Question

a.) What sample size is needed to give a margin of error within +-6 in estimating a population mean with 95% confidence, assuming a previous sample had s=20.

Round your answer up to the nearest integer.                             sample size =

b.) What sample size is needed to give a margin of error within +-13 in estimating a population mean with 99% confidence, assuming a previous sample had s=116

Round your answer up to the nearest integer.                 sample size =

c.) In a study, we see that the average number of close confidants in a random sample of 2006 US adults is 2.2 with a standard deviation of 1.4 . If we want to estimate the number of close confidants with a margin of error within +- 0.008 and with 99% confidence, how large a sample is needed?

Round your answer up to the nearest integer.                            sample size =

Answer 1

Solution:

Part a)

Given:
c = confidence level = 95%
E = Margin of Error = 6
s = Standard deviation= 20

Find sample size n.

Formula:

We need to find z_c value for c=95% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.95) /2 = 1.95 / 2 = 0.9750

Look in z table for Area = 0.9750 or its closest area and find z value.

Area = 0.9750 corresponds to 1.9 and 0.06 , thus z critical value = 1.96

That is : Z_c = 1.96

Since is unknown , we use its sample estimate s= 20

Part b)

Given:

c = confidence level = 99%
E = Margin of Error = 13
s = Standard deviation= 116

Find sample size n.

Formula:

Z_c is z critical value for c = 0.99 confidence level.

Find Area = ( 1+c)/2 = ( 1 + 0.99 ) / 2 = 1.99 /2 = 0.9950

Thus look in z table for Area = 0.9950 or its closest area and find corresponding z critical value.

From above table we can see area 0.9950 is in between 0.9949 and 0.9951 and both are at same distance from 0.9950, Hence corresponding z values are 2.57 and 2.58

Thus average of both z values is 2.575

Thus Z_c = 2.575

Thus

Part c)

Given:

c = confidence level = 99%
E = Margin of Error = 0.008
= Standard deviation= 1.4

Find sample size n.

Formula: