One of the major U.S. tire makers wishes to review its warranty for their rainmaker tire. The warranty is for 40,000 miles. The tire company believes that the tire actually lasts more than 40,000 miles with a standard deviation of 15,000 miles. A sample 49 tires revealed that the mean number of miles is 45,000 miles. Test the hypothesis with a 0.05 significance level. A) The decision rule is to reject if Z 1.645. The calculated value of z is +2.33. B) The decision rule is to reject if Z 1.96. The calculated value of z is +2.33. D) The decision rule is to reject if Z

Question

One of the major U.S. tire makers wishes to review its warranty for their rainmaker tire. The warranty is for 40,000 miles. The tire company believes that the tire actually lasts more than 40,000 miles with a standard deviation of 15,000 miles. A sample 49 tires revealed that the mean number of miles is 45,000 miles. Test the hypothesis with a 0.05 significance level. A) The decision rule is to reject if Z > 1.645. The calculated value of z is +2.33. B) The decision rule is to reject if Z < 1.645. The calculated value of z is -2.33. C) The decision rule is to reject if Z > 1.96. The calculated value of z is +2.33. D) The decision rule is to reject if Z < 1.96. The calculated value of z is +2.33.

Quizplus · Accepted Answer

The hypothesis being tested is whether the tires last more than 40,000 miles, which is a one-tailed test. With a significance level of 0.05 for a one-tailed test, the critical z-value is 1.645. The calculated z-value of +2.33 exceeds this, indicating the null hypothesis (that the tires last 40,000 miles or less) should be rejected in favor of the alternative hypothesis (that the tires last more than 40,000 miles).

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