A survey of small businesses with Web sites found that the average amount spent on a site was $11,500 per year (Fortune, March 5, 2001). Given a sample of 60 businesses and a population standard deviation of $4000, what is the margin of error (o decimals)? Use 95% confidence. What sample size would you recommend if the study required a margin of error of $500 (0 decimals)?

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.

answer; the above statement is (true);

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