Is the cafe's wait time longer than the promised 5 minutes?

Question

A coffee chain promises a mean wait time of 5.0 minutes. A mystery shopper times 12 visits and finds a mean of 5.6 minutes with a sample standard deviation of 1.2 minutes. The population standard deviation is unknown.

Test at the 5% level (two-sided) whether the true mean wait differs from 5.0 minutes, and report the t-statistic.

Accepted Answer

Unknown \sigma, small sample, so use a one-sample t-test with 12 - 1 = 11 degrees of freedom. Hypotheses: H0: \mu = 5.0 versus H1: \mu 
eq 5.0. Standard error: 	ext{SE} = \frac{1.2}{\sqrt{12}} = \frac{1.2}{3.464} = 0.3464 	ext{ min}. Test statistic: t = \frac{5.6 - 5.0}{0.3464} = \frac{0.6}{0.3464} = 1.73. The two-sided 5% critical value for 11 degrees of freedom is t{0.025,\,11} = 2.201. Since 1.73 = 1.73 < 2.201, we fail to reject H0. The two-sided p-value is about 0.11. Conclusion: the observed 0.6-minute overage is within sampling noise; there is not enough evidence to say the mean wai…

Is the cafe's wait time longer than the promised 5 minutes?

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