Having trouble with Standard Error of the Mean

1. Mar 21, 2005

maccaman

Question 8 - From your results above, which would be the best estimate of the true mean
height (170.18cm) of the population of Bioinformatics students? Explain!

I am having a little trouble on this question and what exactly is it. It is asking me for which would be the best estimate of the true mean. Now the results above are as follows:

Sample 1
n = 15
mean = 171.06
s = 9.46
SEM = 2.44

Sample 2
n = 23
mean = 172.13
s = 8.59
SEM = 1.79

I am a little hazy as what i should write, as you would think the lower SEM would result in a closer estimate of the true mean, however the larger SEM actually has a mean closer to the true mean than Sample 2. Any help on this would be greatly appreciated. Thanks

2. Mar 21, 2005

xanthym

Your comments demonstrate some fundamental characteristics of statistical measurements: that error and variation always occur, and that most statistical "interpretations" are only valid ON AVERAGE.

You are certainly correct in stating that the Sample with lower SEM would be expected to have Sample Mean closer to the Population Mean. However, the critical word here is "expected". Only ON AVERAGE will that be true. It is always possible (shown clearly in this example) that a Sample with larger SEM might actually (sometimes) have Sample Mean closer to the Population Mean than another Sample (from the same population) with smaller SEM. This is what statistics is all about, and why care should be used when interpreting results.

For the current situation, most statisticians would consider POOLING the data in order to achieve better overall SEM for the Pooled Sample mean. Consult your textbook for detailed discussion of this technique. For this example, the process would proceed like below:

Sample 1
n = 15
mean = 171.06
s = 9.46
SEM = 2.44

Sample 2
n = 23
mean = 172.13
s = 8.59
SEM = 1.79

STEP #1: Calculate Pooled Sample Mean "m" Value
mpool = {(15)*(171.06) + (23)*(172.13)}/(15 + 23) = (171.71)
STEP #2: Calculate Pooled Sample "s" Value
spool = sqrt{ {(15 - 1)*(9.46)2 + (23 - 1)*(8.59)2}/{15 + 23 - 2} } = (8.94)
STEP #3: Calculate Pooled Sample "SEM" Value
SEMpool = (8.94)/sqrt(15 + 23) = (1.45)

You can see that the Pooled Sample Mean is between those of the original Samples. However, the Pooled SEM is smaller than either of those from the original Samples. This shows another characteristic of Sample Means, namely that their SEMs generally become smaller with increasing Sample size. In this case, the smaller SEM would indicate ON AVERAGE that the Pooled Sample Mean would better approximate the Population Mean.

~~

Last edited: Mar 21, 2005