- April 18, 2017 at 9:12 am #55695
I am having trouble question found on a sample exam.
Need to know fewest samples for 99% confidence when sheep are fed a a new diet, the avg weight gain is 5 lbs over current avg of 20 lbs. weight gain per sheep is normally dist. with a SD of 4.3 lbs. I know the formula is n=(Zsq x st dev sq)/(E)sq
Therefore I came up with Z=2.58 SD=4.3 E=5 plugged them in and came up with 4.19 which of course means I went horribly wrong as the answer is 101.
Where is the flaw in my calculation?
ThanksApril 18, 2017 at 2:40 pm #201209
I don’t see how they are getting 101 either.
If the statement “99% confidence” translates into an alpha of .01 and if the power is assumed to be 80% (i.e. beta = .2 – this is not specified – but is widely accepted as a reasonable test value) and if we treat this as a one sample t-test where the null mean is 20, the test mean is 25, and the estimate of the standard deviation is 4.3 then the total sample size is about 12 with all of the samples pulled from the improved diet group.
If the statement “99% confidence” translates into an alpha of .01 and power = 99% (beta = .01) and if we treat this as a one sample t-test where the null mean is 20, the test mean is 25, and the estimate of the standard deviation is 4.3 then the total sample size is 22 -with all of the samples pulled from the improved diet group.
In the most extreme case, if the 99% statement means we want power = .99 (beta = .01) and alpha = .005 (two sided for a total of 99%) and if we assume we are testing both means and need to pull a sample from both groups and if we assume the pooled standard deviation of the two groups is 4.3 then the total sample is 82 – which is 41 samples per group.
I ran the first analysis by hand and then using a power program – the hand calculated estimate rounded up to 12 and the program gave a value of 12 with the power = .797 which rounded to .8. I used the program to generate the other two.April 21, 2017 at 1:49 pm #201236
Thanks for your time.
They actually say in the answer feedback that the formula I used is the correct one. So:
1: I miss-applied a term to the formula.
2: I mis-calculated the formula using the right terms.
3: They made a mistake in the sample exam.
The sample question is from a very well respected organization but everybody makes mistakes.April 22, 2017 at 7:25 am #201237
So, are you saying items 1-3 in your most recent post are all true or are you saying something else?April 22, 2017 at 8:13 am #201240
I’m thinking at least one point must be true right? Point three encompases they are using the wrong formula altogether. What do you think?April 24, 2017 at 6:38 am #201247
My guess would be the formula. The usual procedure for this kind of sample size calculation is to take the equations for alpha and beta, subtract them from each other, and solve the resulting equation for the sample size.
For the case of alpha = .05 and beta = .1 and the means and standard deviation you supplied (20, 25, 4.3) the simultaneous equations for Ycbar (average critical value for decision making) are as follows:
(Ycbar – 25)/(4.3/sqrt(n)) = -1.645 (Z value based on alpha = .05)
(Ycbar – 20)/(4.3/sqrt(n)) = 1.282 (Z value based on beta = .1)
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