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Given a sample population of 54 men and 61 women, I want to figure out whether or not the average male spends less time during his daily shower than the average female spends during hers. The average male time is 2.9 minutes with a standard deviation of 0.65 minutes. The average female time is 3.31 minutes and the standard deviation is 0.71 minutes.

I am trying to figure out the test statistic and P-value for this test. I was under the impression that when comparing two populations, you should use the following formula to obtain a test statistic:

$$\frac{\hat{p}_x - \hat{p}_y}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_x} +\frac{1}{n_y})}}$$

But because $p$ deals with proportions I'm not sure how to implement it in this case. Any help would be appreciated.

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  • $\begingroup$ You should mark this question with the self-study tag. $\endgroup$ – StatsStudent Apr 15 '15 at 4:26
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This is a t-test question. You want to know if the means of your two samples are different. Your test statistic, t, equals $$t=\frac{\bar y_1-\bar y_2}{\sqrt{s^2_1/n_1+s^2_2/n_2}}=\frac{3.31-2.90}{\sqrt{.71^2/61+.65^2_2/54}}=\frac{.41}{0.13}=3.23$$ $$t(113)=3.23, p<.0017$$ Note this is analogous to what you wrote. P is the mean of a binary variable, and p(1-p) is the variance of a binary variable. This gives you a z statistic rather than a t statistic. The reason is that you don't need to separately estimate the mean and variance with binary variables, so it eases the assumptions a little. The variance is just a function of the mean, i.e. p(1-p). The t-distribution is similar to the normal distribution (what z statistics refer to), and is equal to the normal distribution as the degrees of freedom approach infinity. Anyway, what's important is that you don't have a binary variable, so you need to do a simple t-test.

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