/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 109 Researchers sent 5000 resumes in... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Researchers sent 5000 resumes in response to job 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. The resumes of the first type elicited 250 responses and the resumes of the second type only 167 responses (these numbers are very consistent with information that appeared in a January 15 , 2003 , report by the Associated Press). Does this data strongly suggest that a resume with a "black" name is less likely to result in a response than is a resume with a "white" name?

Short Answer

Expert verified
The data suggests resumes with "black" names are less likely to receive responses, supporting implicit bias evidenced by the response discrepancy.

Step by step solution

01

Understand the Problem

We need to determine if having a "white sounding" name leads to more job ad responses than having a "black sounding" name. To do this, we'll compare the proportions of responses received for each group.
02

Identify Data Points and Hypotheses

From the given problem, the total number of resumes is 5000. "White" names received 250 responses out of 2500 resumes, and "black" names received 167 responses out of 2500 resumes. We can compare these using a hypothesis test. - Null Hypothesis ( H_0 ): There is no difference in response rates between the two groups. - Alternative Hypothesis ( H_a ): The response rate for resumes with "black" names is less than that for "white" names.
03

Calculate Proportions

Calculate the sample proportions for each group:\[ p_1 = \frac{250}{2500} = 0.10 \]\[ p_2 = \frac{167}{2500} = 0.0668 \]where \( p_1 \) is the proportion of responses for "white" names and \( p_2 \) is the proportion for "black" names.
04

Calculate the Test Statistic

For comparison, we use a Z-test for proportions. The formula for the test statistic is:\[ Z = \frac{(p_1 - p_2)}{\sqrt{p(1-p)\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \]where:\[ p = \frac{x_1 + x_2}{n_1 + n_2} \] is the pooled sample proportion and \( n_1 = n_2 = 2500 \) are the sample sizes for the two groups.
05

Calculate the Pooled Sample Proportion

First, calculate the pooled sample proportion:\[ p = \frac{250 + 167}{2500 + 2500} = \frac{417}{5000} = 0.0834 \]
06

Compute the Standard Error

Find the standard error using the pooled proportion:\[ SE = \sqrt{0.0834 \times (1 - 0.0834) \times \left(\frac{1}{2500} + \frac{1}{2500}\right)} = \sqrt{0.0834 \times 0.9166 \times \frac{2}{2500}} \]
07

Calculate the Z-Statistic

Now plug in the values to find the Z-statistic:\[ Z = \frac{(0.10 - 0.0668)}{SE} \]Compute the value after calculating the standard error in the previous step.
08

Make a Decision using Z-Statistic

Compare the computed Z-statistic against standard critical values for significance level \( \alpha \), typically 0.05. A Z-score below the lower critical value suggests significant difference.
09

Conclusion

Based on the calculated Z-statistic and the critical value comparison, conclude whether or not to reject the null hypothesis. If the null hypothesis is rejected, this suggests a statistically significant difference in the response rates.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Proportions
Proportions are essential in statistics when comparing different groups. Here, we need to compare the response rates to resumes based on names. The proportion represents the ratio of responses to the total number of resumes in each group.
To determine the proportion of responses for each group, you divide the number of responses by the total number of resumes sent. For example, the "white sounding" names had a proportion: \( p_1 = \frac{250}{2500} = 0.10 \).
Similarly, for "black sounding" names, the proportion is: \( p_2 = \frac{167}{2500} = 0.0668 \). These proportions enable us to set up a hypothesis test to analyze the likelihood of differences in response rates.
Null Hypothesis
The null hypothesis is a fundamental concept in hypothesis testing. It represents our starting assumption or status quo that there is no effect or difference.
In this exercise, the null hypothesis (\( H_0 \)) assumes that there is no difference in response rates between resumes with "white" and "black" names. This means we believe, initially, that both groups have the same chance of receiving a response.
By default, we assume the null hypothesis is true unless statistical evidence suggests otherwise. The purpose is to find if the observed effect (different response rates) is due to chance or a genuine difference.
Alternative Hypothesis
The alternative hypothesis (\( H_a \)) stands in contrast to the null hypothesis. It expresses what we aim to support with our data.
Here, the alternative hypothesis is that resumes with "black sounding" names receive fewer responses than those with "white sounding" names. Formally, it is expressed as: \( p_2 < p_1 \).
This hypothesis suggests a directional effect—that response rates differ specifically in the way proposed.
Our goal in hypothesis testing is to assess whether there is sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.
Z-test
A Z-test is a statistical procedure used to determine whether two population proportions are significantly different. It's especially useful when sample sizes are large, as in this example where each group sent out 2500 resumes.
The test statistic, \( Z \), is calculated to see if the difference in proportions is due to random sampling variability or a significant effect. The formula for the Z-statistic is:
\[ Z = \frac{(p_1 - p_2)}{\sqrt{p(1-p)\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \]
where \( p \) is the pooled sample proportion, taking into account the total responses received from both groups.
This Z-test will lead us to a decision about the null hypothesis, whether to reject it or not, based on a predetermined significance level such as \( \alpha = 0.05 \). A significant Z-score would suggest the proportions differ substantially.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Obtain or compute the following quantities: a. \(F_{005,5,8}\) b. \(F_{00,8,5}\) c. \(F_{.95,5,8}\) d. \(F_{95,8,5}\) e. The 99 th percentile of the \(F\) distribution with \(v_{1}=10, v_{2}=12\) f. The Ist percentile of the \(F\) distribution with \(v_{1}=10, v_{2}=12\) g. \(P(F \leq 6.16)\) for \(v_{1}=6, v_{2}=4\) h. \(P(.177 \leq F \leq 4.74)\) for \(v_{1}=10, v_{2}=5\)

