The professor swims Here are data on the time (in minutes) Professor Moore takes to swim 2000 yards and his pulse rate (beats per minute) after swimming for a random sample of 23 days:

(a) Is there statistically significant evidence of a negative linear relationship between Professor Moore's swim time and his pulse rate in the population of days on which he swims 2000 yards?

Carry out an appropriate significance test at the $\alpha =0.05$ level.

(b) Calculate and interpret a $95\%$ confidence interval for the slope $\beta$of the population regression line.

Given:

$n=23$

$b=-9.695$

$S{E}_b=1.889$

Determine the hypothesis:

$H_0:\beta =0$

$H_a:\beta <0$

Compute the value of the test statistic:

localid="1650543668177" $$\begin{gathered} t=\frac{b-{\beta }_0}{S{E}_b} \\ =\frac{-9.695-0}{1.889} \\ \approx -5.132 \end{gathered}$$

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Table B containing the t-value in the row localid="1650543673841" $$\begin{gathered} df=n-2 \\ =23-2 \\ =21: \end{gathered}$$

$P<0.0005$

If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:

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

There is sufficient evidence to support the claim that there is a negative linear relationship between Professor Moore's swim time and his pulse rate in the population of days on which he swims 2000 yards.

Given:

$n=23$

$b=-9.695$

$S{E}_b=1.889$

The degrees of freedom is the sample size decreased by 2 :

$$\begin{gathered} df=n-2 \\ =23-2 \\ =21 \end{gathered}$$

The critical t-value can be found in table B in the row of d f=21 and in the column of c=95% :

$t^{*}=2.080$

The boundaries of the confidence interval then become:

localid="1650543691786" $$\begin{gathered} b-t^{*}\times S{E}_b \\ =-9.695-2.080\times 1.889 \\ =-13.6241 \end{gathered}$$

localid="1650543696751" $$\begin{gathered} b+t^{*}\times S{E}_b \\ =-9.695+2.080\times 1.889 \\ =-5.7659 \end{gathered}$$