Question

A preliminary sample of holiday shoppers revealed that the standard deviation of the amount of money they are planning to spend on gifts during the coming holiday season is $80. How many holiday shoppers should be sampled in order to estimate the average amount of money spent on gifts by all holiday shoppers with a margin of error of $7 and with a 99% confidence

Answer 1

Answer:

$n=(\frac{2.58(80)}{7})^2 =869.407 \approx 870$

So the answer for this case would be n=870 rounded up to the nearest integer

Step-by-step explanation:

Information given

$\sigma = 80$ represent the deviation

ME = 7 represent the margin of error

Solution

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$    (a)

And on this case we have that ME =7 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (b)

The critical value for 99% of confidence interval now can be founded using the normal distribution. The significance level would be $\alpha=0.01$ and $\alpha/2 =0.05$. The critical value would be $z_{\alpha/2}=2.58$, replacing into formula (b) we got:

$n=(\frac{2.58(80)}{7})^2 =869.407 \approx 870$

So the answer for this case would be n=870 rounded up to the nearest integer