# 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!

• Does it say what type of distribution the data is collected from? Commented Aug 25, 2015 at 18:19
• No it doesn't, the question is the same as in the post :( Commented Aug 25, 2015 at 18:20
• It looks like you've assumed that the sample is taken from a normal distribution, and then you get a help variable with a t-distribution with some degree of freedom. What degree of freedom have you used for the t-value? Commented Aug 25, 2015 at 18:22
• I looked, you used $t$ table, $25$ degrees of freedom. It should be $24$. Commented Aug 25, 2015 at 18:23
• Unsatisfactory answer, because the distribution of the statistic you used is $t$ with $n-1$ degrees of freedom, where $n$ is the sample size. And make sure to use the two-tailed test, $0.025$ in each tail. Commented Aug 25, 2015 at 18:31

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.

• Thanks David that's great. Any chance you could go into a bit more detail on the unbiased estimate for the population variance please? How do you know it's biased? Why have you done that calculation to give an unbiased result? Commented Aug 25, 2015 at 20:39
• Also, why is it t(24) as opposed to t(25)? Commented Aug 25, 2015 at 20:44
• Essentially, when you only have the sample variance to go on, you need to work out $\frac{n}{n-1}\times \text{sample variance}$ to obtain an unbiased estimate of population variance. There is some algebra behind this which you need to check out, together with the definition of unbiased estimate.... Commented Aug 25, 2015 at 20:44
• It's always $n-1$ when the sample size is $n$, but for a full explanation you need to check put the theory of Student t distribution... Commented Aug 25, 2015 at 20:48
• @DavidQuinn Can you write out the explicit formula for the sample variance? I assumed that the sample variance meant this: $s^2=\frac{1}{n-1}\sum_{i=1}^n (x_i-\overline{x})^2$. Is this an incorrect assumption? Commented Aug 25, 2015 at 21:36

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.