Question

The manager of a fleet of automobiles is testing two brands of radial tires and assigns one tire of each brand at random to the two rear wheels of eight cars and runs the cars until the tires wear out. The data (in kilometers) follow. Find a 99% confidence interval on the difference in the mean life.
Car Brand 1 Brand 2
1 37734 35202
2 45299 41635
3 36240 35500
4 32100 31950
5 37210 38015
6 48360 47800
7 38200 37810
8 33500 33215
(a) Calculate d=
(b) Calculate sD =
(c) Calculate a 99% two-sided confidence interval on the difference in mean life.

Answer 1

Answer:

Brand 1    Brand 2    Difference

37734       35202        2532

45299      41635         3664

36240      35500        740

32100      31950         150

37210       38015       −805

48360     47800        560

38200    37810          390

33500    33215        285

Sum of difference = 2532+ 3664+740+150 −805+  560 +390 +285 = 7516

Mean = $d=\frac{7516}{8}$

Mean = $d=939.5$

a) d= 939.5

$\text{Sample Standard deviation, s} = \sqrt{\dfrac{(x-\bar{x})^2}{n-1}}$

$=\sqrt{\dfrac{(2532-939.5)^2+(3664-939.5)^2+(740-939.5)^2 ...+(285-939.5)^2}{8-1}}$

=1441.21

b)SD= 1441.21

c)Calculate a 99% two-sided confidence interval on the difference in mean life.

confidence level =99%

significance level =α= 0.01

Degree of freedom = n-1 = 8-1 =7

So, $t_{\frac{\alpha}{2}}=3.499$

Formula for confidence interval $= \left( \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}} \right)$

Substitute the values

confidence interval $= 939.5 \pm 3.499 \times \frac{1441.21}{\sqrt{8}} \right)$

confidence interval $= 939.5 - 3.499 \times \frac{1441.21}{\sqrt{8}} \right)$ to  $= 939.5 + 3.499 \times \frac{1441.21}{\sqrt{8}} \right)$

Confidence interval $−843.396$ to  $2722.396$