95% Confidence Interval Problem for a random sample The sample mean of a random sample of $25$ observations is $9.6$ and the sample variance is $22.4$.
Derive a $95$ confidence interval for the population mean.
I calculated the following:
Confidence interval $= x +- ts/root(n)$
where:
$t = 1.708$ (from t distribution table)
  $s = 4.733$ (square root of sample variation)
  $n = 25$  
Using this gives the confidence interval:
$7.983$ <= Population mean <= $11.217$
However in the mark scheme it says this:
$7.606$ <= Population mean <= $11.590.$
Unfortunately it doesn't have any workings and so I really don't know where I'm going wrong. Any pointers would be appreciated, thanks!
UPDATE: I have found some handwritten mark scheme that says:
Unbiased estimator of the population variance = 23.3333
s.e. of sample mean = 0.9661, use t(24)
Now I'm really confused!
 A: Let $x_1, x_2,..., x_{25}$ be a sample from the independent stochastic variables $X_i$, $i=1,2,...,25$. We assume that $X_i$ is $N(\mu,\sigma)$ for $i=1,2,...,25$ for some unknown $\mu$ and $\sigma$. Given is the sample mean $\overline{x}=9.6$, and the sample variance $s^2=22.4$. For future notation, we set $n=25$ as the number of observations.
We then have that the stochastic variable
$$\overline{X}=\frac{\sum_{i=1}^n X_i}{n}$$ 
is $N(\mu,\sigma/\sqrt{n})$, as $X_i$ are independent and have a normal distribution. For the estimation of $\sigma^2$, we have that the stochastic variable
$$\frac{(n-1)S^2}{\sigma^2}=\frac{\sum_{i=1}^n (X_i-\overline{X})^2}{\sigma^2}$$
is $\chi^2(n-1)$, with $(n-1)$ degrees of freedom.
Now $$Z=\frac{\overline{X}-\mu}{S/\sqrt{n}}=\frac{\overline{X}-\mu}{\sqrt{S^2/n}}=\frac{\overline{X}-\mu}{\frac{\sigma}{\sqrt{n}}\sqrt{\frac{S^2}{\sigma^2}}}=\frac{\overline{X}-\mu}{\frac{\sigma}{\sqrt{n}}\sqrt{(\frac{(n-1)S^2}{\sigma^2})/(n-1)}}$$
is $t(n-1)$.
We use this to get the two-sided confidence interval:
$I_\mu^{0.95}=(\overline{x}\mp t_{0.975}(n-1)\cdot s/\sqrt{n})=(9.6\mp 1.95)=(7.65; 11.55)$. 
It seems like they've used the interval 
$I_\mu^{0.95}=(\overline{x}\mp t_{0.975}(n-1)\cdot s/\sqrt{n-1})$ in the answers for some reason, but I can't see why. If anyone sees a clear error with my calculations, please let me know.
Edit: I was mistaken about what "sample variance" meant, and that is why the answer uses the second interval given. By using the following formula:
$$S^2=\frac{\sum_{i=1}^n (X_i-\overline{X})^2}{n-1}$$
you can calculate an unbiased sample variance based on the given data.
A: You should be doing $$\bar{x}\mp t_{0.975}(n-1)\frac{\hat{\sigma}}{\sqrt{n}}$$
In this case, $\bar{x}=9.6$, $t_{0.975}(24)=2.064$, and $\frac{\hat{\sigma}}{\sqrt{n}}=\frac{\sqrt{22.4}}{\sqrt{25}}\times\frac{\sqrt{25}}{\sqrt{\color{red}{25-1}}}$ since we require an unbiased estimate of the population variance, based on the sample variance.
This gives us the supposed answer $7.606$ for the lower bound, but for the upper bound we should get $11.594$
One possible explanation for their value of $11.590$ is the unintended use of $t_{0.975}(25)=2.060$ just for the upper bound, which would give precisely that, so it could be an error on their part.
