First, let's get the notation and definitions right;
The sample mean $\bar X = \frac 1n\sum_{i=1}^n X_i.$
If the population mean $\mu$ is unknown and estimated by $\bar X,$
then the population variance $\sigma^2$ is estimated by the sample variance
$S^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)^2.$
Then
$$\frac{(n-1)S^2}{\sigma^2} = \frac{\sum_{i-1}^n(X_i - \bar X)^2}{\sigma^2}
\sim \mathsf{Chisq}(df = n-1).$$
For your dataset the statistics are:
x = c(22.2, 24.7, 20.9, 26.0, 27.0, 24.8, 26.5, 23.8, 25.6, 23.9)
n = length(x); a = mean(x); s = sd(x)
n; a; s
## 10 # sample size
## 24.54 # sample mean
## 1.912648 # sample SD
Then 95% confidence interval for the population variance $\sigma^2$
is obtained as
$$((n-1)S^2/U,\, (n-1)S^2/L),$$ where $L$ and $U$ cut 2.5% of the
probability from the lower and upper tails, respectively, of
$\mathsf{Chisq(n-1)}.$ Computations of CIs for $\sigma^2$ and $\sigma$ in R statistical software
follow:
UL = qchisq(c(.975, .025), n - 1); UL
## 19.022768 2.700389
CI = (n-1)*s^2 / UL; CI
## 1.730768 12.192315 95% CI for pop var
sqrt(CI)
## 1.315587 3.491750 95% CI for pop SD
Notice that $S = 1.913$ is contained in the CI for $\sigma$ as it
must be, but that $S$ is not at the center of the CI, because the
chi-squared distribution is skewed.
I assume you can use the appropriate quantiles of $\mathsf{Chisq}(9)$
to get 99% confidence intervals.
Addendum per Comments for 99% CIs: Of course, 99% confidence intervals
have to be longer than 95% CIs.
UL = qchisq(c(.995, .005), n - 1); UL
## 23.589351 1.734933 # same as you showed in your question
CI = (n-1)*s^2 / UL; CI
## 1.395715 18.977103 # using correct numerator, this is different
sqrt(CI)
## 1.181404 4.356272
Your use of notation is strange. The hypothesis should be written $$H_0 : \mu = 151 \quad \text{vs.} \quad H_a : \mu < 151.$$ The hypothesis is a statement about the value of the unknown parameter, in this case the mean $\mu$. We make an inference about its value based on the data, from which we calculate a test statistic, which under the null hypothesis is $$T \mid H_0 = \frac{\bar x - 151}{s/\sqrt{n}},$$ where $\bar x$ is the sample mean and $s$ the sample standard deviation. A statistic is never included in the hypothesis because it is something that is calculated from the data; there is no uncertainty about the value we get for it.
Given that $\bar x = 126$ and the value of your test statistic is $T = -2.5$, I get $s \approx 48.9898$ (the exact value of which may differ due to the limited precision you reported for the test statistic).
As for the calculation of a confidence interval, this need not be one-sided even if the hypothesis is one-sided. The two concepts are related, but a confidence interval is nothing more than an interval estimate. So, rather than using the sample mean as a simple point estimate for the true mean, we can incorporate the variability observed in the data to give a more sophisticated estimate of the true mean. You can calculate a two-sided interval or a one-sided interval. The question as stated is not specific about which one is intended.
A two-sided $95\%$ interval is computed as $$\bar x \pm t_{n-1,\alpha/2}^* \frac{s}{\sqrt{n}},$$ where $s/\sqrt{n}$ is the standard error of the mean, and $t_{n-1,\alpha/2}^* $ is the upper $\alpha/2$ quantile of the student $t$ distribution. In your case, it is $$t_{23,0.025}^* = 2.06866.$$ Thus the two-sided interval is $$[105.31, 146.69].$$ The one-sided interval that is in the same direction as the hypothesis test is $$\left(-\infty, \bar x + t_{n-1,\alpha}^* \frac{s}{\sqrt{n}}\right].$$ Here, $t_{23,0.95}^* = 1.71387$ so the upper confidence limit is $143.14$, which is smaller than the upper confidence limit of the two-sided interval. This makes sense because we selected the one-sided upper confidence limit in such a way that the probability the interval does not contain the true mean is $\alpha$; i.e., the upper tail probability of the student $t$ distribution is the full $0.05$, rather than the equal-tailed, two-sided interval.
Best Answer
When doing a confidence interval for a sample mean, you use infinity for the degrees of freedom when you know the population standard deviation $\sigma$, and you use $n-1$ for the degrees of freedom when you don't know $\sigma$ and have to estimate it with the sample standard deviation $s$. Of course, if $n-1$ is large enough there's not much difference between using infinity and using $n-1$. A sample size of $42$ isn't large enough, though; I would say you are right and the answer key is wrong.
It may be helpful to remember the bigger picture: By the central limit theorem, $\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}$ is approximately $N(0,1)$, and so when we know $\sigma$ we use the $N(0,1)$ distribution to obtain the critical value in the confidence interval calculation. It rarely happens in practice that we know $\sigma$, though, and so we usually find ourselves having to estimate it with $s$. In this case, the normal approximation isn't usually good enough, and so instead we use the $t$ distribution with $n-1$ degrees of freedom to obtain the critical value. What ties this together with what I said in the first paragraph is that as the number of degrees of freedom goes to infinity in a $t$ distribution you get the $N(0,1)$ distribution.