Question

A doctor 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?

Answer 1

`n=((1.64(14.7))/(3))^2 =64.57 \approx 65`

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

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

ME represent the margin of error

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

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

2) Solution to the problem

Since the new Confidence is 0.90 or 90%, the value of
`\alpha=0.1` and
`\alpha/2 =0.05`, 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
`z_(\alpha/2)=1.64`

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

`ME=z_(\alpha/2)(\sigma)/(√(n))` (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:

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

Replacing into formula (b) we got:

`n=((1.64(14.7))/(3))^2 =64.57 \approx 65`

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