Question

A researcher is interested in evaluating a certain brand of radial auto tire. Twenty tires are randomly selected from retail outlets throughout the country, and each is placed on a special machine which rotates the tires at a constant speed (equivalent to 55 miles per hour) against the friction equivalent of a 4000 pound auto being driven on a smooth highway. Each tire is run until there is no tread left. The number of miles (in thousands) were as follows: 40, 30, 32, 35, 39, 35, 31, 36, 37, 35, 34, 35, 37, 34, 36, 38, 35, 36, 35, 36.

Setting alpha at .01, test the hypothesis that this sample of tires could represent a population whose mean was 36.50.

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	40	22.09
	30	28.09
	32	10.89
	35	0.09
	39	13.69
	35	0.09
	31	18.49
	36	0.49
	37	2.89
	35	0.09
	34	1.69
	35	0.09
	37	2.89
	34	1.69
	36	0.49
	38	7.29
	35	0.09
	36	0.49
	35	0.09
	36	0.49
Total	706	112.2

Mean X̅ = Σ Xi / n
X̅ = 706 / 20 = 35.3
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 112.2 / 20 -1 ) = 2.4301

To Test :-
H0 :- µ = 36.50
H1 :- µ ≠ 36.50

Test Statistic :-
t = ( X̅ - µ ) / ( S / √(n))
t = ( 35.3 - 36.5 ) / ( 2.4301 / √(20) )
t = -2.2084

Test Criteria :-
Reject null hypothesis if | t | > t(α/2, n-1)
Critical value t(α/2, n-1) = t(0.01 /2, 20-1) = 2.861 ( From t table )
| t | > t(α/2, n-1) = 2.2084 < 2.861
Result :- Fail to reject null hypothesis

Decision based on P value
P - value = P ( t > 2.2084 ) = 0.0397
Reject null hypothesis if P value < α = 0.01 level of significance
P - value = 0.0397 > 0.01 ,hence we fail to reject null hypothesis
Conclusion :- Fail to reject null hypothesis

There is sufficient evidence to support the claim that sample of tyre represent a population whose mean was 36.50.