Question

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.

Short answer

a. The slight deviation is due to rounding errors, the correct is $\left(9016.443,14244.757\right)$.

b. The null hypothesis states that the population parameter is equal to the value given in the claim.

c. The answer is $\mathrm{t}=-2.698$, There is convincing evidence to support the claim that the car age does not increase by $15000$mileage per year.

d. There is sufficient evidence to reject the claim that AP Statistics teachers are typical drivers.

Step-by-step solution

Part (a) step 1: Given Information

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

Part (a) step 2: Simplify

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:

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

Part (b) step 1: Given Information

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

Part (b) step 2: Simplify

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" ${\mathrm{H}}_{\mathrm{\alpha }}:\mathrm{\beta }\ne 15000$

Part (c) step 1: Given Information

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

Part (c) step 2: Simplify

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 $\mathrm{df}=19$.

$0.005<\mathrm{P}<0.01$

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

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

Part (d) step 1: Given Information

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

Part (d) step 2: Simplify

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.