Question

A doctor wants to estimate the mean HDL cholesterol of all​ 20- to​ 29-year-old females. How many subjects are needed to estimate the mean HDL cholesterol within 4 points with 99 % confidence assuming s equals 13.9 based on earlier​ studies? Suppose the doctor would be content with 90 % confidence. How does the decrease in confidence affect the sample size​ required? A​ 99% confidence level requires nothing subjects. ​(Round up to the nearest​ subject.) A 90 % confidence level requires nothing subjects. ​(Round up to the nearest​ subject.)

99 % confidence is

90 confidence is

How does the decrease in confidence affect the sample size​ required?

A. Decreasing the confidence level increases the sample size needed.

B. Decreasing the confidence level decreases the sample size needed.

C. The sample size is the same for all levels of confidence.

Answer 1

Solution :

Given that,

standard deviation =s = 13.9

Margin of error = E = 4

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z_/2 = 2.58 ( Using z table ( see the 0.005 value in standard normal (z) table corresponding z value is 2.58 )  

sample size = n = [Z_/2* / E] ²

n = ( 2.58* 13.9 /4 )²

n =80

Sample size = n =80

(B)

Solution :

Given that,

standard deviation =s = 13.9

Margin of error = E = 4

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645} ( Using z table )

sample size = n = [Z_/2* / E] ²

n = ( 1.645* 13.9 /4 )²

n =33

Sample size = n =33

B. Decreasing the confidence level decreases the sample size needed.