Question

suppose that 100 tyres of a certain brand lasted on the average 21431 miles with a standard deviation of 1295 miles. using a level of significance of 0.05, test the hypothesis mean = 22000 miles against the alternative hypothesis mean < 22000 miles.

Answer 1

$\bar{x}=21431 \\ s=1295 \\ n=100 \\ H_0: \mu = 22000 \\ H_1: \mu < 22000$

Test-statistic:

$t= \frac{\bar{x} - \mu}{s / \sqrt{n}} \\ t = \frac{21431-22000}{1295 / \sqrt{100}} = -4.39 \\ α=0.05 \\ df=100-1 = 99$

Rejection region:

Reject the H₀ if P-value is less than 0.05.

P-value:

The P-value is computed using the excel formula for t-distribution which is =T.DIST(-4.39, 99, TRUE), thus the P-value is computed as $1.42 \times 10^{-5}.$

Conclusion: Since the P-value is less than 0.05 hence we reject the null hypothesis and conclude that mean < 22000 miles at 0.05 level of significance.