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.

Answer 1

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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.

Which brand will you prefer based on this calculation

Car Brand 1 Brand 2

1 36,925 34,318

2 45,300 42,280

3 36,240 35.500

4 32,100 31,950

5 37,210 38,015

6 48,360 47,800

7 38,200 37,810

8 33,500 33,215

Answer:

the 99% confidence interval on the difference in the mean life. =

`- 727.46 < \mu_d < 2464.21`

Explanation:

Car Brand 1 Brand 2 d = Brand (1) - Brand (2)

1 36,925 34,318 2607

2 45,300 42,280 3020

3 36,240 35.500 740

4 32,100 31,950 150

5 37,210 38,015 - 805

6 48,360 47,800 560

7 38,200 37,810 390

8 33,500 33,215 285

The mean of the above value of d is calculated as:

Mean (d') =
`\frac {\sum d}{n}`

=
`(2607 + 3020+740+150+(-805)+ 560+390+285)/(8)`

=
`(6947)/(8)`

= 868.375

Standard deviation

`(S_d) = \sqrt{((2607-868.375)^2+(3020-868.375)^2+(740-868.375)^2+...+(285-868.375)^2)/(8-1)}`

`(S_d) = √(1664185.41)`

`(S_d) = 1290.03`

The degree of freedom is calculated as:

`df = n-1 \\df= 8-1\\df = 7`

Now, from the t - tabulated value;

t-critical value at 99% confidence level at 99% confidence level with 7 degree of freedom =
`\pm \ 3.499`

In order to determine the 99% confidence interval on the difference in mean life ; we have:

`99`% C.I =
`d' + I_(critical) ( {(S_d)/(√(n)) })`

`99`% C.I =
`868.375 \pm3.499 ( {(1290.03)/(√(8)) })`

`99`% C.I =
`868.375 \pm 1595.835`

`99`% C.I = ( -727.46 ; OR; 2464.21 )

Thus; the 99% confidence interval on the difference in the mean life. =

`- 727.46 < \mu_d < 2464.21`

CONCLUSION:

We therefore conclude that the 99% confidence interval contains null values, as such there is no significant difference between the two brands of tire.