Solved – the formula for a normal-theory confidence interval of the mean

A psychologist is collecting data on the time it takes to learn a certain task. For $55$ randomly selected adult subjects, the sample mean is $10.5$ minutes and sample standard deviation is $3.25$ minutes. Construct a $99%$ confidence interval for the mean time required by all adults to learn the task.

a. What formula to use.
b. Plug all values in.

I think I should use a $t$-test formula which would be $(bar x- mu_0 )/(s/sqrt n)$, so plugging in the values would be $(10.5-?)/(3.25/sqrt{55})$.
What is the $mu_0$ value? Am I right?

(Since this question is so old, I suspect it is safe to provide an answer.)

I think you're halfway there. We might often use a $t$-test in a situation like this, but a confidence interval is not quite the same thing as a test. Since the mean and SD are estimated from the data, you need to take that fact into account. Thus, we will use the $t$-distribution to form the confidence interval. The general formula would be:
$$ bar x pm t_{(1-frac{alpha}{2}!, df)} frac{s}{sqrt{N}} $$ The key to using this formula is to find the relevant $t$-value. First, we need to get the df—it is $N-1=54$. Then we look up the quantile that corresponds to the $99.5^{rm th}$ percentile of that particular $t$-distribution in a $t$-table. I find the value $2.67$. Hence, $$ 10.5 pm 2.67 frac{3.25}{sqrt{55}} = 10.5 pm 1.17 Rightarrow (9.33, 11.67) $$

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