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How can I convince my teacher that he is wrong in this task? We were given the following task and my teacher says I have to use a z-test. In my opinion I have to use a t-test.The task is as follows:

A company installs a nuclear power blocker into the power grid to improve the health of its employees. Before installation, employees had an average of 12.7 sick days per year over the years with a standard deviation of 1.2 days. In the two years following installation, the average was 11.4 sick days. Determine whether health has actually improved using a 95% confidence interval assuming that the standard deviation has not changed. Can it be confirmed that the health of the employees has improved?

I calculated the t test because there is a rule that if the sample n, in this case the 2 years, are smaller than 30, then it should you can use the t-test. Since n=2 and less than 30, I thought the t-test was the right choice. Can I use this rule and are there other options too, that proof that the t-test is the right choice? Click for the n<30 then t-test rule

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    $\begingroup$ Welcome to Cross Validated! Why do you use $n=2?$ Are there only two employees? $//$ Think about how the standard deviation factors into whether or not you want to use a t-test. I think your teacher wrote a bad problem, but I do see how your teacher came up with the claim to use z instead of t. $\endgroup$
    – Dave
    Commented Mar 21 at 19:43
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    $\begingroup$ So why is $n=2$, because there are two years? $\endgroup$
    – Dave
    Commented Mar 21 at 20:02
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    $\begingroup$ Out of curiosity, what is a "nuclear power blocker" and how might that improve the health of anyone? Please be aware, too, that sick days are likely far from independent (which means if this were a real problem you wouldn't have enough information to answer and ought to proceed carefully so you don't provide unsupportable opinions). Finally, wouldn't $n$ count employees rather than years? Otherwise you seem to be maintaining that $n=2$ in any t-test, which clearly isn't right. $\endgroup$
    – whuber ♦
    Commented Mar 21 at 20:45
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    $\begingroup$ @Apro9991 and if Whuber and Flom and I say $n=2$ is problematic? $\endgroup$
    – Dave
    Commented Mar 21 at 22:42
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    $\begingroup$ @Apro9991 Then I request that you post a distinct question about that. Unlike some corners of the internet, Cross Validated is strictly Q&A, not a discussion forum or message board. $//$ If my answer here does not address why your teacher wants to use a z-test instead of a t-test, please do ask for clarification. $\endgroup$
    – Dave
    Commented Mar 21 at 23:08

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Your teacher seems to take the stance that the standard deviation of $1.2$ is a known quantity. Since the standard deviation is the same for the "before" and for the "after" groups, your teacher seems to see the problem as using a known standard deviation instead of an estiated standard deviation. In such a case, a z-test would be appropriate over a t-test, regardless of the sample size. The reason for using a z-test over a t-test when the standard deviation is estimated instead of known is because a t distribution is quite similar to a z distribution once the sample size gets large, sometimes with $30$ given as a cutoff. (With how easy it is for software to do a t-test, I would't bother doing otherwise. Do it right.)

I would seek clarification from your teacher about how to read her mind about when quantities are assumed to be known instead of estimated.

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  • $\begingroup$ We don't use software; instead, we have to calculate it manually using a table. On average, there are 12.7 sick days per year with a standard deviation of 1.2 days. Additionally, we're working with a sample size of n = 2. Could you please explain in more detail why a t-test shouldn't be used here? I'm still having trouble understanding it. $\endgroup$
    – Apro9991
    Commented Mar 21 at 20:24
  • $\begingroup$ @Apro9991 I think it’s ridiculous and that the problem isn’t very good in general (not your fault), but your teacher seems to take the stance that the standard deviation is known to be $1.2$. When the standard deviation is known, a z-test is appropriate over a t-test. The typical progression in a course is to learn the z-test for means and then say, “…but we typically have to estimate the standard deviation,” and then get into the t-test. $\endgroup$
    – Dave
    Commented Mar 21 at 21:03
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You can't convince them they were wrong because .. they weren't. At least, not in the part in your question. N is the number of employees, not the number of years. The plant probably has a lot more than two employees.

In fact, if N was 2, then you can't even do a t-test, and the standard deviations will be 0 for each year. Also, years don't take sick days. People do.

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  • $\begingroup$ The formula is following: t = (X-u)/(s/root(n)) . X is the sample mean which is 11.4, s is the stnadard deviation with 1.2 . u is the normal mean with 12.7 and n is the sample which is 2. t = (11.4-12.7)/(1.2/root(2)) . This is the way I calculated it to get the to get the test value. On average, there are 12.7 sick days per year with a standard deviation of 1.2 days. Additionally, we're working with a sample size of n = 2. Could you please explain in more detail why a t-test shouldn't be used here? My teacher also said that the sample size n is 2. $\endgroup$
    – Apro9991
    Commented Mar 21 at 20:25
  • $\begingroup$ Then your teacher was wrong. T is not 2. That would be one before and one after,and the SD would be 0. $\endgroup$
    – Peter Flom
    Commented Mar 21 at 21:16
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    $\begingroup$ @PeterFlom I think the exercise is to treat the two years as one group and then do a one-sample test against a null hypothesis that the new mean is the same $11.4$ as always. If you want to argue that this is a bad problem and that someone with good training in statistics should be able to write a better question about a one-sample test of a mean (even if the nuances that drive us up the wall go over the head of a Stat 100 student), I completely agree. $\endgroup$
    – Dave
    Commented Mar 21 at 23:05
  • $\begingroup$ @Dave If that's the intent of the exercise, it stinks as a problem. But all problems that posit unknown means and known SDs are kind of ridiculous. How do you know the SD without knowing the mean? $\endgroup$
    – Peter Flom
    Commented Mar 22 at 10:08
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As @PeterFlom pointed out, n would be the number of employees in each group (and certainly not the number of years over which data were collected). You don't know the number of employees. So it is impossible to run a t-test with the data you were given.

You also cannot run a z test, because you would need n to compute SEM from SD.

So there is no way to accomplish your goal from the information you were given.

-- Edited based on a comment from @Dave

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    $\begingroup$ How do you calculate the standard error without knowing the sample size? $\endgroup$
    – Dave
    Commented Mar 22 at 0:32
  • $\begingroup$ @Dave Agree. Rewrote my answer accordingly. $\endgroup$
    – Harvey Motulsky
    Commented Mar 22 at 2:33

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