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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 Jan. 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 suggest that resumes with "black" names are less likely to receive a response.

Step by step solution

01

Identify the Null and Alternative Hypotheses

The null hypothesis (\(H_0\)) is that there is no difference in the probabilities of getting a response for resumes with "white" and "black" names. The alternative hypothesis (\(H_1\)) is that the probability of getting a response for resumes with "black" names is less than those with "white" names.
02

Define the Proportions and Sample Sizes

Let \(p_w\) be the proportion of responses for "white" names and \(p_b\) be the proportion for "black" names. From the data, \(p_w = \frac{250}{2500}\) and \(p_b = \frac{167}{2500}\). Also, both sample sizes are 2500.
03

Calculate the Test Statistic

The test statistic for comparing two proportions is calculated as follows:\[z = \frac{(p_w - p_b)}{\sqrt{p(1-p)(\frac{1}{n_1}+\frac{1}{n_2})}}\,\] where \(p = \frac{x_1 + x_2}{n_1 + n_2}\,\) \(x_1 = 250\) and \(x_2 = 167\). Substitute the values to find \(z\).
04

Decision Rule Based on Significance Level

Assume a significance level of \(\alpha = 0.05\). Find the critical value for a one-tailed test at this level. If \(z\) is less than the negative of this critical value, reject \(H_0\).
05

Conclusion on Statistical Significance

After calculating, if the test statistic is less than the critical z-value, it provides evidence against the null hypothesis. This outcome would suggest that resumes with "black" names indeed have a lower probability of receiving a response than those with "white" names.

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Key Concepts

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

Statistical Significance
In any statistical test, determining whether results are statistically significant is a key aspect. When we say that results are statistically significant, it means that they are unlikely to have occurred by random chance alone. In the context of our exercise, we are examining the discrepancy in response rates to resumes with different name types.

Determining statistical significance involves calculating a test statistic and comparing it to a critical value. A result is deemed significant if the observed test statistic falls into a region that suggests the null hypothesis should be rejected. This region is often determined by a chosen significance level (usually \( \alpha = 0.05 \)).

  • If the test result is statistically significant, it suggests that the factors we are examining have a real, consistent effect.
  • If not significant, the observed differences could just be due to chance.
Statistical significance is important because it helps infer whether an observed effect is meaningful beyond mere chance.
Null Hypothesis
The null hypothesis, denoted as \( H_0 \), serves as the starting point in hypothesis testing. It represents a statement of no effect or no difference. In our example problem, the null hypothesis is that the probability of receiving a response is the same for resumes with "white" sounding names and "black" sounding names.

This hypothesis serves as a benchmark for comparison. When conducting a statistical test, if we obtain evidence against \( H_0 \), it suggests that an alternative reality, described by the alternative hypothesis, may better explain our observations.

  • Rejection or failure to reject the null hypothesis informs decision making.
  • Null hypotheses are assumed true unless the evidence strongly indicates otherwise.
Despite its central role, it's crucial to understand that suspecting the null hypothesis doesn't prove any specific alternative true. It simply indicates a need to explore other potential explanations.
Alternative Hypothesis
The alternative hypothesis, denoted \( H_1 \) or \( H_a \), is the counterpart to the null hypothesis. It reflects the statement we're looking to find evidence for in a hypothesis test. In our scenario, the alternative hypothesis suggests that resumes with "black" names are less likely to receive a response than those with "white" names.

This hypothesis is directional in this case because it specifically proposes that one group has a lower rate than the other. While the null hypothesis assumes no difference, the alternative hypothesis highlights what we aim to prove statistically.

  • If the test results lead to rejecting \( H_0 \), the alternative hypothesis gains support.
  • It is important to clearly define \( H_1 \) before performing the test to avoid biases.
The consequence is that accepting the alternative hypothesis provides justification for considering deeper examination or potential policy changes.
Test Statistic
A test statistic is a standard value derived from sample data and used to make an inference about a population parameter. For comparing two proportions, as in our exercise, a z-test is typically employed. The test statistic z reflects how far away our sample result is from what we would expect under the null hypothesis.

In our example, the proportion of responses for "white" names \( (p_w) \) and "black" names \( (p_b) \) are compared using this z-statistic. The formula for the z-test here is:

\[ z = \frac{(p_w - p_b)}{\sqrt{p(1-p)(\frac{1}{n_1}+\frac{1}{n_2})}} \]

where \( p \) is the pooled sample proportion.

This z-value assists in determining whether the observed difference is due to chance. It is compared to a critical value from the standard normal distribution:

  • If the z-value exceeds a certain threshold (critical value), \( H_0 \) is rejected.
  • The decision is usually based on pre-specified significance levels.
The test statistic is a fundamental building block for evaluating hypotheses and drawing conclusions in statistics.

