I have the following problem for which I need solution verification:

We have the sample $(8.3,7,6.5,10,4.4,8.8,8.9,7.5,5.8,8.9)$ which is taken from a normally distributed population $N(\mu, 4)$. I've calculated the sum of the sample values is 76.1.

I am asked to construct a hypothesis testing with significance level $\alpha$=0.01 for $H_O: \mu=7.5, H_A: \mu < 7.5$

I know that my z-score is -2.325. My test statistic is $t=\frac{\sqrt{n}(7.61-7.5)}{\sqrt{\sigma}}=\frac{\sqrt{10}(7.61-7.5)}{2}=-0.61664$.

Thus t does not fall in the rejection region ($t > -0.2325$) and from there we conclude that we cannot reject the null hypothesis.

Is my solution correct and good enough?

  • 1
    $\begingroup$ Error in data or computed sample mean. Please check and revise your question, and say what you corrected. (Otherwise, I anticipate down-votes. None of them mine yet.) See my Answer below. $\endgroup$
    – BruceET
    Aug 22, 2017 at 18:18

1 Answer 1


No, something is wrong here: Your sample mean doesn't match the posted data.

Here is a printout of the relevant test from Minitab statistical software for the data you posted. I assume your notation for the normal distribution means $\sigma^2 = 4$ so that the population SD is $\sigma = 2.$

Data entered:

    8.3    7.0    6.5   10.0    4.4    8.8    8.9    7.5    5.8    8.9

Test of μ = 7.5 vs < 7.5
The assumed standard deviation = 2

Variable   N   Mean  StDev  SE Mean  95% Upper Bound     Z      P
Nik       10  7.610  1.704    0.632            8.650  0.17  0.569

According to this you have the wrong sample mean. Intuitively, the question is whether $\bar X = 7.61$ is sufficiently smaller than the hypothetical mean $\mu_0 7.5$ to cast doubt on that hypothetical mean. But $\bar X$ for the data you show is not smaller then 7.5. So you are 'testing in the wrong tail'. The P-value exceeds $.01 = 1\%,$ which also indicates that you cannot reject $H_0.$

In case you typed the data into your Question incorrectly [something I might have done!] and the sample mean really is $\bar X = 7.11,$ here is the Minitab output for the mean you posted. The Z-score in the output matches what you computed.

Test of μ = 7.5 vs < 7.5
The assumed standard deviation = 2

 N   Mean  SE Mean  95% Upper Bound      Z      P
10  7.110    0.632            8.150  -0.62  0.269

Now the test makes sense, but you still can't reject $H_0.$

  • $\begingroup$ Thanks! No, the mean really is 7.61 and not 7.1. I should have double checked that. So basically if my test statistic is greater than the $H_0$ mean I have to do a right tail test? $\endgroup$
    – Nikola
    Aug 22, 2017 at 18:34
  • 2
    $\begingroup$ Strictly speaking the alternative $<, >$ or $\ne$ should be decided before you see the data. The 'direction' of the alternative should match the reason you're doing the experiment. (Not easy to know in unmotivated drill exercises.) $\endgroup$
    – BruceET
    Aug 22, 2017 at 18:38

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