SOLUTION: Please help with my homework question on Interval Estimations. The book is Statistics for Business and Economics. A sample survey of 54 discount brokers showed that the mean price

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Question 165502: Please help with my homework question on Interval Estimations. The book is Statistics for Business and Economics. A sample survey of 54 discount brokers showed that the mean price charged for a trade of 100 shares at $50 per share was $33.77. The survey is conducted annually. With the historical data available, assume a known population standard deviation of $15.
a. Using the sample data, what is the margin of error associated with a 95% confidence interval?
b. Develop a 95% confidence interval for the mean price charged by discount brokers for a trade of 100 shares at $50 per share.
Answer by gonzo(654) (Show Source):
You can put this solution on YOUR website!
i don't have your book but i've done some research online and have come up with the following:
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n = 54 = sample size.
standard deviation of population is 15.
you are looking for a confidence level of 95%
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the formula for determining the margin of error associated with a 95% confidence interval looks very much like it is:
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E is the margin of error you are looking for.
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(alpha/2) is 1 - confidence level you are looking for divided by 2. since the confidence level you are looking for is 95%, (alpha/2) is (1-.95)/2 = .05/2 = .025.
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Z of (.025) would be the value of Z where the right hand tail of the normal distribution begins.
.025 of the area under the curve is to the right of that point.
that equates to a Z value of 1.96 which means that you are 1.96 standard deviations to the right of the mean.
in the standard Z table, the mean is 0 and the standard deviation is 1.
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SD is the standard deviation of the population = 15.
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n is the sample size = 54.
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the formula becomes:
E = 1.96 * (15/sqrt(54)
This comes out to be: 4.000833247.
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looks like your margin of error is +/- 4.000833247
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the confidence interval is twice the margin of error.
if your mean is 33.77, then your margin of error is +/- 4.000833247 from that.
that makes your confidence interval from 29.77 to 37.77 at 95% confidence level.
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that's what i think. i'm reasonably sure this is correct. check with others to make sure.