Question

many high school students take the ap tests in different subject areas. in one year, of the 210,269 students who took the calculus ab exam 104,874 of them were female and 105,395 of them were male. estimate using a 95% level the difference in proportion of female students taking the calculus ab exam versus male students taking the calculus ab exam. round to four decimal places. with % confidence, it can be concluded that the difference in proportion of female students taking the calculus ab exam versus male students taking the calculus ab exam (p1 - p2) is between and. question help: message instructor post to forum

Explanation:

Step1: Calculate sample proportions

Let $n_1 = n_2=210269$, $x_1 = 104874$, $x_2 = 105395$. The sample proportion of female students $p_1=\frac{x_1}{n_1}=\frac{104874}{210269}\approx0.4987$, and the sample proportion of male students $p_2=\frac{x_2}{n_1}=\frac{105395}{210269}\approx0.5012$.

Step2: Determine z - value for 95% confidence

For a 95% confidence level, the critical value $z_{\alpha/2}=1.96$.

Step3: Calculate the margin of error

The formula for the margin of error $E$ for the difference in two - proportions is $E = z_{\alpha/2}\sqrt{\frac{p_1(1 - p_1)}{n_1}+\frac{p_2(1 - p_2)}{n_2}}$.
Substitute the values:

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Step4: Calculate the confidence interval

The difference in sample proportions is $\hat{p}_1-\hat{p}_2=0.4987 - 0.5012=- 0.0025$.
The confidence interval is $(\hat{p}_1-\hat{p}_2)-ESubstitute the values: $-0.0025-0.0030 < p_1 - p_2<-0.0025 + 0.0030$.
So, $-0.0055 < p_1 - p_2<0.0005$.

Answer:

With 95% confidence, it can be concluded that the difference in proportion of female students taking the calculus AB exam versus male students taking the calculus AB exam ($p_1 - p_2$) is between - 0.0055 and 0.0005.