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Chapter 11: Problem 65

The article "A 'White' Name Found to Help in Job Search" (Associated Press, January 15,2003 ) described an experiment to investigate if it helps to have a "white-sounding" first name when looking for a job. Researchers sent 5000 resumes in response to ads that appeared in the Boston Globe and Chicago Tribune. The resumes were identical except that 2500 of them had "white- sounding" first names, such as Brett and Emily, whereas the other 2500 had "black-sounding" names such as Tamika and Rasheed. Resumes of the first type clicited 250 responses and resumes of the second type only 167 responses. Do these data support the theory that the proportion receiving responses is higher for those resumes with "white-sounding first" names?

Short Answer

Expert verified
Yes, the data support the theory that the proportion receiving responses is higher for those resumes with 'white-sounding' first names.

Step by step solution

01

Identify the hypothesis

Firstly, the hypotheses for the problem need to be defined. The null hypothesis (\(H_0\)) is that the proportions of responses for the white-sounding names (\(p1\)) and the black-sounding names (\(p2\)) are equal. The alternative hypothesis (\(H_1\)) is that the proportion of responses for white-sounding names is higher, i.e. \(p1 > p2\). So, \(H_0: p1 = p2\) and \(H_1: p1 > p2\).
02

Calculate the sample proportions and the difference between them

Next, calculate the sample proportions. For the white-sounding names, the sample proportion is \(p1 = 250/2500 = 0.10\). For the black-sounding names, the sample proportion is \(p2 = 167/2500 = 0.0668\). The difference between the sample proportions is \(p1 - p2 = 0.10 - 0.0668 = 0.0332\).
03

Conduct the hypothesis test

To test the hypothesis, calculate the standard error for the difference in proportions under the null hypothesis. The combined sample proportion (\(p\)) is \((r1 + r2) / (n1 + n2) = (250 + 167) / (2500 + 2500) = 0.0834\), where \(r1\) is the number of responses for white-sounding names, \(r2\) is the number of responses for black-sounding names, \(n1\) is the total number of white-sounding name resumes and \(n2\) is the total number of black-sounding name resumes. The standard error (\(SE\)) is \(\sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] } = \sqrt{ 0.0834 * (1 - 0.0834) * [ (1/2500) + (1/2500) ] } = 0.0108\). So the test statistic (\(z\)) is \((p1 - p2) / SE = 0.0332 / 0.0108 = 3.07\). Using a standard normal table or calculator, the P-value is found to be about 0.001 (for a one-sided test). This P-value is less than a typical significance level (e.g., 0.05), thus we reject the null hypothesis and conclude that the proportion receiving responses is higher for those resumes with 'white-sounding' first names.

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Key Concepts

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

Null and Alternative Hypotheses
Understanding null and alternative hypotheses is foundational to hypothesis testing in statistics. The null hypothesis (\( H_0 \) represents a statement of no effect or no difference, and it serves as a baseline or a claim to be tested against. In our exercise, the null hypothesis is that the proportion of job responses is the same between resumes with 'white-sounding' and 'black-sounding' names (\( p1 = p2 \)).

The alternative hypothesis (\( H_1 \) posits the opposite 鈥� it's what you expect to find if the null hypothesis is not true. In this scenario, the alternative hypothesis is that the proportion of responses is higher for 'white-sounding' names than for 'black-sounding' names (\( p1 > p2 \)).

The key to a well-performed hypothesis test is correctly identifying these hypotheses because they set the stage for the entire analysis. Their formulation should be clear, concise, and based on the research question or theory being tested.
Sample Proportions
In the context of hypothesis testing, sample proportions represent the ratio of individuals within a sample meeting a certain criterion to the total number in the sample. In this exercise, sample proportions indicate the response rates to the resumes: \( p1 = 250/2500 \) for 'white-sounding' names and \( p2 = 167/2500 \) for 'black-sounding' names.

Calculating these sample proportions is vital since they act as estimates of the true populations' proportions. When we compare \( p1 - p2 \) and find a difference (\( 0.10 - 0.0668 = 0.0332 \)), it suggests a disparity that may or may not be due to chance. Therefore, sample proportions are essential in gauging the direction and magnitude of any apparent effect in experimental or observational studies.
Z-test
A z-test is a statistical test used to determine whether two population means are different when the variances are known and the sample size is large. For our resumes example, we're dealing with proportions rather than means, but the z-test concept still applies because we're using the sample information to infer about the population.

To conduct the z-test, we use the sample information and the standard error to calculate the z-statistic. In this case, \( z = (p1 - p2) / SE = 3.07 \). Once calculated, this z-value is compared to values in a standard normal distribution to find the P-value, which helps decide whether or not to reject the null hypothesis. A P-value lower than the chosen significance level (e.g., 0.05) suggests that there is sufficient evidence to reject the null hypothesis.
Standard Error
The standard error is a measure of the variability or precision of a sample statistic. For hypothesis testing involving proportions, the standard error reflects how much we expect sample proportions to vary due to random sampling variability.

