Chapter 12.1, Problem 20E

(a)

To determine

To explain 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 and carry out an appropriate test at the $\alpha =0.05$ level.

Answer to Problem 20E

There is sufficient evidence to support the claim 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.

Explanation of Solution

Now, we will use the excel technology to find output by first selecting the data and then going to the data analysis option and when the drop box appear select the t-test, thus, using the technology we will find the following output as:

  

It is given in output that:

  $\begin{array}{l}n=23\\ b=-9.695\\ {\text{SE}}_b=1.889\end{array}$

Thus, we define the hypothesis by:

  $\begin{array}{l}{H}_0:\beta =0\\ H_a:\beta <0\end{array}$

Thus, the value of the test statistics is as:

  $\begin{array}{l}t=\frac{b-\beta }{{\text{SE}}_b}\\ =\frac{-9.695-0}{1.889}\\ =-5.132\end{array}$

And the degrees of freedom is:

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

Thus, the P-value is as:

  $P<0.0005$

As we know that if the P-value is less than or equal to the significance level, then the null hypothesis is rejected as:

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

Thus, we conclude that there is sufficient evidence to support the claim 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.

(b)

To determine

To calculate and interpret the $95\%$ confidence interval for the slope of the population regression line.

Answer to Problem 20E

  $(-13.6241,-5.7659)$ .

Explanation of Solution

Now, we will use the excel technology to find output by first selecting the data and then going to the data analysis option and when the drop box appear select the t-test, thus, using the technology we will find the following output as:

  

It is given in output that:

  $\begin{array}{l}n=23\\ b=-9.695\\ {\text{SE}}_b=1.889\end{array}$

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

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

The critical t -value can be found in table B in the row of $df=21$ and in the column of $c=95\%$ as:

  $t^{*}=2.080$

Thus, the confidence interval is as:

  $\begin{array}{l}b-t^{*}\times {\text{SE}}_b\\ =-9.695-2.080\times 1.889\\ =-13.6241\\ b+t^{*}\times {\text{SE}}_b\\ =-9.695+2.080\times 1.889\\ =-5.7659\end{array}$

We are $95\%$ confident that the slope of the true regression line is between $(-13.6241,-5.7659)$ .