Difference between two proportions in a Confidence Interval

Question

Ten engineering schools in the United States were surveyed. The sample contained $250$ electrical engineers, $80$ being women; $175$ chemical engineers, $40$ being women. Compute a $90\%$ confidence interval for the difference between the proportions of women in these two yields of engineering. Is there a significant difference between the two proportions?

After all my calculations I get this interval $[0.01863, 0.16137]$ I interpret it as a big difference because the interval is from .02% to .16% is this an erroneous interpretation?

Answer 1

First, as a check, I computed the CI and got $(0.0201, 0.1627).$ Formulas differ a bit from text to text, but you might want to check your computations.

Second, as you suppose, you can reject the null hypothesis that the two population proportions of women are equal against the two-sided alternative that the proportions differ at the 10% level of significance. This because your 90% CI does not contain 0.

A formal test of $H_o: \theta_1 = \theta_2$ against $H_a: \theta_1 \ne \theta_2$, rejects if the Z-statistic has

$$|Z| = \frac{|\hat \theta_1 - \hat \theta_2|} {\sqrt{\frac{\hat \theta_1(1-\hat \theta_1)}{n_1} + \frac{\hat \theta_2(1-\hat \theta_2)}{n_2}}} = \frac{0.0914}{ 0.0433} = 2.11. $$ Thus, the P-value of the test is about 0.0349, so you could reject at the 4% level, but not the 3% level of significance.

Answer 2

I wonder if the question is a bit different from examining the difference in the proportion of women in the electrical engineering field and the proportion of women in the chemical engineering field as shown by @Bruce Trumbo.

In other words because the wording suggests a single sample of size 425 (and not a sample of 250 electrical engineers and a separate sample of 175 chemical engineers), might this not be a comparison of two multinomial proportions? The question could be interpreted as finding the difference in the proportion of women chemical engineers and the proportion of women electrical engineers in the population of chemical and electrical engineers in the 10 engineering schools.

If so, we have $n = 425$, and $\hat p_{WE} =80/425$ and $\hat p_{WC} =40/425$. An approximate test would be constructing a test statistic

$$z=(\hat p_{WE} -\hat p_{WC})/\sqrt{\hat p_{WE}(1-\hat p_{WE})/n+\hat p_{WC}(1-\hat p_{WC})/n+2\hat p_{WE} \hat p_{WC}/n}$$

and an approximate 90% confidence interval would be

$$(\hat p_{WE} -\hat p_{WC}\pm 1.645\sqrt{\hat p_{WE}(1-\hat p_{WE})/n+\hat p_{WC}(1-\hat p_{WC})/n+2\hat p_{WE} \hat p_{WC}/n}) $$

with the result for this sample

$$(0.05239,0.13585)$$

Does this suggest a "big" difference? This is up to you. It is a subject-matter choice and not the choice of a statistician. For this particular application, how big is big? That needs to be decided before the analysis of the data.