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

n=17

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)(\sigma)/(√(n))` (1)

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) 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=((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