With only 8 data points, why not just use 1.96?

Question

You have just 8 observations, a sample mean, and a sample standard deviation, and you want a 95% confidence interval for the mean. A colleague plugs in $z = 1.96$ "because that's the 95% number."

Explain why the t-distribution is required here, what goes wrong with using $z$ , and roughly how much the critical value changes.

Accepted Answer

Why t, not z. The z = 1.96 value assumes you know the true standard deviation \sigma. You don't, you estimated it from the same 8 points. That estimate is itself noisy, and with only 7 degrees of freedom it can be badly off. The t-distribution accounts for this extra uncertainty: it has the same bell shape as the normal but fatter tails, so its critical values are larger. As the sample grows, the estimate of \sigma sharpens, the tails thin, and t converges to z. How much it changes. For a 95% two-sided interval: - n = 8 (df = 7): the t critical value is about 2.365, not 1.96. - n = 21 (df = 2…

With only 8 data points, why not just use 1.96?

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