In a survey of families in which both parents work, one of the questions asked was, "Have you refused a job, promotion, or transfer because it would mean less time with your family?" A total of 200 men and 200 women were asked this question. "Yes" was the response given by 34% of the men and 26% of the women. Based on this survey, can we conclude that there is a difference in the proportion of men and women responding "yes" at the 0.05 level of significance? (Use Men - Women.)

(a) Find z. (Give your answer correct to two decimal places.)


(ii) Find the p-value. (Give your answer correct to four decimal places.)

z = 1.75

p-value =.0809

If p-value is greater than alpha
Fail to reject Ho
If p-value is less than alpha
Reject Ho

1. Men 2. Women

n1 = 200. n2 = 200

x1 = .34(200) =64. x2 = .26(200) = 52
phat1 = .34. phat2 = .26

pbar = (x1 + x2)/(n1+n2)

pbar = (68+52)/(200+200)

pbar = .30

qbar = 1-pbar = .70

z = (phat1-phat2)/sqrt(phar*qbar/n1 + pbar*qbar/n2)

z = (.34-.26)/sqrt(.3*.7/200 + .3*.7/200)

z = -+ 1.7457

ti-83
2nd vars

2-normalcdf(-10, -1. 7457) = = .04043

two tail so you multiply 2
2( .04043) = 0.08086
Four decimal places = .0809

To determine if there is a difference in the proportion of men and women responding "yes" at the 0.05 level of significance, we can use a two-sample z-test.

(a) To calculate the z-score, we need the sample proportions for both men and women and the respective sample sizes. Let's denote the proportions as p1 for men and p2 for women.

These proportions can be calculated as follows:
p1 = 0.34 (34% of men responded "yes")
p2 = 0.26 (26% of women responded "yes")

The sample sizes are:
n1 = 200 (number of men)
n2 = 200 (number of women)

To find the z-score, we use the formula:
z = (p1-p2) / sqrt(p * (1-p) * (1/n1 + 1/n2))

Where p is the pooled sample proportion, calculated as:
p = (p1*n1 + p2*n2) / (n1 + n2)

Let's calculate the z-score:

p = (0.34 * 200 + 0.26 * 200) / (200 + 200) = 0.30

z = (0.34 - 0.26) / sqrt(0.30 * (1-0.30) * (1/200 + 1/200)) = 0.08 / sqrt(0.30 * 0.70 * (1/200 + 1/200)) = 0.08 / sqrt(0.30 * 0.70 * 0.01) = 0.08 / sqrt(0.0021) ≈ 2.07

Therefore, the z-score is approximately 2.07.

(ii) To find the p-value associated with this z-score, we need to determine the area under the standard normal curve that is greater than the absolute value of the z-score (since the alternative hypothesis is two-sided).

Using a standard normal distribution table or a calculator, the p-value is the probability of observing a z-score greater than 2.07 (in absolute value) or less than -2.07.

The p-value is approximately 0.0192 (to four decimal places).

Therefore, the p-value is 0.0192.

Conclusion:
Since the p-value (0.0192) is less than the significance level of 0.05, we can conclude that there is a significant difference in the proportion of men and women responding "yes" at the 0.05 level of significance.

To determine if there is a difference in the proportion of men and women responding "yes" at the 0.05 level of significance, we can use a hypothesis test.

Hypotheses:
- Null Hypothesis (H0): The proportion of men and women responding "yes" is the same.
- Alternative Hypothesis (H1): The proportion of men and women responding "yes" is different.

To find z, we first need to calculate the standard error of the difference in proportions.

Standard error formula:
SE = sqrt[(p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2)]

where:
p1 = proportion of men responding "yes" = 34% = 0.34 (in decimal form)
p2 = proportion of women responding "yes" = 26% = 0.26 (in decimal form)
n1 = sample size of men = 200
n2 = sample size of women = 200

Substituting the values, we get:
SE = sqrt[(0.34 * (1 - 0.34) / 200) + (0.26 * (1 - 0.26) / 200)]

Calculating the SE gives us:
SE ≈ sqrt(0.000867 + 0.000958)
SE ≈ sqrt(0.001825)
SE ≈ 0.0427

To find z, we need to calculate the test statistic using the formula:
z = (p1 - p2) / SE

Substituting the values, we get:
z = (0.34 - 0.26) / 0.0427

Calculating the z-score gives us:
z ≈ 0.08 / 0.0427
z ≈ 1.874

Now, we can find the p-value associated with the calculated z-score.

Using a standard normal distribution table or statistical software, we can find that the p-value for z = 1.874 is approximately 0.0304.

Therefore, the p-value is 0.0304 (rounded to four decimal places).

Next, we compare the p-value to the significance level (0.05). The p-value (0.0304) is less than the significance level (0.05).

Since the p-value is less than the significance level, we reject the null hypothesis (H0).

Conclusion:
Based on the survey, we can conclude that there is a significant difference in the proportion of men and women responding "yes" at the 0.05 level of significance.