Question

A sociologist wishes to estimate the average number of automobile thefts in a large city per day within 2 automobiles. He wishes to be 99% confident, and from a previous study the standard deviation was found to be 4.2. How many days should he select to survey

Answer 1

`n=((2.58(4.2))/(2))^2 =29.35 \approx 30`

So the answer for this case would be n=30 rounded up to the nearest integer

Explanation:

Information given

`ME = 2` the margin of error desired

`\sigma =4.2` standard deviation from previous studies

The margin of error is given by this formula:

`ME=z_(\alpha/2)(\sigma)/(√(n))` (a)

And on this case we have that ME =2 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 is
`\alpha=0.01` and the critical value would be
`z_(\alpha/2)=2.58`, replacing into formula (b) we got:

`n=((2.58(4.2))/(2))^2 =29.35 \approx 30`

So the answer for this case would be n=30 rounded up to the nearest integer