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I was making an attempt to solve the following problem sets, however, my answer 15.913 seemed quite different from that in the answer key 16.84. I have checked my answer a few times and re-doing the question as well, but I was consistently pulled back to the same outcome of 15.913. Appreciate some guidance and advice please.

Question

A recent study of home technologies reported the number of hours (to the nearest hour) of personal computer usage per week for a sample of 10 persons. Excluded from the study were people who worked out of their home and used the computer as a part of their work.

8, 5, 6, 8, 6, 1, 5, 6, 9, 4

In a study with sample size of 30 people who worked out of their home and used the computer as a part of their work, the sample mean and standard deviation of the number of hours personal computer usage per week are 42.0 and 6.4, respectively. If we wants to combine the results of these two samples, find the standard deviation of the number of hours personal computer usage per week of in the combined sample.

My Attempt

New Sample

n=10, sample mean = 5.8, Sum(xi) = 58, Sum(xi^2) = 384

Old Sample

n=30, sample mean = 42, s= 6.4, Sum(xi) = (30)(40) = 1260, Sum(xi^2) = (30)(40^2) = 52920

Combined Sample

n=40

sample mean = (1260+58)/40 = 32.95

Sum(xi) = 1260 + 58 = 1318

Sum(xi^2) = 384 + 52920 = 53304

s = SqRt{[53304 - (40)(32.95^2)]/[40-1]} = 15.91314

The answer offered in the answer key 16.84 seemed quite different from my 15.91314. Appreciate any advice if there might be some steps which I may be doing wrongly.

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  • $\begingroup$ The answer key is definitely correct. $\endgroup$ – Glen_b Apr 22 '14 at 12:58
  • $\begingroup$ See here and here $\endgroup$ – Glen_b Apr 22 '14 at 13:33
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There is an error in the sum of xi² for the old sample. I think you mix the sum of squares and the square of sum.

Old Sample

n=30, sample mean = 42, s= 6.4, Sum(xi) = (30)( 42 ) = 1260, (Sum(xi) )^2 /30 = (30)(42^2) = 52920

You should try to demonstrate this formula : $$ \sigma^2 = \left[ \frac{n_1}{n} \sigma_1^2 + \frac{n_2}{n} \sigma_2^2 \right] + \left[ \frac{n_1}{n} \bar{x}_1^2 + \frac{n_2}{n} \bar{x}_2^2 - \bar{x}^2 \right] $$ where 1 represent the new sample and 2 the old one

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I believe that your concept is alright. However, you should explore how you obtained the sum of $x_i^2$. Your method of calculation may be incorrect. So I would advise you to explore how you should use standard deviation to obtain the desired result.

$s^2 = \frac{\sum{[x_i- \frac{\sum x_i}{n}}]^2}{n-1} $

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