90% t interval for mean delivery time

Question

A courier records n = 15 delivery times with a sample mean of \bar{x} = 34 minutes and a sample standard deviation of s = 5 minutes. The population standard deviation is unknown. Construct a 90% confidence interval for the true mean delivery time.

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Setup. With \sigma unknown, use the t interval \bar{x} \pm t^{}\,s/\sqrt{n} with 	ext{df} = n - 1 = 14. For 90% confidence, t^{} = 1.761. Compute. The standard error is \frac{5}{\sqrt{15}} = \frac{5}{3.873} = 1.291. Margin of error = 1.761 	imes 1.291 = 2.27, so 34 \pm 2.27 = [31.73,\ 36.27]. Interpret. Two things narrow this interval relative to earlier ones: the larger sample of 15 lowers the standard error, and the 90% level uses a smaller t^{} than 95% would. Takeaway. Both the sample size and the confidence level move the t interval's width. Larger n and lower confidence each make it n…

90% t interval for mean delivery time

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