Problem 29 - Basic Statistics for Business and Economics with Student

Question

As part of a recent survey among dual-wage-earner couples, an industrial psychologist found that 990 men out of the 1,500 surveyed believed the division of household duties was fair. A sample of 1,600 women found 970 believed the division of household duties was fair. At the .01 significance level, is it reasonable to conclude that the proportion of men who believe the division of household duties is fair is larger? What is the \(p\) -value?

Step-by-Step Solution

Answer

Yes, it's reasonable to conclude that more men than women find it fair. The p-value is essentially 0.

Step 1: State the Hypotheses

We begin by stating the null and alternative hypotheses for this hypothesis test. The null hypothesis is that the proportions of men and women who believe the division of household duties is fair are equal, while the alternative hypothesis is that a greater proportion of men than women believe the division is fair. Formally, we write: \[ H_0: p_1 = p_2 \]\[ H_a: p_1 > p_2 \]where \( p_1 \) is the proportion of men and \( p_2 \) is the proportion of women.

Step 2: Calculate Sample Proportions

Next, we calculate the sample proportions for both men and women. The sample proportion for men \( p_1 \) and for women \( p_2 \) is computed as follows:\[ p_1 = \frac{990}{1500} = 0.66 \]\[ p_2 = \frac{970}{1600} = 0.60625 \]

Step 3: Compute Test Statistic

To compare the proportions, we use the formula for the test statistic for two proportions:\[ z = \frac{(p_1 - p_2)}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \]Where \( \hat{p} \) is the pooled sample proportion calculated by:\[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{990 + 970}{1500 + 1600} = 0.6324 \]Substitute \( \hat{p} \) into the z-formula:\[ z = \frac{0.66 - 0.60625}{\sqrt{0.6324 (1 - 0.6324) (\frac{1}{1500} + \frac{1}{1600})}} \]\[ z \approx 3.65 \]

Step 4: Determine Critical Value and Compare

We find the critical z-value for a one-tailed test at the 0.01 significance level. This critical value is approximately 2.33. Since the calculated z-value (3.65) is greater than 2.33, we reject the null hypothesis.

Step 5: Calculate p-value

Using standard normal distribution tables or a calculator, we find the p-value corresponding to a z-value of 3.65. This p-value is very small (much less than 0.01), which supports rejecting the null hypothesis.

Key Concepts

Proportion Comparison

In statistical analysis, **proportion comparison** is the process used to determine if there are significant differences between two groups based on categorical data. This comparison is particularly useful in situations where we want to understand if something, like a particular belief or trait, is more common in one group compared to another. In the context of our exercise, the goal is to compare the proportions of men and women who think that the division of household duties is fair.

To do this, we calculate the proportion of each group who holds the belief. It involves dividing the number of individuals with the trait by the total number of individuals in the group. For instance, in our exercise:

• The proportion of men: \( p_1 = \frac{990}{1500} = 0.66 \)
• The proportion of women: \( p_2 = \frac{970}{1600} = 0.60625 \)

Comparing these proportions helps us decide if more men than women believe in the fairness of household duties. We use hypothesis testing to statistically verify whether this difference in proportions is significant.

z-test

The **z-test** for comparing two proportions is a common statistical method used to see if there is a significant difference between two sample proportions. This test is suitable when both sample sizes are large, which is the case in our exercise with 1,500 men and 1,600 women.

We start by setting up our hypotheses:

• Null hypothesis \( (H_0) \): The proportions are equal, \( p_1 = p_2 \)
• Alternative hypothesis \( (H_a) \): The proportion of men is greater than that of women, \( p_1 > p_2 \)

The z-test helps compute a test statistic to measure the extent of the difference between the proportions. The formula is: \[z = \frac{(p_1 - p_2)}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \] where \( \hat{p} \) is the pooled sample proportion. Calculating the z-value provides a statistic that can be compared to a threshold (critical value) to make decisions about our hypotheses.

p-value

The **p-value** is a crucial concept in hypothesis testing. It tells us the probability of observing a test statistic as extreme as the one we calculated, assuming the null hypothesis is true. In simple terms, it's a measure of how much evidence we have against the null hypothesis.

In our exercise, the calculated z-value turned out to be 3.65. The p-value is then derived from this z-value. A very low p-value indicates strong evidence against the null hypothesis. Generally:

• If the p-value is less than the significance level (e.g., 0.01), we reject the null hypothesis.
• If it is greater, we fail to reject the null hypothesis.

In our test, the p-value was significantly lower than 0.01, confirming that the proportion of men who believe the division of household duties is fair is indeed greater than the proportion of women, leading us to reject the null hypothesis and accept the alternative hypothesis.