If the sample size is large enough, then the distribution of the difference between to sample proportions (\(\hat p_1 - \hat p_2\)) has the following properties
\[ SE = \sqrt{ \frac{p_1 (1-p_1)}{n_1} + \frac{p_2 (1-p_2)}{n_2} } \] where \(p_1\) and \(p_2\) are the actual proportions in the population and \(n_1\) and \(n_2\) are the two sample sizes.
Unfortunately, we don’t ever know \(p_1\) and \(p_2\). So we need to approximate them.
\[ SE \approx \sqrt{ \frac{\hat{p}_1 (1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2 (1-\hat{p}_2)}{n_2} } \]
\[ SE \approx \sqrt{ \frac{\hat{p} (1-\hat{p})}{n_1} + \frac{\hat{p} (1-\hat{p})}{n_2} } = \sqrt{ \hat{p} (1-\hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2}\right) } \]
\(\hat p\) is estimated using all of the data (from both groups)
\[ \hat p = \frac{\mbox{number of successes}}{\mbox{number of cases}} = \frac{ x_1 + x_2}{n_1 + n_2} = \frac{ \hat{p}_1 n_1 + \hat{p}_2 n_2}{n_1 + n_2} \]
This SE formula is an approximation (because we don’t know the true value of \(p\)). The approximation is better if \(n\) is large and if \(p\) is closer to 0.5. Our rule of thumb will be that the SE methods for a proportion are good enough to use if
In each situation (a) check that the approximation is good enough to use, (b) compute the standard error and (c) compute the requested confidence interval or p-value (if neither is specified, do one or the other or both as is appropriate to the situation).
If the sample size is not large enough, don’t do steps b and c. (We could use randomization or bootstrap in these situations.)
A study compared two treatment protocols for type I diabetes – standard and intensive. Subjects were randomly assigned to receive one of the two treatment protocols. In the intensive treatment group, the treatment aimed at agressively controling blood glucose levels as close to normal as possible through more frequent monitoring blood levels and adjusting insulin as needed. Patients were followed for six years after the beginning of their treatments.
Of 348 patients receiving intensive care, 23 developed retinopathy. Of 378 patients receiving standard care, 91 developed retinopathy. Is this evidence that one treatment is better than the other?
Refering back to the type I diabetes study, contstruct a confidence interval for the difference in the proportions of subjects developing retinopathy.
Researchers conducted a study that looked at the sex of snow geese chicks compared to the laying order of the eggs. The considered only nests in which 4 eggs had been laid. The first two eggs they called “early” and the third and fourth they called “late”. Of the 52 live goslings from early eggs, 19 were female. Of the 43 live goslings from late eggs, 31 were female. Is this enough evidence to conclude that the sex ratio differs between early and late hatchlings?
Refering back to the previous example, compute a confidence interval for the difference in the proportions of female hatchlings in early and late eggs.
| \ | duct tape | liquid nitrogen |
|---|---|---|
| complete remission | 22 | 15 |
| incomplete remission | 4 | 10 |
In a study of published papers in two medical journals 135 out of 190 papers authored without a statistician were rejected without a detailed review; 293 of the 514 papers authored with a statistician were sent back without detailed review.
What proportion of papers to these journals include statisticians as co-authors?
Are papers without statisticians more likley to be rejected without detailed review?