Question

In a random sample of students who took the SAT test, 427 had paid for coaching courses and the remaining 2733 had not. Calculate the 95% confidence interval for the proportion of students who get coaching on the SAT .

Answer 1

Answer:

The 95% confidence interval for the proportion of students who get coaching on the SAT is (0.1232, 0.147).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

427 had paid for coaching courses and the remaining 2733 had not.

This means that $n = 427 + 2733 = 3160, \pi = \frac{427}{3160} = 0.1351$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1351 - 1.96\sqrt{\frac{0.1351*0.8649}{3160}} = 0.1232$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1351 + 1.96\sqrt{\frac{0.1351*0.8649}{3160}} = 0.147$

The 95% confidence interval for the proportion of students who get coaching on the SAT is (0.1232, 0.147).