Question

A university financial aid office polled a random sample of 670 male undergraduate students and 617 female undergraduate students. Each of the students was asked whether or not they were employed during the previous summer. 388 of the male students and 323 of the female students said that they had worked during the previous summer. Give a 90% confidence interval for the difference between the proportions of male and female students who were employed during the summer. Construct the 90 % confidence interval.

Answer 1

Answer:

The 90% confidence interval for the difference between the proportions of male and female students who were employed during the summer is (0.01, 0.1012).

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.

Male undergraduates:

670, 388 were employed. So

$p_M = \frac{388}{670} = 0.5791$

$s_M = \sqrt{\frac{0.5791*0.4209}{670}} = 0.0191$

Female undergraduates:

Of 617, 323 were employed. So

$p_F = \frac{323}{617} = 0.5235$

$s_F = \sqrt{\frac{0.5235*0.4765}{617}} = 0.0201$

Distribution of the difference:

$p = p_M - p_F = 0.5791 - 0.5235 = 0.0556$

$s = sqrt{s_M^2+s_F^2} = \sqrt{0.0201^2 + 0.0191^2} = 0.0277$

Confidence interval:

The confidence interval is given by:

$p \pm zs$

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:

$p - zs = 0.0556 - 1.645*0.0277 = 0.01$

The upper bound of the interval is:

$p + zs = 0.0556 + 1.645*0.0277 = 0.1012$

The 90% confidence interval for the difference between the proportions of male and female students who were employed during the summer is (0.01, 0.1012).