Question

a high school wants to determine the difference between the proportions of seniors enrolled in ap classes and those enrolled in regular classes who plan on going to college next year. a srs of 85 seniors enrolled in ap classes is taken and 70 plan on enrolling in college next year, whereas, in an srs of 72 seniors enrolled in regular classes, 63 plan on going to college next year. construct a 95% confidence interval to estimate the difference in the proportion of seniors enrolled in ap classes and those enrolled in regular classes who plan on going to college next year. a (-0.089, 0.163) b (-0.066, 0.112) c (-0.115, 0.165) d (-0.038, -0.135) e (-0.145, -0.007)

Explanation:

Step1: Define sample proportions

Let $p_1$ = proportion of AP seniors going to college, $p_2$ = proportion of regular seniors going to college.
$n_1=85$, $x_1=70$, so $\hat{p}_1=\frac{70}{85}\approx0.8235$
$n_2=72$, $x_2=63$, so $\hat{p}_2=\frac{63}{72}=0.875$
Difference: $\hat{p}_1-\hat{p}_2=0.8235-0.875=-0.0515$

Step2: Find critical z-value

For 95% confidence, $z^*=1.96$

Step3: Calculate standard error

$$ SE=\sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} $$

$$ =\sqrt{\frac{0.8235(1-0.8235)}{85}+\frac{0.875(1-0.875)}{72}} $$

$$ =\sqrt{\frac{0.8235\times0.1765}{85}+\frac{0.875\times0.125}{72}} $$

$$ =\sqrt{\frac{0.1453}{85}+\frac{0.1094}{72}}\approx\sqrt{0.00171+0.00152}\approx\sqrt{0.00323}\approx0.0568 $$

Step4: Compute margin of error

$ME=z^*\times SE=1.96\times0.0568\approx0.1113$

Step5: Find confidence interval

Lower bound: $-0.0515-0.1113\approx-0.1628$
Upper bound: $-0.0515+0.1113\approx0.0598$
Rounding gives (-0.163, 0.060), which matches option A.

Answer:

A. (-0.163, 0.060)