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Most popular questions from this chapter

An article in the November 1983 Consumer Reports compared various types of batteries. The average lifetimes of Duracell Alkaline AA batteries and Eveready Energizer Alkaline AA batteries were given as 4.1 hours and \(4.5\) hours, respectively. Suppose these are the population average lifetimes. a. Let \(\bar{X}\) be the sample average lifetime of 100 Duracell batteries and \(\bar{Y}\) be the sample average lifetime of 100 Eveready batteries. What is the mean value of \(\bar{X}-\bar{Y}\) (i.e., where is the distribution of \(\bar{X}-\bar{Y}\) centered)? How does your answer depend on the specified sample sizes? b. Suppose the population standard deviations of lifetime are \(1.8\) hours for Duracell batteries and \(2.0\) hours for Eveready batteries. With the sample sizes given in part (a), what is the variance of the statistic \(\bar{X}-\bar{Y}\), and what is its standard deviation? c. For the sample sizes given in part (a), draw a picture of the approximate distribution curve of \(\bar{X}-\bar{Y}\) (include a measurement scale on the horizontal axis). Would the shape of the curve necessarily be the same for sample sizes of 10 batteries of each type? Explain.

Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of toxaphene exposure on animals, groups of rats were given toxaphene in their diet. The article "Reproduction Study of Toxaphene in the Rat"" \((J\). of Emiron. Sei. Health, 1988: 101-126) reports weight gains (in grams) for rats given a low dose (4 ppm) and for control rats whose diet did not include the insecticide. The sample standard deviation for 23 female control rats was \(32 \mathrm{~g}\) and for 20 female low-dose rats was \(54 \mathrm{~g}\). Does this data suggest that there is more variability in low-dose weight gains than in control weight gains? Assuming normality, carry out a test of hypotheses at significance level .05.

Determine the number of degrees of freedom for the twosample \(t\) test or CI in each of the following situations: a. \(m=10, n=10, s_{1}=5.0, s_{2}=6.0\) b. \(m=10, n=15, s_{1}=5.0, s_{2}=6.0\) c. \(m=10, n=15, s_{1}=2.0, s_{2}=6.0\) d. \(m=12, n=24, s_{1}=5.0, s_{2}=6.0\)

A mechanical engineer wishes to compare strength properties of steel beams with similar beams made with a particular alloy. The same number of beams, \(n\), of each type will be tested. Each beam will be set in a horizontal position with a support on each end, a force of \(2500 \mathrm{lb}\) will be applied at the center, and the deflection will be measured. From past experience with such beams, the engineer is willing to assume that the true standard deviation of deflection for both types of beam is .05 in. Because the alloy is more expensive, the engineer wishes to test at level .01 whether it has smaller average deflection than the steel beam. What value of \(n\) is appropriate if the desired type II error probability is \(.05\) when the difference in true average deflection favors the alloy by \(.04\) in.?

Damage to grapes from bird predation is a serious problem for grape growers. The article "Experimental Method to Investigate and Monitor Bird Behavior and Damage to Vineyards" (Amer. \(J\). of Enology and Viticulture, 2004: 288-291) reported on an experiment involving a bird-feeder table, time-lapse video, and artificial foods. Information was collected for two different bird species at both the experimental location and at a natural vineyard setting. Consider the following data on time \((\mathrm{sec})\) spent on a single visit to the location. $$ \begin{array}{llrrc} \text { Species } & \text { Location } & n & \bar{x} & \text { SE mean } \\ \hline \text { Blackbirds } & \text { Exptl } & 65 & 13.4 & 2.05 \\ \text { Blackbirds } & \text { Natural } & 50 & 9.7 & 1.76 \\ \text { Silvereyes } & \text { Exptl } & 34 & 49.4 & 4.78 \\ \text { Silvereyes } & \text { Natural } & 46 & 38.4 & 5.06 \\ \hline \end{array} $$ a. Calculate an upper confidence bound for the true average time that blackbirds spend on a single visit at the experimental location. b. Does it appear that true average time spent by blackbirds at the experimental location exceeds the true average time birds of this type spend at the natural location? Carry out a test of appropriate hypotheses. c. Estimate the difference between the true average time blackbirds spend at the natural location and true average time that silvereyes spend at the natural location, and do so in a way that conveys information about reliability and precision. [Note: The sample medians reported in the article all seemed significantly smaller than the means, suggesting substantial population distribution skewness. The authors actually used the distribution-free test procedure presented in Section 2 of Chapter 15.]

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