

Subject:
Business statistic exercise
Category: Science > Math Asked by: thepilkga List Price: $5.00 
Posted:
11 Aug 2002 16:23 PDT
Expires: 10 Sep 2002 16:23 PDT Question ID: 53378 
A manufacturer of pain relievers claims that it takes an average of 12.75 min. for a person to be relieved of headache pain after taking it's pain reliever. The time it takes to get relief is normally distributed with a standard deviation of 0.5 min. A sample of (12) people is taken and the data arae shown below: 12.9, 13.2, 12.7, 13.1, 13.0, 13.1, 13.0, 12.6, 13.1, 13.0, 13.1, 12.8. (a) Find the sample mean. (b) Find the standard error of X. If the manufacturer claims that the mean is 12.75 min. find the Zscore of the sample mean. (d) What do you think of the manufacturer's claim based on the Zscore? 

Subject:
Re: Business statistic exercise
Answered By: jasonm1ga on 11 Aug 2002 16:51 PDT Rated: 
Hi thepilk, Here are your answers, with short explanations. a) Sample mean: This is the sum of all of the observations divided by the number of observations. In this case, that is: 155.6/12 = 12.97 or about 13 (rounded to significant digits) b) Standard deviation (error): This is the square root of the variance. The sample variance is: s^2 = sum(x  M)^2/(N1) where M is the mean and N is the number of observations. So this one is: s^2 = (.01+.04+.09+.01+0+.01+0+.16+.01+0+.01+.04)/(121) = .0345 Hence s = 0.186 is the standard deviation. c) zscore: Zscore is an indication of how far an observation lies from the mean, in terms of the variance. z = (xM)/s z = (12.7512.97)/0.186 = 1.184 d) Since about 68% of the observations should be within 1 standard deviation away from the mean, and this zscore is more than one standard deviation away from the mean, the manufacturer claim is unlikely to be true. However, the sample size of this data set is small, so one cannot be very confident in the results.  

thepilkga
rated this answer:
It was a 4 part question, and 2 of his answers were challenged by a later Researcher. The original guy agreed those 2 answers were wrong. 

Subject:
Re: Business statistic exercise
From: jeremymilesga on 12 Aug 2002 11:24 PDT 
I think that Jasonm1ga has made an error here. The standard error of the mean is not the same as the standard deviation. The standard error is given by SD/sqrt(N). The standard error is therefore equal to 0.18257/sqrt(12) = 0.053. The error arises partly because people use +/ to mean SE or SD, and they don't specify. The equations are given at: http://davidmlane.com/hyperstat/A103735.html Search term: standard error of the mean ://www.google.com/search?q=standard+error+of+the+mean I would also cast a little doubt on the interpretation of the conclusion to part D. If a maufacturer's claim is deemed to be 'unlikely to be true' because it has a 32% (i.e. about 1 in 3) chance of being wrong, we are going to say that people are wrong an awful lot of the time. And whilst they may well be wrong, we are probably going to be embarrassed if we claim that 1 in 3 correct claims is wrong. (Although this sort of decision becomes rapidly complex.) My final point is that the confidence of the probability judgment, when the null hypothesis is rejected, is not dependent upon sample size. If the null hypothesis (that the manufacturer is correct) is rejected, the sample size does not come into it. Indeed, we might state with more confidence that the manufacturer is wrong, because we managed to find that they were wrong, even though we didn't try very hard (i.e. we had a small sample). (Incidentally, the analysis also shows the problem that not using sufficient digits might cause  when this sort of calculation has several steps, which follow one another, rounding errors tend to add up. The error here was small, but could ave been large had the SD been used in more steps.) 
Subject:
Re: Business statistic exercise
From: jasonm1ga on 12 Aug 2002 16:33 PDT 
You are absolutely correct on the Standard Error part. Short lapse of judgement. 
If you feel that you have found inappropriate content, please let us know by emailing us at answerssupport@google.com with the question ID listed above. Thank you. 
Search Google Answers for 
Google Home  Answers FAQ  Terms of Service  Privacy Policy 