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

Answer:

$n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65$

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

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  

ME represent the margin of error

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (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:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$    (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:

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

Replacing into formula (b) we got:

$n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65$

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