Question

What aspects of rowing technique distinguish between novice and skilled competitive rowers? Researchers compared two randomly selected groups of female competitive rowers: a group of skilled rowers and a group of novices. The researchers measured many mechanical aspects of rowing style as the subjects rowed on a Stanford Rowing Ergometer. One important variable is the angular velocity of the knee, which describes the rate at which the knee joint opens as the legs push the body back on the sliding seat. The data show no outliers or strong skewness. Here is the SAS computer output:

(a) The researchers believed that the knee velocity would be higher for skilled rowers. Use the computer output to carry out an appropriate test of this belief. (Note that SAS provides two-sided P-values.) What do you conclude?

(b) Use technology to construct and interpret a$90\%$ confidence interval for the mean difference between the knee velocities of skilled and novice female rowers.

(c) If you had used Table B to construct the confidence interval described in part (b), how would the two results compare? Justify your answer without doing any calculations

Short answer

(a) Yes, the knee velocity is higher for skilled rowers according to the test

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

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

Step-by-step solution

Part (a) Step 1: Given information

The data is given as

Part (a) Step 2: Explanation

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

$H_0:{\mu }_1={\mu }_2$

$H_a:{\mu }_1>{\mu }_2$

Determine the test statistic

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

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:

$P<0.05\Rightarrow Reject{H}_0$

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

Part (b) Step 2: Given information

The given data is

Part (b) Step 2: Explanation

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 ${\mu }_1-{\mu }_2$are

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

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

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

Part (c) Step 1: Given information

The given data is

Part (c) Step 2: Explanation

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.