# solution verification for hypothesis testing problem

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?

• 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. Aug 22, 2017 at 18:18

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:

Nik
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 $H_0.$

• 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? Aug 22, 2017 at 18:34
• 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.) Aug 22, 2017 at 18:38