Question

At a large Midwestern university, a simple random sample of 100 entering freshmen in 1993 found that 20 of the sampled freshmen finished in the bottom third of their high school class. Admission standards at the university were tightened in 1995. In 1997, a simple random sample of 100 entering freshmen found that only 10 finished in the bottom third of their high school class. Let p 1 and p 2 be the proportions of all entering freshmen in 1993 and 1997, respectively, who graduated in the bottom third of their high school class. What is a 90% plus four confidence interval for p 1 – p 2?

Answer 1

Answer:

The 90% confidence interval for the difference of proportions is (0.01775,0.18225).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

p1 -> 1993

20 out of 100, so:

$p_1 = \frac{20}{100} = 0.2$

$s_1 = \sqrt{\frac{0.2*0.8}{100}} = 0.04$

p2 -> 1997

10 out of 100, so:

$p_2 = \frac{10}{100} = 0.1$

$s_2 = \sqrt{\frac{0.1*0.9}{100}} = 0.03$

Distribution of p1 – p2:

$p = p_1 - p_2 = 0.2 - 0.1 = 0.1$

$s = \sqrt{s_1^2+s_2^2} = \sqrt{0.04^2 + 0.03^2} = 0.05$

Confidence interval:

$p \pm zs$

In which

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

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.  

The lower bound of the interval is:

$p - zs = 0.1 - 1.645*0.05 = 0.01775$

The upper bound of the interval is:

$p + zs = 0.1 + 1.645*0.05 = 0.18225$

The 90% confidence interval for the difference of proportions is (0.01775,0.18225).