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Multiple choice: Select the best answer for Exercises 29 to 32. A sample survey interviews SRSs of 500 female college students and 550 male college students. Each student is asked whether he or she worked for pay last summer. In all, 410 of the women and 484 of the men say 鈥淵es.鈥 Exercises 29 to 31 are based on this survey. The 95\(\%\) confidence interval for the difference \(p_{M}-p_{F}\) in the proportions of college men and women who worked last summer is about (a) \(0.06 \pm 0.00095\) (b) \(0.06 \pm 0.043\) (c) \(0.06 \pm 0.036\) (d) \(-0.06 \pm 0.043\) (e) \(-0.06 \pm 0.036\)

Short Answer

Expert verified
The answer is (b) \(0.06 \pm 0.043\).

Step by step solution

01

Define Variables and Find the Proportions

Let \( p_M \) be the proportion of college men who worked last summer, and \( p_F \) be the proportion of college women who worked last summer. We have the data: 484 men out of 550 worked and 410 women out of 500 worked.Calculate the sample proportions: \[ \hat{p}_M = \frac{484}{550} \approx 0.88 \]\[ \hat{p}_F = \frac{410}{500} \approx 0.82 \]
02

Calculate the Difference in Sample Proportions

The difference in sample proportions is given by:\[ \hat{p}_M - \hat{p}_F = 0.88 - 0.82 = 0.06 \]
03

Calculate the Standard Error of the Difference

The standard error for the difference in proportions is calculated using:\[ SE(\hat{p}_M - \hat{p}_F) = \sqrt{\frac{\hat{p}_M (1 - \hat{p}_M)}{n_M} + \frac{\hat{p}_F (1 - \hat{p}_F)}{n_F}} \]Substitute the values:\[ SE(\hat{p}_M - \hat{p}_F) = \sqrt{\frac{0.88 \times 0.12}{550} + \frac{0.82 \times 0.18}{500}} \approx 0.029 \]
04

Construct the 95% Confidence Interval

To construct a 95% confidence interval, use the estimated difference and the standard error:\[ CI = (\hat{p}_M - \hat{p}_F) \pm z^* \times SE(\hat{p}_M - \hat{p}_F) \] where \( z^* \) is the z-score for a 95% confidence interval, which is approximately 1.96.\[ CI = 0.06 \pm 1.96 \times 0.029 \approx 0.06 \pm 0.057 \]
05

Select the Closest Answer

The calculated confidence interval is approximately \(0.06 \pm 0.057\), which is closest to option (b) \(0.06 \pm 0.043\).

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Key Concepts

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

Sample Proportions
Sample proportions are essentially fractions or percentages that provide insights into specific characteristics within a sample. When dealing with survey data, like in the original exercise, calculating the sample proportions helps us understand the behavior or attribute of interest across different groups. Consider a simple survey that seeks to determine how many people worked last summer. If we have 484 men out of 550 who worked for pay, the sample proportion of men is calculated as:\[ \hat{p}_M = \frac{484}{550} \approx 0.88 \]Similarly, for women, if 410 out of 500 worked last summer, the proportion is:\[ \hat{p}_F = \frac{410}{500} \approx 0.82 \]These calculations provide a snapshot of the population's behavior, based on the collected sample data. The primary purpose of using sample proportions is to make generalizations about the whole population with a limited dataset.
Standard Error
Standard error (SE) is a statistical term that reflects the variability or spread of a sampling distribution. It tells us how much the sample statistic, such as a sample proportion, is expected to vary from the actual population parameter.When calculating the difference in proportions of two groups, it's important to estimate the SE to understand how precise that estimate is. The formula for the standard error of the difference in sample proportions is:\[ SE(\hat{p}_M - \hat{p}_F) = \sqrt{\frac{\hat{p}_M (1 - \hat{p}_M)}{n_M} + \frac{\hat{p}_F (1 - \hat{p}_F)}{n_F}} \]Using the sample proportions \(\hat{p}_M\) for men and \(\hat{p}_F\) for women, along with their sample sizes \(n_M\) and \(n_F\), we can calculate the SE as a part of constructing confidence intervals. It helps in measuring the sampling variability, thereby giving us an indication of how much the sample proportions' difference might fluctuate.
Z-Score
The z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. In simple terms, it tells us how many standard deviations away a particular score is from the mean.In the context of constructing confidence intervals, the z-score determines how far we need to stretch our interval from the mean difference based on the standard error. For a typical 95% confidence interval, the z-score is approximately 1.96.Knowing that the z-score at 95% confidence level comes from standard normal distribution assumptions, we use this multiplier for the standard error to provide a range within which we expect the true population difference to fall. In the example exercise, a 95% confidence interval is built by:\[ CI = (\hat{p}_M - \hat{p}_F) \pm z^* \times SE(\hat{p}_M - \hat{p}_F) \]The z-score helps ensure that the confidence interval accurately reflects the data's variability and offers us a reliable estimation.
Statistical Inference
Statistical inference is the process of using data from a sample to draw conclusions about the population from which the sample was drawn. It's a fundamental aspect of statistics that allows us to make educated guesses about population parameters based on sample data. One of the most common forms of statistical inference is the confidence interval. In our scenario with college students, we use sample proportions from male and female students to infer the true proportion of all college males and females who may have worked last summer. By determining the difference of the sample proportions and subsequently constructing a confidence interval, we are making an inference about the population's characteristics. This includes estimating where the true proportion difference could lie within a certain level of certainty (95% in this case). Ultimately, statistical inference provides a structured, scientific basis for decision-making based on incomplete data.

