t interval for mean exam score

Question

A random sample of n = 12 students has a mean exam score of \bar{x} = 78 with sample standard deviation s = 9. The population standard deviation is unknown. Construct a 95% confidence interval for the true mean score.

Accepted Answer

Setup. With \sigma unknown, use the t interval \bar{x} \pm t^{}\,s/\sqrt{n} with 	ext{df} = n - 1 = 11. For 95% confidence, t^{} = 2.201. Compute. The standard error is \frac{9}{\sqrt{12}} = \frac{9}{3.464} = 2.598. Margin of error = 2.201 	imes 2.598 = 5.72, so 78 \pm 5.72 = [72.28,\ 83.72]. Interpret. The interval is fairly wide because a sample of 12, with a spread of 9 points, cannot pin the mean down tightly. Takeaway. For the t interval you need three sample quantities: the mean, the standard deviation, and the size, which sets both the standard error and the degrees of freedom.

t interval for mean exam score

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