Answer: Margin of error 1012.144.
I would recommend a sample size of 246.
Explanation:
Given data:
Mean= M = $11,500
Standard deviation =s = $4,000
sample size=n= 60
s x = standard error of mean
=s / square root of n= 516.4 = ( 4000 / square root of 60)
Confidence level= 95%
Therefore, Significance level= a (alpha) = 5% = 1- 0.95
No of trails= 2
This is so because it’s a 2 tailed test because we need to find the margin of error.
Since sample size= 60 >= 30
we will use normal distribution
Z at the 0.05 level of significance 2 tailed test = 1.96
Margin of error =z*s x
= 1.96*516.4
= 1012.144.
if the study requires us to assume a margin of error of $500, we will then need to increase the sample size
•Therefore,
Standard deviation =s = $4,000
confidence interval= 95%
Z corresponding to 95% and two tailed test is = 1.96
We have to ensure that Z* s x < $500
or s x < =500/Z
= 500/1.96
= or s x < 255.102041
But
s x=standard error of mean=s / square root of n
s = 4000
or n=(s ^2)/(s x^2)
= 4000^2/255.102041^2
= 246.
Therefore, the sample size should be increased to 246.
