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, $$ 10.5 \pm 2.67\ \frac{3.25}{\sqrt{55}} = 10.5 \pm 1.17 \Rightarrow (9.33,\ 11.67) $$


Let's start saying you can use this tool to check whether you answer is correct or not.

You should get a confidence interval of $\pm 1.13$.

That said, the correct formula to use is:

$$\bar x \pm z \frac{s}{\sqrt n}$$

Where the $\bar x$ is your average, the $z$ is a constant, $s$ is the standard deviation and $n$ is your sample size. The $z$ varies depending on the confidence interval you are trying to calculate. In your case, because 99%, $z$ is equal to $2.58$.

Plug all the numbers in the equation and you should get 1.130633169350857 which rounded gives you the $\pm 1.13$ I mentioned before. And that's it.

  • 2
    $\begingroup$ It is strange to read a pair of sentences in which the first claims "$z$ is a constant" and the second describes how "$z$ varies." It would be hard for both to be true simultaneously! Could you be more explicit about what $z$ represents, what it depends on, and how it is calculated? $\endgroup$ – whuber Nov 19 '14 at 17:36
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    $\begingroup$ I just want to add one more: the constant should not be 2.58. It is 2.58 if the sample size is large. And 55 is not large, we should probably use $t_{\alpha/2}$ instead. $\endgroup$ – Penguin_Knight Nov 19 '14 at 17:47
  • $\begingroup$ Hi @whuber, interesting question. I checked different books and 'z' is often referred to as a constant so it is not a terminology I introduced. Now that you had me think about it, I believe the term coefficient should be used instead as more appropriate. $\endgroup$ – Edgar Derby Nov 20 '14 at 9:56
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    $\begingroup$ @AnnoysParrot, check t-table. z-table does not have degrees of freedom but t-distribution does. Here is a general table. And here is an online calculator. If you punch in df of 1000 and alpha of 0.01, it'd be your answer 2.58; it's because t-distribution approaches normal distribution when df is high. But since the n is only 55, the df is now 54, yielding a 2.67. Also refer to Wikipedia page on "Student's t-distribution." $\endgroup$ – Penguin_Knight Nov 20 '14 at 14:36
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    $\begingroup$ @ChrisC, thank you. Your comment was well taken. Regards. $\endgroup$ – Penguin_Knight Jun 8 '15 at 20:07

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