Question

Many high school students take the AP tests in different subject areas. In 2007, of the 144,796 students who took the biology exam 84,199 of them were female. In that same year, of the 211,693 students who took the calculus AB exam 102,598 of them were female ("AP exam scores," 2013). Estimate the difference in the proportion of female students taking the biology exam and female students taking the calculus AB exam using a 90% confidence level.

Answer 1

Answer:

$(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942$  

$(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978$  

And the 90% confidence interval would be given (0.0942;0.09978).  

We are confident at 90% that the difference between the two proportions is between $0.0942 \leq p_A -p_B \leq 0.09978$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$p_A$ represent the real population proportion female for Biology

$\hat p_A =\frac{84199}{144796}=0.582$ represent the estimated proportion female for biology

$n_A=144796$ is the sample size for A

$p_B$ represent the real population proportion female for calculus AB

$\hat p_B =\frac{102598}{211693}=0.485$ represent the estimated proportion female for Calculus AB

$n_B=211693$ is the sample size required for B

$z$ represent the critical value for the margin of error  

The population proportion have the following distribution  

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$  

The confidence interval for the difference of two proportions would be given by this formula  

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$  

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.64$  

And replacing into the confidence interval formula we got:  

$(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942$  

$(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978$  

And the 90% confidence interval would be given (0.0942;0.09978).  

We are confident at 90% that the difference between the two proportions is between $0.0942 \leq p_A -p_B \leq 0.09978$