Let mu_1 denote true average tread life for a premium brand of P205/65R15 radial

Question

Let mu_1 denote true average tread life for a premium brand of P205/65R15 radial tire, and let mu_2 denote the true average tread life for an economy brand of the same size. Test H_0: mu_1 - mu_2 = 5000 versus H_1: mu_1 - mu_2 > 5000 at level 0.01, using the following data: m = 35, x bar = 42,100, s_1 = 2500, n = 35, y bar = 36,900, and s_2 = 1500. Calculate the test statistic and determine the P_value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value = State the conclusion in the problem context. Reject H_0. The data does not suggest that the difference in average tread life exceeds 5000. Fail to reject H_0. The data does not suggest that the difference in average tread life exceeds 5000. Reject H_0. The data suggests that the difference in average tread life exceeds 5000. Fail to reject H_0. The data suggests that the difference in average tread life exceeds 5000.

Explanation / Answer

The statistical software output for the given problem is:

Two sample Z hypothesis test:
1 : Mean of population 1 (Std. dev. = 2500)
2 : Mean of population 2 (Std. dev. = 1500)
1 - 2 : Difference between two means
H0 : 1 - 2 = 5000
HA : 1 - 2 > 5000

Hypothesis test results:

Hence,

z = 0.41

p - value = 0.3424

Since this is greater than 0.01, we do not reject the null hypothesis.

Option B is correct.

Difference n1 n2 Sample mean Std. err. Z-stat P-value 1 - 2 35 35 5200 492.80538 0.40583972 0.3424

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