Question

a sociologist wishes to estimate the average number of automobile thefts in a large city per day within two automobiles. He wishes to be 95% confident, and from a previous study the standard deviation was found to be 4.2. How kany days should he select to survey?

Answer 1

Answer:

n=17

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

2) Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

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:

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

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025,0,1)", and we got $z_{\alpha/2}=1.960$, replacing into formula (b) we got:

$n=(\frac{1.960(4.2)}{2})^2 =16.94 \approx 17$

So the answer for this case would be n=17 rounded up to the nearest integer