Adoctor wants to estimate the mean hdl cholesterol of all 20- to 29-year-old females. the number of subjects needed to estimate the mean hdl cholesterol within 3 points with 99% confidence is 160 subjects. suppose the doctor decides that he could be content with only 90% confidence. assuming s=14.7 based on earlier studies, how would this decrease in confidence affect the sample size required?

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects

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".

represent the sample mean for the sample  

population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

ME represent the margin of error

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

  (1)

2) Solution to the problem

Since the new Confidence is 0.90 or 90%, the value of and , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that

Assuming that the deviation is known we can express the margin of error is given by this formula:

   (a)

And on this case we have that ME =\pm 3 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

  (b)

Replacing into formula (b) we got:

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects