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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?

Short Answer

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

Step by step solution

01

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.
02

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:\[ p_1 = \frac{990}{1500} = 0.66 \]\[ p_2 = \frac{970}{1600} = 0.60625 \]
03

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:\[ z = \frac{0.66 - 0.60625}{\sqrt{0.6324 (1 - 0.6324) (\frac{1}{1500} + \frac{1}{1600})}} \]\[ z \approx 3.65 \]
04

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.
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

The research department at the home office of New Hampshire Insurance conducts ongoing research on the causes of automobile accidents, the characteristics of the drivers, and so on. A random sample of 400 policies written on single persons revealed 120 had at least one accident in the previous three-year period. Similarly, a sample of 600 policies written on married persons revealed that 150 had been in at least one accident. At the .05 significance level, is there a significant difference in the proportions of single and married persons having an accident during a three-year period? Determine the \(p\) -value.

Fairfield Homes is developing two parcels near Pigeon Fork, Tennessee. In order to test different advertising approaches, it uses different media to reach potential buyers. The mean annual family income for 15 people making inquiries at the first development is \(\$ 150,000\) with a standard deviation of \(\$ 40,000\). A corresponding sample of 25 people at the second development had a mean of \(\$ 180,000,\) with a standard deviation of \(\$ 30,000\). Assume the population standard deviations are the same. At the .05 significance level, can Fairfield conclude that the population means are different?

A recent study compared the time spent together by single- and dual-earner couples. According to the records kept by the wives during the study, the mean amount of time spent together watching television among the single-earner couples was 61 minutes per day, with a standard deviation of 15.5 minutes. For the dual-earner couples, the mean number of minutes spent watching television was 48.4 minutes, with a standard deviation of 18.1 minutes. At the .01 significance level, can we conclude that the single-earner couples on average spend more time watching television together? There were 15 single-earner and 12 dual-earner couples studied.

A nationwide sample of influential Republicans and Democrats was asked as a part of a comprehensive survey whether they favored lowering environmental standards so that high-sulfur coal could be burned in coal-fired power plants. The results were: $$ \begin{array}{|lcc|} \hline & \text { Republicans } & \text { Democrats } \\ \hline \text { Number sampled } & 1,000 & 800 \\ \text { Number in favor } & 200 & 168 \end{array} $$ At the .02 level of significance, can we conclude that there is a larger proportion of Democrats in favor of lowering the standards? Determine the \(p\) -value.

A sample of 40 observations is selected from one population with a population standard deviation of \(5 .\) The sample mean is \(102 .\) A sample of 50 observations is selected from a second population with a population standard deviation of \(6 .\) The sample mean is \(99 .\) Conduct the following test of hypothesis using the .04 significance level. $$ \begin{array}{l} H_{0}: \mu_{1}=\mu_{2} \\ H_{1}: \mu_{1} \neq \mu_{2} \end{array} $$ a. Is this a one-tailed or a two-tailed test? b. State the decision rule. c. Compute the value of the test statistic. d. What is your decision regarding \(H_{0}\) e. What is the \(p\) -value?

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