hypothesis testing with central limit theorem

Question

Company A claims that the average number of biscuits in a tin of BiskO shortbread biscuits is $27$. The fair business regulators believes that the Company A’s claims are inaccurate, and that the true average number of biscuits in a tin is less than $27$. To prove their claim, the number of biscuits in $50$ tins of BiskO were counted and the sample mean $\overline X$ was $25.3$ biscuits. Assume that the standard deviation of number biscuits in a tin of BiskO is known to be $2.6$.

(a) Construct appropriate null and alternative hypotheses for this situation. Make sure the null hypothesis is simple.

(b) Perform the test using the hypotheses you outlined above at level $\alpha = 0.05$ based on the sample data. Make note of any approximations used. What do you conclude about Company A’s claim?

So I am a little stuck here. I was having a conversation with others about it and with our lecturer about part a and our answer is that we will use the central limit theorem.

$H_0$ is $\mu=27$ and

$H_A$ is $\mu < 27$

But when it comes to part b, I'm unsure on how to apply the central limit theorem . I can't see it anywhere in my lecture notes so any advice on how to perform this would be great.

Answer 1

I have put the relevant numbers into the 'one-sample z test' procedure of a recent release of Minitab software.

According to the Central Limit Theorem, we are assuming that the average number of biscuits per tin has a normal distribution with unknown population mean $\mu$ and known standard deviation $\sigma = 2.6,$ so that the standard error is $\sigma/\sqrt{n}.$ [Standard error is the usual name for $SD(\bar X).]$

We wish to test $H_0: \mu = 27$ against the alternative $H_a: \mu < 27,$ at the 5% level of significance $\alpha.$ You have correctly stated these (except for using $u$ instead of Greek letter $\mu,$ which I fixed.) Now, here are some clues for doing part (b).

Results from Minitab are shown below:

One-Sample Z 

Test of μ = 27 vs < 27
The assumed standard deviation = 2.6

 N    Mean  SE Mean  95% Upper Bound      Z      P
50  26.300    0.368           26.905  -1.90  0.028

In your text or class notes, find the formula for computing the test statistic $Z$ and do the appropriate computation to obtain the value shown in the output above. [Some of the "Relevant" links in the margin of this page may also be helpful.]

For this one-sided test at the 5% level, the critical value is $c = -1.645.$ That is, we reject the null hypothesis because $Z \le -1.645.$ How was this critical value $c$ obtained? (Use a printed table of the standard normal CDF, a statistical calculator, or a statistical computer program.)

Most software programs show P-values. One rejects at the 5% level if the P-value is less than $0.05 = 5\%.$ So, also according to that criterion, we would Reject $H_0.$

[The output labeled '95% Upper Bound' is for a one-sided confidence interval, and is not directly relevant to your question.]

If the null hypothesis is true then $Z$ is approximately distributed as a standard normal random variable. The figure below (made using R) shows the standard normal density function. The critical value $c$ is at the vertical red dotted line. The observed value of the z statistic is at the heavy black line. The P-value is the area under the density curve to the left of the heavy black line.