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: marginal utility of the fourth brownie is 0.

explanation: marginal utility is the addtional utility gained from the consumption of one more unit of a good. it is given by,

so, marginal utility at the 4th unit will be given by

since, the values are not given here, you can simply find marginal utility at fourth unit by subtracting total utility at 3rd unit from total utility at 4th unit.