A random sample of $21{\mathrm{AP}}^{\circledR }$Statistics teachers was asked to report the age (in years) and mileage of their primary vehicles. Here is a scatterplot of the data:

Here is some computer output from a least-squares regression analysis of these data. Assume that the conditions for regression inference are met

$$\begin{gathered} \mathrm{Variable}\mathrm{Coef}\mathrm{SE}\mathrm{Coef}\mathrm{t}-\mathrm{ratio}\mathrm{prob} \\ \mathrm{Constant}7288.5465911.110.2826 \\ \mathrm{Car}\mathrm{age}11630.612499.31<0.0001 \\ \mathrm{S}=19280\mathrm{R}-\mathrm{Sq}=82.0\%\mathrm{RSq}\left(\mathrm{adj}\right)=81.1\% \end{gathered}$$

a. Verify that the $95\%$confidence interval for the slope of the population regression line is $(9016.4,14,244.8)$.

b. A national automotive group claims that the typical driver puts $15,000$miles per year on his or her main vehicle. We want to test whether ${\mathrm{AP}}^{\circledR }$Statistics teachers are typical drivers. Explain why an appropriate pair of hypotheses for this test is${\mathrm{H}}_0:{\mathrm{\beta }}_1=15,000\mathrm{versus}{H}_{\mathrm{\alpha }}:{\mathrm{\beta }}_1\ne 15,000.$

c. Compute the standardized test statistic and P-value for the test in part (b). What conclusion would you draw at the $\mathrm{\alpha }=0.05$significance level?

d. Does the confidence interval in part (a) lead to the same conclusion as the test in part (c)? Explain your answer.

We need to verify the$95\%$confidence interval for the slope of the population regression line is$(9016.4,14,244.8)$.

Consider:

$$\begin{gathered} \mathrm{n}=21 \\ \mathrm{b}=11630.6 \\ {\mathrm{SE}}_{\mathrm{b}}=1249 \end{gathered}$$

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

$\mathrm{df}=\mathrm{n}-2=211-2=19$

The critical t-value can be found in table B in the row of $\mathrm{df}=19$and in the column of $\mathrm{c}=95\%$:

$\mathrm{t}\text{'}=2.093$

The boundaries of the confidence interval then become:

$$\begin{gathered} b-t\text{'}\times SEb=11630.6-2.093\times 1249=9016.443 \\ b+t\text{'}\times SEb=11630.6+2.093\times 1249=14244.757 \end{gathered}$$

We need to explain whether an appropriate pair of hypotheses for this test is${\mathrm{H}}_0:{\mathrm{\beta }}_1=15,000\mathrm{versus}{H}_{\mathrm{\alpha }}:{\mathrm{\beta }}_1\ne 15,000.$

Here,

Claim: the typical driver puts $15000$miles per year on his or her main vehicle.

This means that the mileage is expected to be about $15000$miles per year, which corresponds with a slope of $15000$. The null hypothesis states that the population parameter is equal to the value given in the claim:

role="math" localid="1654334694777" ${\mathrm{H}}_0:\mathrm{\beta }=15000$

The alternative hypothesis states the opposite of the null hypothesis"

role="math" localid="1654335081788" ${\mathrm{H}}_{\mathrm{\alpha }}:\mathrm{\beta }\ne 15000$

We need to find a conclusion would you draw at the $\mathrm{\alpha }=0.05$significance level.

Consider:

$$\begin{gathered} \mathrm{df}=19 \\ \mathrm{\alpha }=\mathrm{Significance}\mathrm{level}=0.05 \end{gathered}$$

$$\begin{gathered} H_0:\beta =15000 \\ {H}_1:\beta \ne 15000 \end{gathered}$$

The estimate of role="math" localid="1654335438803" $\mathrm{\beta }$is given in the row "Car age" and in the column "Coef" of the given computer output:

role="math" localid="1654335362617" $\mathrm{b}=11630.6$

The estimated standard error of$\mathrm{\beta }$is gave in the row "Car age" and in the column "SE Coef" of the given computer output:

$S{E}_b=1249$

Compute the value of the test statistic:

$\mathrm{t}=\frac{{\mathrm{b}}_1-{\mathrm{\beta }}_1}{{\mathrm{SE}}_{{\mathrm{b}}_1}}=\frac{11630.6-15000}{1249}\approx -2.698$

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 the Student's T table in the appendix containing the t -value in the row $\mathrm{df}=19$.

$0.005<\mathrm{P}<0.01$

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

We need to explain the confidence interval in part (a) lead to the same conclusion as the test in part (c).

Here,

$$\begin{gathered} {\mathrm{H}}_0:\mathrm{\beta }=15000 \\ {\mathrm{H}}_{\mathrm{\alpha }}:\mathrm{\beta }\ne 15000 \end{gathered}$$

Confidence interval found in part a:

$(9016.443,14244.757)$

The confidence interval does not contain 15000 and thus it is to obtain =15000, which that there is sufficient evidence to reject the claim that AP Statistic teachers are typical drivers.