Question

The Metropolitan Bus Company claims that the mean waiting time for a bus during rush hour is less than 10 minutes. A random sample of 20 waiting times has a mean of 8.6 minutes with a sample standard deviation of 2.1 minutes. At ? = 0.01, test the bus company's claim. Assume the distribution is normally distributed.

-		critical value z0 = -2.326; standardized test statistic ? -2.981; reject H0; There is sufficient evidence to support the Metropolitan Bus Company's claim.
	-	critical value t0 = -2.539; standardized test statistic ? -2.981; reject H0; There is sufficient evidence to support the Metropolitan Bus Company's claim.

Answer 1

The claims of the Metropolitan Bus Company is that "the mean waiting time for a bus during rush hour is less than 10 minutes".

Sample size = n = 20

sample mean = = 8.6 minutes

sample standard deviation = s= 2.1 minutes.

level of significance ? = 0.01,

We want to test the bus company's claim.

here we assume that the distribution is normally distributed.

Since population standard deviation is not given and we used sample satnadrd deviation instead of population standard deviation. So we need to do t test for testing the above claim

Critical value of t (t_c ) for = 0.01 and degrees of freedom = df = n - 1 = 20 - 1 = 19

t_c = "=TINV(0.02,19)" = 2.329

Since the test is left tailed so we need to assign negative sign.

formula of t test statistic is

So the correct answer is "critical value t0 = -2.539; standardized test statistic ? -2.981; reject H0; There is sufficient evidence to support the Metropolitan Bus Company's claim."