Anyone good at statistics?????

Question

A business wants to estimate the true mean annual income of its customers. The business needs to be within $500 of the true mean. The business estimates the true population standard deviation is around $2,300. If the confidence level is 95%, find the required sample size in order to meet the desired accuracy.

Answer 1

EDITED.

I believe the above answer is wrong.

Aside. Proof of the formula that is used in below.

The confidence interval for the true mean is

[ _x(+-)z(α/2)*s/√n]

To obtain the sample size we need to set the error bound in the interval to B, then we get

z(α/2)*s/√n = B

1/√n = B/z(α/2)*s

B*√n = z(α/2)*s

    Ã¢ÂˆÂšn = z(α/2)*s/B

Squaring both sides of the equation, then

   n = [z(α/2)*s/B]^2

Let n be the sample size required to estimate the true mean annual income of its customers, B = Error bound of the true mean = $500, δ= true population standard deviation = $2300 and α be the level of significance.

Given that 100(1-α)% = 95%

                    1-α = 0.95

                      ÃŽÂ± = 0.05

n = [z(α/2)*s/B]^2

n = [z(0.05/2)*2300/500]^2

n = [z0.025 * 2300/500]^2

n = [1.96*2300/500]^2

n = [1.96*23/5]^2

n = (9.016)^2

n = 81.288256

n ≈ 81

A sample size of n=81 is required to meet the above desired accuracy.

Hope this helps.

Answer 2

ANSWER: indeterminate.  Confidence Interval cannot be determined.

Why??

"business needs"  "$500 of the true mean" and "standard deviation" "$2,300"  

True mean (real positive number) must exceed standard deviation for confidence interval to be determined in an income problem.