91影视

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:

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

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

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" ${\mathrm{H}}_0:\mathrm{尾}=15000$

The alternative hypothesis states the opposite of the null hypothesis"

role="math" localid="1654335081788" ${\mathrm{H}}_{\mathrm{伪}}:\mathrm{尾}鈮�15000$

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

Consider:

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

$$\begin{gathered} H_0:尾=15000 \\ {H}_1:尾鈮�15000 \end{gathered}$$

The estimate of role="math" localid="1654335438803" $\mathrm{尾}$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$\mathrm{尾}$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:

$\mathrm{t}=\frac{{\mathrm{b}}_1鈭�{\mathrm{尾}}_1}{{\mathrm{SE}}_{{\mathrm{b}}_1}}=\frac{11630.6鈭�15000}{1249}鈮�鈭�2.698$

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 $\mathrm{df}=19$.

$0.005<\mathrm{P}<0.01$

$\mathrm{P}<0.05鈬�\mathrm{Reject}{\mathrm{H}}_0$

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

Here,

$$\begin{gathered} {\mathrm{H}}_0:\mathrm{尾}=15000 \\ {\mathrm{H}}_{\mathrm{伪}}:\mathrm{尾}鈮�15000 \end{gathered}$$

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.