91Ó°ÊÓ

Given:

$$\begin{gathered} n=21 \\ b=11630.6 \\ S{E}_b=1249 \end{gathered}$$

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

$$\begin{gathered} df=n-2 \\ =21-2 \\ =19 \end{gathered}$$

The critical t-value can be found in table B in the row of d f=19 and in the column of c=95% :

$t^{*}=2.093$

The boundaries of the confidence interval then become:

localid="1650543247381" $$\begin{gathered} b-t^{*}×S{E}_b \\ =11630.6-2.093×1249 \\ =9016.443 \end{gathered}$$

localid="1650543258240" $$\begin{gathered} b+t^{*}×S{E}_b \\ =11630.6+2.093×1249 \\ =14244.757 \end{gathered}$$

The slight deviation is due to rounding errors.

Need for an appropriate pair of hypotheses for this test.

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:

$H_0:\mathrm{β}=15000$

The alternative hypothesis states the opposite of the null hypothesis:

$H_a:\mathrm{β}≠15000$

Need to find this significance test based on your interval in part (a).

$H_0:\mathrm{β}=15000$

$H_a:\mathrm{β}≠15000$

Confidence interval found in part a:

$(9016.443,14244.757)$

The confidence interval does not contain 15000 and thus it is unlikely to obtain $\mathrm{β}=15000$, which means that there is sufficient evidence to reject the claim that AP Statistics teachers are typical drivers.