Question

A researcher wishes to estimate within 2 points the average systolic blood pressure of female college students. If she wishes to be 90% confidence, how large a sample should she select if the population standard deviation of female systolic blood pressure is 4.8? Please show steps, please and thank you.

Answer 1

n=16

Explanation:

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)

`\sigma` represent the population standard deviation

n represent the sample size (variable of interest)

The margin of error is given by this formula:

`ME=z_(\alpha/2)(\sigma)/(√(n))` (1)

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 (1) we got:

`n=((z_(\alpha/2) \sigma)/(ME))^2` (2)

The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.05,0,1)", and we got
`z_(\alpha/2)=1.640`, replacing into formula (2) we got:

`n=((1.64(4.8)/(2))^2 =15.492 \approx 16`

So the answer for this case would be n=16 rounded up to the nearest integer