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

Answer:

$n=(\frac{2.58(4.2)}{2})^2 =29.35 \approx 30$

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

Step-by-step 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:

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

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

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

$n=(\frac{2.58(4.2)}{2})^2 =29.35 \approx 30$

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