Chapter 12.1, Problem 21E

(a)

To determine

To Explain: the 95% confidence interval for the slope of the population regression line is (9016.4, 14244.8).

Answer to Problem 21E

(9016.443, 14244.757)

Explanation of Solution

Given:

  

  

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

Formula used:

The boundaries of the confidence interval are

  $\begin{array}{l}b-t^{\ast }\times S{E}_b\\ b+t^{\ast }\times S{E}_b\end{array}$

Calculation:

The degrees of freedom

  $df=n-2=21-2=19$

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

  $t^{\ast }=2.093$

The boundaries of the confidence interval are

  $\begin{array}{l}b-t^{\ast }\times S{E}_b=11630.6-2.093\times 1249=9016.443\\ b+t^{\ast }\times S{E}_b=11630.6+2.093\times 1249=14244.757\end{array}$

(9016.443, 14244.757)

The slight deviation is there due to rounding errors.

(b)

To determine

To Explain: that an appropriate pair of hypotheses for this test is $H_0:\beta =15,000$ v/s $H_1:\beta \ne 15,000$ .

Answer to Problem 21E

  $\begin{array}{l}{H}_0:\beta =15000\\ H_a:\beta \ne 15000\end{array}$

Explanation of Solution

Given:

  

  

Claim: the typical driver puts 15000 miles per year on its own main vehicle.

This implies that mileage is predicted to be about 15000 miles per year, which associates with a slope of 15000. The null hypothesis statement is that the population parameter is equal to the value mention in the claim:

  $H_0:\beta =15000$

The alternative hypothesis statement is the opposite of the null hypothesis:

  $H_a:\beta \ne 15000$

(c)

To determine

To find: the standardized test statistic and P-value on the basis of part (b) and conclusion at the $\alpha =0.05$ .

Answer to Problem 21E

  $\begin{array}{l}t=-2.698\\ 0.005<P<0.01\end{array}$

There is enough convincing evident to help the claim that the car age does not increase by 15,000 mileages per year.

Explanation of Solution

Given:

  

  

  $\begin{array}{l}{H}_0:\beta =15,000\\ H_1:\beta \ne 15,000\\ \alpha =0.05\\ df=19\end{array}$

Formula used:

The test statistics

  $t=\frac{b-{\beta }_0}{S{E}_b}$

Calculation:

The calculation of $\beta$ is mention in the row “Car age” and in the column “Coef” of the computer output:

  $b=11630.6$

The standard error of the calculation of $\beta$ is mention in the row “Car age” and in the column “SE Coef” of the computer output:

The test static:

  $\begin{array}{l}t=\frac{b-{\beta }_0}{S{E}_b}\\ =\frac{11630.6-15000}{1249}\\ =-2.698\end{array}$

The P-value is the probability of getting the value of the test static, when that the null hypothesis is true. $df=19$ (two tails):

  $0.005<P<0.01$

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

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

There is enough convincing evident to help the claim that the car age does not increase by 15,000 mileages per year.

(d)

To determine

To find: that the conclusion of part (a) and part (c) is same or not.

Answer to Problem 21E

There is enough proof to reject the claim that AP Statistic teachers are typical drivers.

Explanation of Solution

Given:

  

  

  $\begin{array}{l}{H}_0:\beta =15,000\\ H_1:\beta \ne 15,000\end{array}$

Confidence interval found in part a:

(9016.443, 14244.757)

The confidence interval is not having 15000 and therefore it is unlikely to get $\beta =15,000$ , which implies that there is enough proof to reject the claim that AP Statistic teachers are typical drivers.