Mathematics, 16.07.2021 06:20 tommyaberman

The Wall Street Journal recently ran an article on perception of sexual harassment on the job between men and women. The article claimed that women perceived the problem to be much more prevalent than did men. One question asked to both men and women was: "Do you think sexual harassment is a major problem in the American workplace?" Some 24% of the men compared to 62% of the women responded "Yes." Suppose that 150 women and 200 men were interviewed. Construct a 95% confidence interval estimate of the difference between the proportion of women and men who think sexual harassment is a major problem in the American workplace. What is the upper limit of the confidence interval?

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The 95% confidence interval estimate of the difference between the proportion of women and men who think sexual harassment is a major problem in the American workplace is (0.2824, 0.4776). The upper limit of the confidence interval is 0.4776.

Step-by-step explanation:

Before solving this question, 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.

Proportion of women:

62% of 150, so:

p_W = 0.62

$s_W = \sqrt{\frac{0.62*0.38}{150}} = 0.0396$

Proportion of men:

24% of 200, so:

p_M = 0.24

$s_M = \sqrt{\frac{0.24*0.76}{200}} = 0.0302$

Distribution of the difference:

p = p_W - p_M = 0.62 - 0.24 = 0.38

$s = \sqrt{s_W^2+s_M^2} = \sqrt{0.0396^2+0.0302^2} = 0.0498$

Confidence interval:

$p \pm zs$

In which

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

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:

p - zs = 0.38 - 1.96*0.0498 = 0.2824

The upper limit of this interval is:

p + zs = 0.38 + 1.96*0.0498 = 0.4776

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The Wall Street Journal recently ran an article on perception of sexual harassment on the job betwee...