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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!

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  • $\begingroup$ Does it say what type of distribution the data is collected from? $\endgroup$
    – Scounged
    Commented Aug 25, 2015 at 18:19
  • $\begingroup$ No it doesn't, the question is the same as in the post :( $\endgroup$
    – James Baker
    Commented Aug 25, 2015 at 18:20
  • $\begingroup$ 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? $\endgroup$
    – Scounged
    Commented Aug 25, 2015 at 18:22
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    $\begingroup$ I looked, you used $t$ table, $25$ degrees of freedom. It should be $24$. $\endgroup$
    – André Nicolas
    Commented Aug 25, 2015 at 18:23
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    $\begingroup$ 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. $\endgroup$
    – André Nicolas
    Commented Aug 25, 2015 at 18:31

2 Answers 2

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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.

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  • $\begingroup$ 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? $\endgroup$
    – James Baker
    Commented Aug 25, 2015 at 20:39
  • $\begingroup$ Also, why is it t(24) as opposed to t(25)? $\endgroup$
    – James Baker
    Commented Aug 25, 2015 at 20:44
  • $\begingroup$ 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.... $\endgroup$
    – David Quinn
    Commented Aug 25, 2015 at 20:44
  • $\begingroup$ 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... $\endgroup$
    – David Quinn
    Commented Aug 25, 2015 at 20:48
  • $\begingroup$ @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? $\endgroup$
    – Scounged
    Commented Aug 25, 2015 at 21:36
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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.

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