In this resume response study, the standard error helps us understand the difference in response rates due to random chance. It鈥檚 calculated using the combined sample proportion and the sizes of both samples, as shown by the formula \( SE = \sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] } = 0.0108 \). This standard error is crucial because it's used in the denominator of our test statistic (z-value), determining the level of confidence we have in our results. The smaller the standard error, the more confident we can be in the stability of our sample proportion estimate.

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Most popular questions from this chapter

In a study of malpractice claims where a settlement had been reached, two random samples were selected: a random sample of 515 closed malpractice claims that were found not to involve medical errors and a random sample of 889 claims that were found to involve errors (New England journal of Medicine \(12006 \mathrm{l}\) : \(2024-2033) .\) The following statement appeared in the referenced paper: "When claims not involving errors were compensated, payments were significantly lower on average than were payments for claims involving errors \((\$ 313,205\) vs. \(\$ 521,560, P=0.004) . "\) a. What hypotheses must the researchers have tested in order to reach the stated conclusion? b. Which of the following could have been the value of the test statistic for the hypothesis test? Explain your reasoning. i. \(t=5.00\) ii. \(t=2.65\) iii. \(t=2.33\) iv. \(t=1.47\)

The study described in the paper "Marketing Actions Can Modulate Neural Representation of Experienced Pleasantness" (Proceedings of the National Academy of Science [2008]: \(1050-1054\) ) investigated whether price affects people's judgment. Twenty people each tasted six cabernet sauvignon wines and rated how they liked them on a scale of 1 to 6 . Prior to tasting each wine, participants were told the price of the wine. Of the six wines rasted, two were actually the same wine, but for one tasting the participant was told that the wine cost \(\$ 10\) per bottle and for the other tasting the participant was told that the wine cost \(\$ 90\) per bottle. The participants were randomly assigned either to taste the \(\$ 90\) wine first and the \(\$ 10\) wine second, or the \(\$ 10\) wine first and the \(\$ 90\) wine second. Differences (computed by subtracting the rating for the tasting in which the participant thought the wine cost \(\$ 10\) from the rating for the tasting in which the participant thought the wine cost \(\$ 90\) were compured. The differences that follow are consistent with summary quantities given in the paper. Difference \((\$ 90-\$ 10)\) \(\begin{array}{llllllllllllllllllll}2 & 4 & 1 & 2 & 1 & 0 & 0 & 3 & 0 & 2 & 1 & 3 & 3 & 1 & 4 & 1 & 2 & 2 & 1 & -1\end{array}\) Carry out a hypothesis test to determine if the mean rating assigned to the wine when the cost is described as \(\$ 90\) is greater than the mean rating assigned to the wine when the cost is described as \(\$ 10 .\) Use \(\alpha=.01\).

Common Sense Media surveyed 1000 teens and 1000 parents of teens to learn about how teens are using social networking sites such as Facebook and MySpace ("Teens Show. Tell Too Much Online." San Francisco Chronicle, August 10,2009 ). The two samples were independently selected and were chosen in a way that makes it reasonable to regard them as representative of American teens and parents of American teens. a. When asked if they check their online social networking sites more than 10 times a day, 220 of the teens surveyed said yes. When parents of teens were asked if their teen checked his or her site more than 10 times a day, 40 said yes. Use a significance level of \(.01\) to carry out a hypothesis test to determine if there is convincing evidence that the proportion of all parents who think their teen checks a social networking site more than 10 times a day is less than the proportion of all teens who report that they check more than 10 times a day. b. The article also reported that 390 of the teens surveyed said they had posted something on their networking site that they later regretted. Would you use the two-sample \(z\) test of this section to test the hypothesis that more than one-third of all teens have posted something on a social networking site that they later regretted? Explain why or why not. c. Using an appropriate test procedure, carry out a test of the hypothesis given in Part (b). Use \(\alpha=.05\) for this test.

An electronic implant that stimulates the auditory nerve has been used to restore partial hearing to a number of deaf people. In a study of implant acceptability (Los Angeles Times. January 29,1985 ), 250 adults born deaf and 250 adults who went deaf after learning to speak were followed for a period of time after receiving an implant. Of those deaf from birth, 75 had removed the implant, whereas only 25 of those who went deaf after learning to speak had done so. Does this suggest that the true proportion who remove the implants differs for those who were born deaf and those who went deaf after learning to speak? 'Test the relevant hypotheses using a .01 significance level.

A hotel chain is interested in evaluating reservation processes. Guests can reserve a room by using either a telephone system or an online system that is accessed through the hotel's web site. Independent random samples of 80 guests who reserved a room by phone and 60 guests who reserved a room online were selected. Of those who reserved by phone, 57 reported that they were satisfied with the reservation process. Of those who reserved online, 50 reported that they were satisfied. Based on these data, is it reasonable to conclude that the proportion who are satisfied is higher for those who reserve a room online? Test the appropriate hypotheses using \(\alpha=.05 .\)
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