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The article "A 'White' Name Found to Help in Job Search" (Associated Press, January 15,2003 ) described in experiment to investigate if it helps to have a "whitesounding" 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 elicited 250 responses and resumes of the second type only 167 responses. Do these data support the theory that the proportion receiving positive responses is higher for those resumes with "whitesounding first" names?

Short Answer

Expert verified
The P-value obtained in this case is less than 0.05, which means the assumption that 'whitesounding' names receive more positive responses is statistically significant in the given data.

Step by step solution

01

Set Initial Hypothesis

Let's denote \(p_1\) as the proportion of positive responses for 'whitesounding' names and \(p_2\) as the proportion of positive responses for 'blacksounding' names. Our null hypothesis \(H_0\) is that the proportions are equal: \(p_1 = p_2\). Our alternate hypothesis \(H_a\) is that the proportion of positive responses for 'whitesounding' names is higher than for 'blacksounding' names: \(p_1 > p_2\).
02

Calculate Proportions

Next, we calculate the proportions of positive responses. For 'whitesounding' names (nw=2500) with 250 positive responses, it's \(p_1 = 250 / 2500 = 0.1\). For 'blacksounding' names (nb=2500) with 167 positive responses, it's \(p_2 = 167 / 2500 = 0.0668\).
03

Perform Z-test

To test our hypothesis, we use a one-sided Z-test. We first calculate the pooled sample proportion \(p = (x_1 + x_2) / (n_1 + n_2) = (250 + 167) / (2500 + 2500) = 0.0834\) , and then the standard error \(SE = \sqrt{ p *(1 - p )*(1/n_1 + 1/n_2)} = \sqrt{0.0834 * (1 - 0.0834) * (1/2500 + 1/2500)} = 0.0086\). The z-score is calculated as \(z = (p_1 – p_2) / SE = (0.1 - 0.0668) / 0.0086 = 3.86\).
04

Find P-value

After finding the z-score, we look up this value in the z-table (or use a statistical calculator or software) to obtain the P-value, which represents the probability that the difference in responses is due to chance. If the P-value is lower than the chosen significance level (for example, 0.05), we reject the null hypothesis.
05

Interpret Result

The z-value of 3.86 corresponds to a P-value less than 0.05, which means we reject the null hypothesis \(H_0\), supporting the theory that the proportion receiving positive responses is higher for resumes with 'whitesounding' names, based on the data given.

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Key Concepts

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

Proportion Hypothesis Test
When we talk about a proportion hypothesis test in statistics, we are referring to a method used to determine if there is a significant difference between the proportion of a specific outcome in two groups. These tests are essential in fields like marketing, medicine, and social sciences, where comparing the effectiveness or preferences between groups is common.

For instance, in the exercise provided, researchers are interested in whether the proportion of positive job responses is different between resumes with 'white-sounding' names and those with 'black-sounding' names. To test this, they use a hypothesis test for two proportions. Initially, they assume that there is no difference in the proportions (null hypothesis), and then they use data collected to decide whether this assumption seems plausible or should be rejected.

To improve understanding, it is useful to consider real-world examples where such tests apply. Whether it's assessing the rate of disease recovery with two different treatments or comparing click-through rates on two versions of a website (A/B testing), proportion hypothesis tests are widely applicable and fundamentally important for data-driven decision-making.
Z-Test
The z-test is a statistical test used to determine whether there are significant differences between sample observations and the population parameters. Its name comes from its reliance on the standard normal distribution, also known as the Z-distribution. In the context of a proportion hypothesis test, the z-test is particularly used when sample sizes are large enough, typically when both np and n(1-p) are greater than 5, where n is the sample size and p is the proportion.

In the provided exercise, the z-test compares the difference between two proportions (\( p_1 - p_2 \) ) to the expected standard error if the null hypothesis were true. To understand this better, it's helpful to know the z-score represents the number of standard errors a point is from the mean. If the z-score is large (either positive or negative), it suggests that the observed data are unlikely under the null hypothesis, indicating potential statistical significance.

An important tip for students tackling z-tests is to ensure they understand the prerequisites for utilizing it, such as knowing the samples are independent and that the distribution of sample estimates is approximately normally distributed, which generally holds true under the central limit theorem for large samples.
Statistical Significance
Understanding statistical significance is critical in hypothesis testing. It is a measure of the strength of the evidence against the null hypothesis. A result is considered statistically significant if it is unlikely to have occurred by chance alone, under the assumption that the null hypothesis is true. This is quantified using a P-value, which is the probability of obtaining test results at least as extreme as the observed results, assuming the null hypothesis is correct.

Statistical significance does not make a statement about the magnitude of an effect, only that the effect exists. For example, in our exercise, a P-value less than 0.05 led to the rejection of the null hypothesis, suggesting that resumes with 'white-sounding' names are likely to receive more positive job responses compared to those with 'black-sounding' names.

The threshold for determining statistical significance, often set at 0.05 or 5%, can be arbitrary and should be chosen with consideration of the context and consequences of Type I errors (false positives). It is also important to understand that statistical significance alone does not imply practical significance; the actual difference between groups must also be considered for making informed decisions.

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Most popular questions from this chapter

An Associated Press article (San Luis Obispo Telegram-Tribune, September 23,1995 ) examined the changing attitudes of Catholic priests. National surveys of priests aged 26 to 35 were conducted in 1985 and again in 1993\. The priests surveyed were asked whether they agreed with the following statement: Celibacy should be a matter of personal choice for priests. In \(1985,69 \%\) of those surveyed agreed; in \(1993,38 \%\) agreed. Suppose that the samples were randomly selected and that the sample sizes were both 200 . Is there evidence that the proportion of priests who agreed that celibacy should be a matter of personal choice declined from 1985 to 1993 ? Use \(\alpha=.05\).

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 that were born deaf and those that went deaf after learning to speak? Test the relevant hypotheses using a .01 significance level.

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 [2006]: 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. \(t=1.47\)

According to the Associated Press (San Luis Obispo Telegram-Tribune, June 23,1995 ), a study by Italian researchers indicated that low cholesterol and depression were linked. The researchers found that among 331 randomly selected patients hospitalized because they had attempted suicide, the mean cholesterol level was 198 . The mean cholesterol level of 331 randomly selected patients admitted to the hospital for other reasons was 217 . The sample standard deviations were not reported, but suppose that they were 20 for the group who had attempted suicide and 24 for the other group. Do these data provide sufficient evidence to conclude that the mean cholesterol level is lower for those who have attempted suicide? Test the relevant hypotheses using \(\alpha=.05\).

Do teenage boys worry more than teenage girls? This is one of the questions addressed by the authors of the article "The Relationship of Self-Esteem and Attributional Style to Young People's Worries" (Journal of Psychology [1987]: 207-215). A scale called the Worries Scale was administered to a group of teenagers, and the results are summarized in the accompanying table. \begin{tabular}{lccc} & & Sample & Sample \\ Gender & \(n\) & Mean Score & sd \\ \hline Girls & 108 & \(62.05\) & \(9.5\) \\ Boys & 78 & \(67.59\) & \(9.7\) \\ & & & \\ \hline \end{tabular} Is there sufficient evidence to conclude that teenage boys score higher on the Worries Scale than teenage girls? Use a significance level of \(\alpha=.05\).

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