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

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

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{(\pi(1-\pi))/(n)}`

In which

z is the z-score that has a p-value of
`1 - (\alpha)/(2)`.

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

This means that
`n = 427 + 2733 = 3160, \pi = (427)/(3160) = 0.1351`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a p-value of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

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

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.1351 + 1.96\sqrt{(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).