Applying the central limit while having the sample standard deviation

Question

I have the following exercise:

In a research report, Richard H. Weindruch of the UCLA Medical School claims that mice with an average life span of $32$ months will live to be about $40$ months old when $40$% of the calories in their diet are replaced by vitamins and protein. Is there any reason to believe that $μ < 40$ if $64$ mice that are placed on this diet have an average life of $38$ months with a standard deviation of $5.8$ months? Use a P-value in your conclusion.

From my personal understanding, in applying the central limit theorem we dividethe difference of the means by the sample standard deviation which is given as $5.8$, however in the solution they did the following:

$z=\frac{38-40}{\frac{5.8}{\sqrt{64}}}$, what I don't understand is that why we divided $5.8$ by $\sqrt{64}$, shouldn't we do that iff this value was the standard deviation for the population which isn't the case here.

Answer 1

Many comments, some confusion, maybe no resolution:

Seems to me this is a left-sided, 1-sample t test of $H_0: \mu=40$ against $H_1: \mu < 40$ based on sample mean and standard deviation $\bar X =38, S = 5.8,$ respectively, for $n=64$ random observations from a normal sample. Output from a recent release of Minitab, which accepts summarized data:

One-Sample T 

Test of μ = 40 vs < 40


 N    Mean  StDev  SE Mean  95% Upper Bound      T      P
64  38.000  5.800    0.725           39.210  -2.76  0.004

I will leave it to you to show how to obtain the $T$ statistic (following @Henry's Comments) and to show that you reject $H_0$ at the 5% level (or even the 1% level).

It is easy to see that $\bar X = 38 < 40.$ You have evidence that it is significantly less than 40, in a statistical sense.

Note: DF$=64-1=63$ for the $T$-statistic. P-value from R statistical software, where pt is the CDF of a t-distribution:

pt(-2.76, 63)
[1] 0.003780076