OK – OK – so I am the very first to admit, even as a Staff Engineer, I am crummy at math. I am working on a Statistics/Probability question and would like some feedback from some of the math guru’s in the audience. Here is the question I have a population that makes an average of $24.07 per hour The standard deviation is $4.80 The sample size is 120 individuals What is the possiblility that the sample mean will be within 50 cents of the population mean? I figured that the range is $23.57 to $24.57 23.57 – 24.47 / 4.80 = -0.2083 or -20.83% chance that the sample mean will be within 50 cents of the population mean. This looks too easy – am I doing something wrong or leaving something out?
Okay. I'm not any kind of math guru, but this is how I would approach it. You need to find the t value for both your upper $.50 limit ($24.57) and your lower limit ($23.57). You can find a "very close" (because of the relatively large n) z value(s) to substitute for the t value(s) by subtracting the average from you limit(s), and then dividing by the standard deviation. Then plug the numbers into your z table, subtract the lower from the higher, and there is a very close approximation of the probability. Or, you could use excel (or a stats program) to find out the exact probability. I think it should be somewhere around 8%. Tony P.S. I sincerely hope this isn't misleading, but if it is, I hope someone corrects me. Thanks.
Once N > 30 then the sample mean is usually a good estimate of the population mean. Using the Central Limit Theorem Std error = (std Dev)/(N^0.5) = .4382 Z1 = (23.57 - 24.07)/.4382 = -1.141 Z2 = (24.57 – 24.07)/.4382 = +1.141 Probability equals the area under the normal curve = .8729 - .1271 = .7458 (74.58%) Ref: Walpole & Meyers. Probability and Statistics for Engineers and Scientists. Macmillan. Fourth Ed. (Chapter 6.8). (My MSQA Statistics textbook)
You're right, Ian. I misread the problem, and was looking for the percentage of the sample population within $.50 either way. Sorry. Tony
Actually Ian's answer was correct. I went through the problem several times before coming up with the right answer -- and Ian confirmed it. Thanks for all of the help! W.