I am going through the Descriptive Statistics class on Udacity.

One of the in class problems discussed is as follows:

Retain or reject the following null hypothesis:

  1. H_null : pop. mean doesn't change significantly following an intervention. ie mu = mu_int

  2. H_alt: pop. mean (mu) changes after an intervention. ie mu =/= mu_int

Alpha level chosen is 0.05 for a two-tailed test and this corresponds to absolute normalized z-score of 1.96

Original population parameters are:

mu = 7.47 and standard deviation, sigma = 2.41

For a sample of size n = 30, when the intervention is applied, the mean, x_bar = 8.3

The z-score for this sample is calculated to be:

(x_bar-mu)/(sigma/sqrt(30)) = 1.89

On the basis of this result, I understand that we can retain the null hypothesis, H_null

Now here is my doubt:

The problem also gives us the information that, the population mean after applying the intervention is calculated to be:

mu_int = 7.8

ie in my understanding this is the new population parameter after applying the intervention.

Now the instructor compares this new mean to the old one in order to calculate the significance of the change. So she calculates the z-score for mu_int using the same formula:

z-score(mu_int) = (mu_int - mu)/(sigma/sqrt(30))

Q: Why is the division by sqrt(30) valid when mu_int applies to the whole population rather than a sample of size n = 30?


1 Answer 1


If both mu and mu_int are population parameters, then significance testing makes no sense to begin with. We know that the population parameter has changed.

However, we typically don't know the population parameter and need to estimate it. In this case, you would need to divide the estimated standard deviation by the square root of the sample size to get the standard error in a z transformation.

It might be worthwhile to discuss this on any course forums there are on Udacity. Maybe the instructor was imprecise, or maybe you misunderstood.

  • $\begingroup$ I saw the video a couple of times and have put down the entire problem exactly as it was presented. Also after doing the significance test, (mu_int-mu)/(sigma/sqrt(30)), the z-score for mu_int turned out to be 0.75. Since this is less 1.96 (for alpha=0.05), the conclusion was that we were correct in rejecting the null hypothesis. $\endgroup$
    – kchak
    Oct 25, 2017 at 9:55
  • $\begingroup$ It sounds like the instructor misspoke, confusing a population parameter with its estimate. $\endgroup$ Oct 25, 2017 at 10:00

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