A metropolitan transportation authority has set a bus mechanical reliability goal of 3,800 bus miles. Bus mechanical reliability is measured specifically as the number of bus miles between mechanical road calls. Suppose a sample of 100 of 100 buses resulted in a sample mean of 3850 bus miles and a sample standard deviation of 275 bus miles. Please show your work.

A. T-stat=

B. The critical values is/are?

C. Is there sufficient evidence to reject the null hypothesis using a=0.05.

(A) T-stat = 1.82

(B) Critical value is 1.645

(C) Yes, there is sufficient evidence to reject the null hypothesis using a 0.05 significance level.

Step-by-step explanation:

(A) T-stat (z) = (sample mean - population mean) ÷ sd/√n

sample mean = 3850 bus miles

population mean = 3800 bus miles

sd = 275 bus miles

n = 100

z = (3850 - 3800) ÷ 275/√100 = 50 ÷ 275/10 = 50 ÷ 27.5 = 1.82

(B) The test is a one-tailed test. The critical value is obtained from the standard normal distribution table. The critical value using a 0.05 significance level is 1.645

(C) Null hypothesis: The bus mechanical reliability is 3850 bus miles.

Alternate hypothesis: The bus mechanical reliability is less than 3850 bus miles.

Conclusion:

There is sufficient evidence to reject the null hypothesis because the test statistic (1.82) is greater than the critical value (1.645)

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