Researchers were interested in comparing two methods for estimating tire wear. The first method used the amount of weight lost by a tire. The second method used the amount of wear in the grooves of the tire. A random sample of $16$tires was obtained. Both methods were used to estimate the total distance traveled by each tire. The table below provides the two estimates (in thousands of miles) for each tire.

(a) Construct and interpret a $95\%$confidence interval for the mean difference $\mu$in the estimates from these two methods in the population of tires.

(b) Does your interval in part (a) give convincing evidence of a difference in the two methods of estimating tire wear? Justify your answer.

role="math" localid="1649913699288" $\begin{array}{ccccccc}\text{Tire}& \text{Weight}& \text{Groove}& & \text{Tire}& \text{Weight}& \text{Groove}\\ 1& 45.9& 35.7& & 9& 30.4& 23.1\\ 2& 41.9& 39.2& & 10& 27.3& 23.7\\ 3& 37.5& 31.1& & 11& 20.4& 20.9\\ 4& 33.4& 28.1& & 12& 24.5& 16.1\\ 5& 31.0& 24.0& & 13& 20.9& 19.9\\ 6& 30.5& 28.7& & 14& 18.9& 15.2\\ 7& 30.9& 25.9& & 15& 13.7& 11.5\\ 8& 31.9& 23.3& & 16& 11.4& 11.2\\ & & & & & & \end{array}$

$\begin{array}{rrr}\text{Sample 1}& \text{Sample 2}& \text{Difference D}\\ 45.9& 35.7& 10.2\\ 41.9& 39.2& 2.7\\ 37.5& 31.1& 6.4\\ 33.4& 28.1& 5.3\\ 31& 24& 7\\ 30.5& 28.7& 1.8\\ 30.9& 25.9& 5\\ 31.9& 23.3& 8.6\\ 30.4& 23.1& 7.3\\ 27.3& 23.7& 3.6\\ 20.4& 20.9& -0.5\\ 24.5& 16.1& 8.4\\ 20.9& 19.9& 1\\ 18.9& 15.2& 3.7\\ 13.7& 11.5& 2.2\\ 11.4& 11.2& 0.2\\ Mean& & 4.5563\\ Sd& & 3.2255\end{array}$

The mean is the sum of all values divided by the number of values:

$$\begin{gathered} \overline{x}=\frac{10.2+2.7+\dots +2.2+0.2}{16} \\ \approx 4.5563 \end{gathered}$$

$n$is the number of values in the data set.

The variance is the sum of squared deviations from the mean divided by$n-1$:

localid="1650087093425" $$\begin{gathered} s^2=\frac{(10.2-4.5563{)}^2+\dots ..+(0.2-4.5563{)}^2}{16-1} \\ \approx 10.4040 \end{gathered}$$

The standard deviation is the square root of the variance:

localid="1650087102851" $$\begin{gathered} s=\sqrt{10.4040} \\ \approx 3.2255 \end{gathered}$$

Determine the t-value by looking in the row starting with degrees of freedom $n-1=16-1=15$and in the row with $c=95\%$in table B :

$t^{*}=2.131$

The margin of error is then:

localid="1650087135808" $$\begin{gathered} E=t^{*}·\frac{s}{\sqrt{n}} \\ =2.131\times \frac{3.2255}{\sqrt{16}} \\ \approx 1.7184 \end{gathered}$$

Then the confidence interval becomes:

localid="1650087172012" $$\begin{gathered} 2.8379=4.5563-1.7184 \\ =\overline{x}-E<\mu <\overline{x}+E \\ =4.5563+1.7184 \\ =6.2747 \end{gathered}$$

$\begin{array}{rrr}\text{Sample 1}& \text{Sample 2}& \text{Difference D}\\ 45.9& 35.7& 10.2\\ 41.9& 39.2& 2.7\\ 37.5& 31.1& 6.4\\ 33.4& 28.1& 5.3\\ 31& 24& 7\\ 30.5& 28.7& 1.8\\ 30.9& 25.9& 5\\ 31.9& 23.3& 8.6\\ 30.4& 23.1& 7.3\\ 27.3& 23.7& 3.6\\ 20.4& 20.9& -0.5\\ 24.5& 16.1& 8.4\\ 20.9& 19.9& 1\\ 18.9& 15.2& 3.7\\ 13.7& 11.5& 2.2\\ 11.4& 11.2& 0.2\\ Mean& & 4.5563\\ Sd& & 3.2255\end{array}$

Confidence interval found in exercise part (a):

$2.8925<\mu <6.2201$

The confidence interval does not contain $0$and thus there is convincing evidence of a difference in the two methods of estimating tire wear.