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Most popular questions from this chapter

How tall? 6.2 ) The heights of young men follow a Normal distribution with mean 69.3 inches and standard deviation 2.8 inches. The heights of young women follow a Normal distribution with mean 64.5 inches and standard deviation 2.5 inches. (a) Let \(M=\) the height of a randomly selected young man and \(W=\) the height of a randomly selected young woman. Describe the shape, center, and spread of the distribution of \(M-W\) (b) Find the probability that a randomly selected young man is at least 2 inches taller than a randomly selected young woman. Show your work.

Down the toilet A company that makes hotel toilets claims that its new pressure-assisted toilet reduces the average amount of water used by more than 0.5 gallon per flush when compared to its current model. To test this claim, the company randomly selects 30 toilets of each type and measures the amount of water that is used when each toilet is flushed once. For the current-model toilets, the mean amount of water used is 1.64 gal with a standard deviation of 0.29 gal. For the new toilets, the mean amount of water used is 1.09 gal with a standard deviation of 0.18 gal. (a) Carry out an appropriate significance test. What conclusion would you draw? (Note that the null hypothesis is not \(H_{0} : \mu_{1}-\mu_{2}=0\) ) (b) Based on your conclusion in part (a), could you have made a Type I error or a Type II error? Justify your answer.

Young adults living at home A surprising number of young adults (ages 19 to 25) still live in their parents鈥 homes. A random sample by the National Institutes of Health included 2253 men and 2629 women in this age group.\(^{11}\) The survey found that 986 of the men and 923 of the women lived with their parents. (a) Construct and interpret a 99% confidence interval for the difference in population proportions (men minus women). (b) Does your interval from part (a) give convincing evidence of a difference between the population proportions? Explain.

Listening to rap Is rap music more popular among young blacks than among young whites? A sample survey compared 634 randomly chosen blacks aged 15 to 25 with 567 randomly selected whites in the same age group. It found that 368 of the blacks and 130 of the whites listened to rap music every day.\(^{10}\) Construct and interpret a 95% confidence interval for the difference between the proportions of black and white young people who listen to rap every day.

What鈥檚 wrong? A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state鈥檚 driver鈥檚 license exam. An incoming class of 100 students is randomly assigned to two groups, each of size 50. One group is taught by Instructor A; the other is taught by Instructor B. At the end of the course, 30 of Instructor A鈥檚 students and 22 of Instructor B鈥檚 students pass the state exam. Do these results give convincing evidence that Instructor A is more effective? Min Jae carried out the significance test shown below to answer this question. Unfortunately, he made some mistakes along the way. Identify as many mistakes as you can, and tell how to correct each one. State: I want to perform a test of $$H_{0} : p_{1}-p_{2}=0$$ $$H_{a} : p_{1}-p_{2}>0$$ where \(p_{1}=\) the proportion of Instructor A's students that passed the state exam and \(p_{2}=\) the proportion of Instructor B's students that passed the state exam. Since no significance level was stated, I'll use \(\sigma=0.05\) Plan: If conditions are met, I'll do a two-sample \(z\) test for comparing two proportions. \(\bullet\) Random The data came from two random samples of 50 students. \(\bullet\) Normal The counts of successes and failures in the two groups - \(30,20,22\) , and \(28-\) are all at least \(10 .\) \(\bullet\) Independent There are at least 1000 students who take this driving school's class. Do: From the data, \(\hat{p}_{1}=\frac{20}{50}=0.40\) and \(\hat{p}_{2}=\frac{30}{50}=0.60 .\) So the pooled proportion of successes is $$\hat{p}_{C}=\frac{22+30}{50+50}=0.52$$ \(\bullet\) Test statistic $$z=\frac{(0.40-0.60)-0}{\sqrt{\frac{0.52(0.48)}{100}+\frac{0.52(0.48)}{100}}}=-2.83$$ Conclude: The P-value, \(0.9977,\) is greater than \(\alpha=\) \(0.05,\) so we fail to reject the null hypothesis. There is not convincing evidence that Instructor A's pass rate is higher than Instructor B's.

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