Question

. Suppose (as is roughly true) that 88% of college men and 82% of college women were employed last summer. A sample survey interviews SRSs of 400 college men and 400 college women. The two samples are of course independent. (a) What is the approximate distribution of the proportion pF of women who worked last summer? What is the approximate distribution of the proportion pM of men who worked? (b) The survey wants to compare men and women. What is the approximate distribution of the difference in the proportions who worked, pM − p F ? Explain the reasoning behind your answer. (c) What is the probability that in the sample a higher proportion of women than men worked last summer?

Answer 1

(a) The approximate distribution of the proportion pf of women who worked = 0.82, the standard distribution = 0.00738

The approximate distribution of the proportion pM of men who worked = 0.88, the standard distribution ≈ 0.01625

(b) pM - pF = 0.06

While pF - pM = -0.06

The difference in the standard deviation ≈ 0.01025

(c) The probability that a higher proportion of women than men worked last year is 0

Explanation:

(a) The given information are;

The percentage of college men that were employed last summer = 88%

The percentage of college women that were employed last summer = 82%

The number of college men interviewed in the survey = 400

The number of college women interviewed in the survey = 400

Therefore, given that the proportion of women that worked = 0.82, we have for the binomial distribution;

p = 0.82

q = 1 - 0.82 = 0.18

n = 400

Therefore;

p × n = 0.82 × 400 = 328 > 10

q × n = 0.18 × 400 = 72 > 10

Therefore, the binomial distribution is approximately normal

We have;

`The \ mean = p = 0.82\\\\The \ standard \ deviation, \sigma = \sqrt{ \frac{p * q}{{n} } }= \sqrt{ \frac{0.82 * 0.18}{{400} }} \approx 0.01921\\`

Therefore, the approximate distribution of the proportion pf of women who worked = 0.82, the standard distribution ≈ 0.01921

Similarly, given that the proportion of male that worked = 0.88, we have for the binomial distribution;

p = 0.88

q = 1 - 0.88 = 0.12

n = 400

Therefore;

p × n = 0.88 × 400 = 352 > 10

q × n = 0.12 × 400 = 42 > 10

Therefore, the binomial distribution is approximately normal

We have;

`The \ mean = p = 0.88\\\\The \ standard \ deviation, \sigma = \sqrt{ \frac{p * q}{{n} } }= \sqrt{ \frac{0.88 * 0.12}{{400} }} \approx 0.01625\\`

Therefore, the approximate distribution of the proportion pM of men who worked = 0.88, the standard distribution ≈ 0.01625

(b) Given two normal random variables, we have

The distribution of the difference the two normal random variable = A normal random variable

The mean of the difference = The difference of the two means = pM - pF = 0.88 - 0.82 = 0.06

While pF - pM = -0.06

The difference in the standard deviation, giving only the real values, is given as follows;

`The \ difference \ in \ standard \ deviation = \sqrt{ \frac{p_1 * q_1}{{n_1} } -\frac{p_2 * q_2}{{n_2} } }\\\\= \sqrt{\frac{0.82 * 0.18}{{400} }-\frac{0.88 * 0.12}{{400} }} \approx 0.01025\\`

(c) When there is no difference between the the proportion of men and women that worked last summer, the probability that there is a difference = 0

Therefore, taking 0 as the standard score, we have;

`z = (0 - (-0.06))/(0.01025) \approx 5.86`

Given that the maximum values for a cumulative distribution table is approximately 4, we have that the probability that a higher proportion of women than men worked last year is 0.