Question

Question 12 5 pts A survey with a 90% confidence level is being designed for a population that has a standard deviation of 6.5. The team wanted a margin of error equal to 10 but then elected to cut it in half to 5. How will this change affect the number of people to be included in the survey? HTML Editor BIVA A-TEX X, VV 12pt O words 30 EUR 9

Answer 1

Solution:
Given in the question
Confidence level = 0.9
Level of significance($\alpha$) = 1 - Confidence level = 1 - 0.9 = 0.1
$\alpha$ /2 = 0.1/2 = 0.05, From Z table we found Z$\alpha$/2 = 1.645
Population standard deviation($\sigma$) = 6.5
Margin of error (E)= 10
So sample size can be calculated as
Sample size (n) = (Zalpha/2 * $\sigma$ /E)^2
Here we can see that sample size is inversly proportion to the sqare of Margin of error(E)
I.e. n ~ (1/E)^2
if E is divide by 2 than
n ~ (1/(E/2))^2
n ~ 4*(1/E)^2
If Margin of error will decreases by half than sample size will increase by 4 times.