Statistical Interference

Question

A controller of a store wishes to estimate the average amount spent each month by individuals holding credit cards to within ±€6 of the true amount. Based on previous experience it is known that the standard deviation is €21. Determine the sample size necessary to estimate the mean to within ±€6 of the true value with

(a) 95% confidence

(b) 99% confidence

What formula do I use to answer this question and what type of distribution table do I use as I am not used to having to find the sample size. Hope someone can help. Thanks

Answer 1

$$MOE = z_{\frac{\alpha}{2}}. \frac{s}{\sqrt{n}}$$

$MOE = 6$, $s = 21$, $z_{{0.025}}$

Look up $z_{critical}$ from the charts and compute n

Where MOE is the margin of error, s is the sample standard deviation and n is the number of samples.

Edit:

95% cofidence:

$6 = 1.96.\frac{21}{\sqrt{n}}$

$\sqrt{n} = \frac{1.96\times 21}{6}$

$\sqrt{n} = 6.86$

$n = 47$

99% confidence:

$6 = 2.575.\frac{21}{\sqrt{n}}$

$\sqrt{n} = \frac{2.575\times 21}{6}$

$\sqrt{n} = 9.0125$

$n = 81$

Hope this is clear

Goodluck

Answer 2

Let $X_1,X_2,...,X_n$ be iid observations. Then by Central Limit Theorem, (since variance exists, we can use CLT), $$\sqrt{n}\dfrac{\overline{X_n}-\mu}{\sigma}\to\mathcal N(0,1)$$

Here $\mu$ is the mean and $\sigma$ is the s.d. of the distribution. You want to estimate $\mu$ and $\sigma=21$ is known (so you know the population s.d.).

Then a $100(1-\alpha)\%$ C.I. for $\mu$ is given by $$\left[\overline{X_n}-\dfrac{\sigma}{\sqrt{n}}z_{\alpha/2},\overline{X_n}+\dfrac{\sigma}{\sqrt{n}}z_{\alpha/2}\right]$$

Here by $z_{\alpha/2}$ we mean the number $r$ such that if $X\sim \mathcal N(0,1)$ then $P[X>r]=\alpha/2$.

You want to solve the equation $\dfrac{\sigma}{\sqrt{n}}z_{\alpha/2}=6$ for $n$ in the two cases, as you know $\sigma=21$ and given your $\alpha=0.05$ and $0.01$ respectively in the two cases, you can calculate your $z_{\alpha/2}$ values in the two cases.

Solving for $n$ should be simple now!