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this problem is from the wackerly, but i have a particular question, in this problem we have to estimate the variance with $S^2_p$ or can we use the information standard deviations that they gave us.

Here is the problem:

Two new drugs were given to patients with hypertension. The first drug lowered the blood pressure of 16 patients an average of 11 points, with a standard deviation of 6 points. The second drug lowered the blood pressure of 20 other patients an average of 12 points, with a standard deviation of 8 points. Determine a 95% confidence interval for the difference in the mean reductions in blood pressure, assuming that the measurements are normally distributed with equal variances.

neto333
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  • I know that if the variance is known then you have to estimate it, but in this case they are giving you the probational variance no the sample variance right? – neto333 Oct 11 '16 at 23:55
  • It's unclear what your question is. And this should probably be moved to: http://stats.stackexchange.com/ – Jonathan H Oct 12 '16 at 00:01

1 Answers1

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Assuming the given sample standard deviations $S_1,S_2$ have respective denominators $16-1$ and $20-1$, the pooled sample variance is \begin{align} S_p^2 & = \frac{(16-1)S_1^2 + (20-1)S_2^2}{(16-1)+(20-1)} = \frac{(15\times6^2) + (19\times 8^2)}{(16-1)+(20-1)} \\[12pt] & = \frac{540 + 1216}{34} = \frac{1756}{34} \approx 7.186589^2. \end{align}

Recall that $((16-1)+(20-1)) \dfrac{S_p^2}{\sigma^2} \sim \chi^2_{(16-1)+(20-1)}.$

You need to use $S_p$ with a t-distribution with $(16-1)+(20-1) = 34$ degrees of freedom.

Michael Hardy
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  • But why you have use the $S_p$? – neto333 Oct 12 '16 at 01:30
  • Because $S_p^2$ is an unbiased estimator of the populations variance and $$ ((16-1)+(20-1)) \dfrac{S_p^2}{\sigma^2} \sim \chi^2_{(16-1)+(20-1)} $$ and $S_p^2$ is independent of the pair of sample means. Thus you have $$ \frac{\left( \dfrac{(\bar X_1 - \mu_1) - (\bar X_2 - \mu_2)}{\sigma/\sqrt{ \frac 1 {16} + \frac 1 {20} }} \right)}{\left( \left. \dfrac{S_p}{\sigma} \right/ \sqrt{(16-1)+(20-1)} \right)} \sim \frac{Z}{\chi/\sqrt d} \sim t_{(16-1)+(20-1).} $$ Notice that the $\sigma$ cancels out in the expression above, so you have an observable quantity. $\qquad$ – Michael Hardy Oct 12 '16 at 01:41
  • $\ldots,$where $d$ is of course the degrees of freedom: $d=(16-1)+(20-1)$, and $Z$ and $\chi$ are independent. The fraction $$ \frac Z {\chi/\sqrt d} $$ is often taken to be the definition of the t-distribution with $d$ degrees of freedom (where the numerator and denominator are independent). $\qquad$ – Michael Hardy Oct 12 '16 at 01:44
  • But why you have to estimate the population variance if they are giving you the standard deviation? – neto333 Oct 12 '16 at 02:28
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    @neto333 : That have not given you the population standard deviation. If you had that, then what was done above would be replaced by something simpler. What you have is two sample standard deviations. Is that not clear to you from the way the problem is stated? It says "assuming that the measurements are normally distributed with equal variances", and yet the two given standard deviations are not equal to each other. They are not the population standard deviation. $\qquad$ – Michael Hardy Oct 12 '16 at 16:46