Persons having Raynaud's syndrome are apt to suffer a sudden impairment of blood circulation in fingers and toes. In an experiment to study the extent of this impairment, each subject immersed a forefinger in water and the resulting heat output \(\left(\mathrm{cal} / \mathrm{cm}^{2} / \mathrm{min}\right)\) was measured. For \(m=10\) subjects with the syndrome, the average heat output was \(\bar{x}=.64\), and for \(n=10\) nonsufferers, the average output was \(2.05\). Let \(\mu_{1}\) and \(\mu_{2}\) denote the true average heat outputs for the two types of subjects. Assume that the two distributions of heat output are normal with \(\sigma_{1}=.2\) and \(\sigma_{2}=.4\). a. Consider testing \(H_{0}: \mu_{1}-\mu_{2}=-1.0\) versus \(H_{\mathrm{a}}: \mu_{1}-\mu_{2}<-1.0\) at level .01. Describe in words what \(H_{a}\) says, and then carry out the test. b. Compute the \(P\)-value for the value of \(Z\) obtained in part (a). c. What is the probability of a type II error when the actual difference between \(\mu_{1}\) and \(\mu_{2}\) is \(\mu_{1}-\) \(\mu_{2}=-1.2 ?\) d. Assuming that \(m=n\), what sample sizes are required to ensure that \(\beta=.1\) when \(\mu_{1}-\) \(\mu_{2}=-1.2\) ?

Is someone who switches brands because of a financial inducement less likely to remain loyal than someone who switches without inducement? Let \(p_{1}\) and \(p_{2}\) denote the true proportions of switchers to a certain brand with and without inducement, respectively, who subsequently make a repeat purchase. Test \(H_{0}: p_{1}-p_{2}=0\) versus \(H_{a}: p_{1}-p_{2}<0\) using \(\alpha=.01\) and the following data: $$ \begin{array}{ll} m=200 & \text { number of successes }=30 \\ n=600 & \text { number of successes }=180 \end{array} $$ (Similar data is given in "Impact of Deals and Deal Retraction on Brand Switching," J. Marketing. \(1980: 62-70 .\) )

Sometimes experiments involving success or failure responses are run in a paired or before/ after manner. Suppose that before a major policy speech by a political candidate, \(n\) individuals are selected and asked whether \((S)\) or not \((F)\) they favor the candidate. Then after the speech the same \(n\) people are asked the same question. The responses can be entered in a table as follows: where \(X_{1}+X_{2}+X_{3}+X_{4}=n\). Let \(p_{1}, p_{2}, p_{3}\), and \(P_{4}\) denote the four cell probabilities, so that \(P_{1}=P(S\) before and \(S\) after \()\), and so on. We wish to test the hypothesis that the true proportion of supporters \((S)\) after the speech has not increased against the alternative that it has increased. a. State the two hypotheses of interest in terms of \(p_{1}, p_{2}, p_{3}\), and \(p_{4}\). b. Construct an estimator for the after/before difference in success probabilities. c. When \(n\) is large, it can be shown that the rv \(\left(X_{i}-X_{j}\right) / n\) has approximately a normal distribution with variance \(\left[p_{i}+p_{j}-\left(p_{i}-p_{j}\right)^{2}\right] / n\). Use this to construct a test statistic with approximately a standard normal distribution when \(H_{0}\) is true (the result is called MeNemar's test). d. If \(x_{1}=350, x_{2}=150, x_{3}=200\), and \(x_{4}=300\), what do you conclude?

The article "The Accuracy of Stated Energy Contents of Reduced-Energy, Commercially Prepared Foods" ( \(J\). Am. Diet. Assoc., 2010: 116-123) presented the accompanying data on vendor-stated gross energy and measured value (both in kcal) for 10 different supermarket convenience meals): $$ \begin{array}{lcccccccccc} \text { Meal } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \text { Stated } & 180 & 220 & 190 & 230 & 200 & 370 & 250 & 240 & 80 & 180 \\\ \text { Measured } & 212 & 319 & 231 & 306 & 211 & 431 & 288 & 265 & 145 & 228 \end{array} $$ Obtain a \(95 \%\) confidence interval for the difference of population means. By roughly what percentage are the actual calories higher than the stated value? Note that the article calls this a convenience sample and suggests that therefore it should have limited value for inference. However, even if the ten meals were a random sample from their local store, there could still be a problem in drawing conclusions about a purchase at your store.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.