THERE ARE ESTIMATES OF THE MARGIN OF ERROR FOR CONFIDENCE LEVELS OTHER THAN 95%. FOR A 90% CONFIDENCE LEVEL AND A SAMPLE SIZE N, THE MARGIN OF ERROR IS APPROXIMATELY 82/N%.SUPPOSE WE WISH TO HAVE A MARGIN OF ERROR OF 1% WITH A 90% CONFIDENCE LEVEL. APPROXIMATELY HOW MANY PEOPLE SHOULD WE INTERVIEW?

Question

THERE ARE ESTIMATES OF THE MARGIN OF ERROR FOR CONFIDENCE LEVELS OTHER THAN 95%. FOR A 90% CONFIDENCE LEVEL AND A SAMPLE SIZE N, THE MARGIN OF ERROR IS APPROXIMATELY 82/N%.SUPPOSE WE WISH TO HAVE A MARGIN OF ERROR OF 1% WITH A 90% CONFIDENCE LEVEL. APPROXIMATELY HOW MANY PEOPLE SHOULD WE INTERVIEW?

Answer 1

ME=Z√(0.25/n)

For 90% confidence level, and ME=1%
Z=1.644854 [ from tables for two tale at 95%]
=>
0.01=1.644854√(0.25/n)
=>
n=(1.644854/0.01)^2*0.5
=6763.86 ~ 6764 persons

Answer 2

To determine the sample size needed to achieve a margin of error of 1% with a 90% confidence level, we can use the formula provided: Margin of Error = 82/N%.

Let's rearrange the formula to solve for N:

Margin of Error = 82/N%
1% = 82/N%

Now, cross-multiply and solve for N:

N * 1% = 82
N = 82 / 1%
N = 82 / 0.01
N ≈ 8200

Therefore, approximately 8200 people should be interviewed to achieve a margin of error of 1% with a 90% confidence level.