```Question 333315
Many people will make lots of assumptions to simplify a problem like this and in many cases you will get answers close enough, but you should be aware of the proper way to do things
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1. You are dealing with a test of two independent samples, with unknown population variances.
2. You did not state the underlying distribution that these samples came from, but as long as you are analyzing averages and the samples are larger than 30, you can rely on the Central limit theorem for license to utilize Normal approximations to whatever underlying distributions you started with.
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Now just because you can rely on normal approximations does not mean you can just jump into the Z normal distribution, there is the t normal distribution.
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Since the population variances are not known, this indicates that you should use a "t" distribution.  Althought, the samples sufficiently large, both the t distribution and Z distribution will merge and become indistinguishable.
but...... most people believe that this happens at samples even close to 30. In fact, the two distributions do not merge close enough until the samples are in the magnitude of approximately 200 each.
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So in your case you really should stick with the t distribution.
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To do things correctly, there are two basic types of t distribution.  One for when the "unknown" population variances can be assummed equal and one where the variances cannot be assumed equal.  (There is a F statistic for testing this, and I carried it out and determined that the population variances can be deemed to be equal,  but I am not going to cover that here)
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So the particular t test that you use is the one where you can pool the sample standard deviations (since the population variances are deemed to be equal)
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Confidence interval for population mean differences
{{{xbar[1] - xbar[2]}}} -/+ {{{ t*Sqrt(S[p]^2*(1/n[1]+1/n[2]))}}}
and {{{S[p]^2}}} =pooled_variance ={{{ ((n[1]-1)*S[1]^2+(n[2]-1)*S[2]^2)/(n[1]+n[2]-2)}}}
t = critical t value with 0.05 tail probability and 50+35-2 =83 degrees of freedom = 1.663,   NOTE: for 90% confidence you will have 5% at the tails
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plugging in your information
{{{S[p]^2}}} ={{{(34*12.8^2+49*14.6^2)/83 = 192.96}}}
{{{(75.1-72.1)}}} -/+ {{{1.663*Sqrt(192.96*(1/50+1/35))}}}

3.0 -/+ 5.09
(-2.09, 8.09)is the 90% confidence interval for the population mean differences
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