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

`n=((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

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)(\sigma)/(√(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=((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=((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