-1,8-1) \\ =7 \end{gathered}$$

The P-value is the chance of getting the test statistic's result or a number that is more severe. The P-value is the number (or interval) in Table B's column title that corresponds to the t-value in the row $df=20$

$0.005<P<0.01$

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

$P<0.05⇒Reject{H}_0$

There is ample data to support the claim that expert rowers' knee velocity would be higher, so it is true.

The given data is

From the given data in part (a)

Determine the degrees of freedom

$$\begin{gathered} df=\min \left(n_1-1,n_2-1\right) \\ =\min (10-1,8-1) \\ =7 \end{gathered}$$

Find $t*$with $df=7$and $c=90\%$using table B

$t^{*}=1.895$

The confidence interval with endpoints for ${\mathrm{μ}}_1-{\mathrm{μ}}_2$are

localid="1650519363931" $$\begin{gathered} \left({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2\right)-t_{\mathrm{Î\pm }/2}·\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=(4.182-3.010)-1.895·\sqrt{\frac{0.{479}^2}{10}+\frac{0.{959}^2}{8}} \\ ≈0.4683 \end{gathered}$$

localid="1650519384761" $$\begin{gathered} \left({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2\right)+t_{\mathrm{Î\pm }/2}·\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=(4.182-3.010)+1.895·\sqrt{\frac{0.{479}^2}{10}+\frac{0.{959}^2}{8}} \\ ≈1.8757 \end{gathered}$$

It is $90\%$confident that the mean difference is between$(0.4683,1.8757).$

The given data is

Technology has more degrees of freedom than table B, resulting in a smaller t-value $t*$for Technology.

Since Technology has a smaller $t*$ value, the confidence interval by using Technology is narrower than the confidence interval by using the table B.