Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 "whitesounding" first name when looking for a job. Researchers sent resumes in response to 5,000 ads that appeared in the Boston Globe and Chicago Tribune. The resumes were identical except that 2,500 of them used "white-sounding" first names, such as Brett and Emily, whereas the other 2,500 used "black- sounding" names such as Tamika and Rasheed. The 5,000 job ads were assigned at random to either the white-sounding name group or the blacksounding name group. Resumes with white-sounding names received 250 responses while resumes with black sounding names received only 167 responses. Do these data support the claim that the proportion receiving a response is significantly higher for resumes with "white-sounding" first names? (Hint: See Example 14.2 )

Short Answer

Expert verified
The answer depends on the calculated z value. If it's greater than 1.645, the conclusion will be that the response rate for resumes with 'white-sounding' names is significantly higher than for resumes with 'black-sounding' names.

Step by step solution

01

Formulate Hypotheses

The Null Hypothesis: The responses for both groups are equal or (\( p_1 = p_2 \)). The Alternative Hypothesis: The response rate for resumes with white sounding names is significantly higher than for those with black sounding names (\( p_1 > p_2 \)). \( p_1 \) denotes the proportion for white sounding names, while \( p_2 \) denotes the proportion for black sounding names.
02

Calculate Proportions and Difference

Let's calculate the proportions for both groups. For white sounding names, 250 responses out of 2500 resumes gave a response rate of \( p_1 = 250/2500 = 0.1 \). For black sounding names, 167 responses out of 2500 resumes gave a response rate of \( p_2 = 167/2500 = 0.0668 \). The difference between both proportions is \( p_1 - p_2 = 0.1 - 0.0668 = 0.0332 \).
03

Calculate Test Statistic

The z-test statistic formula is: \( z = (p_1 - p_2)/ \sqrt{p(1-p)(1/n_1 + 1/n_2)} \), where \( p \) is the pooled proportion calculated as \( (x_1+x_2)/(n_1 + n_2) \). Here, \( x_1 = 250, x_2 = 167, n_1 = n_2 = 2500 \). Calculate p and substitute all values to get the z value.
04

Compare Test Statistic to Critical Value

For a significance level of 0.05 (typical), the critical z value (one-tailed) is about 1.645. If the calculated z value is greater than this critical value, we reject the null hypothesis.
05

Conclude

If the z value is greater than the critical value, we reject our null hypothesis and conclude that the response rate for resumes with 'white-sounding' names is significantly higher. If not, we do not have enough evidence to support this claim.

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 Vaia!

Key Concepts

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

Null and Alternative Hypotheses
When performing hypothesis testing in statistics, we begin by formulating two contrasting statements: the null hypothesis and the alternative hypothesis. The null hypothesis, typically denoted as \( H_0 \), represents a hypothesis of no effect or no difference. It is the statement being tested and usually proposes that there is no significant difference between specified populations, any observed difference being due to sampling or experimental error. For example, if we were investigating the response rates to resumes with different-sounding names, the null hypothesis would claim that the proportion of responses \( p_1 \) for 'white-sounding' names is equal to the proportion \( p_2 \) for 'black-sounding' names, expressed as \( p_1 = p_2 \).

Conversely, the alternative hypothesis, denoted as \( H_a \) or \( H_1 \), is a statement that indicates the presence of an effect or difference. In our case, the alternative hypothesis states that there is a significant difference in response rates, positing that the response rate for 'white-sounding' names is higher than for 'black-sounding' names, expressed as \( p_1 > p_2 \). Determining whether to reject the null hypothesis or not is based on the data and appropriate statistical tests.
Z-Test Statistic
The z-test statistic is a numerical measurement that is used to evaluate the difference between an observed statistic and its hypothesized population parameter when the sample mean and population standard deviation are known. It is calculated using a standard formula, which in the case of comparing two proportions is given by: \[ z = (p_1 - p_2) / \sqrt{p(1-p)(1/n_1 + 1/n_2)} \]

In this formula, \( p_1 \) and \( p_2 \) are the sample proportions, while \( p \) is the pooled sample proportion representing a weighted average of the two sample proportions. The terms \( n_1 \) and \( n_2 \) denote the sample sizes for each group respectively. The z-test statistic provides a way of determining how many standard deviations an element is from the mean. A larger absolute value of the z-test statistic indicates a larger difference between the compared proportions.
Statistical Significance
Statistical significance is a determination of whether the observed results in a study are likely due to something other than mere random chance. This concept plays a central role in hypothesis testing. 'Significance level' is the threshold against which we measure the p-value of our test results and is denoted by \( \alpha \), commonly set at 0.05 or 5%. This means we would expect to see the observed outcome by chance only 5% of the time or less if the null hypothesis were true.

To assess statistical significance, we compare the calculated z-test statistic to a critical value, which corresponds to the cut-off for what we consider rare enough to be considered statistically significant under the null hypothesis. If the calculated statistic exceeds this critical value in the direction implied by the alternative hypothesis, we reject the null hypothesis, deeming the results statistically significant. This forms the basis for drawing conclusions in the context of hypothesis testing.
Proportions Comparison
Comparison of proportions is often employed when two distinct groups are observed, and we aim to ascertain whether there is a statistically significant difference in the proportion of individuals with certain characteristics between the two groups. For example, comparing the response rates to resumes with different-sounding names involves calculating the proportion of positive responses for both 'white-sounding' and 'black-sounding' names.

In the scenario provided, the calculated proportions were \( p_1 = 0.1 \) for 'white-sounding' names and \( p_2 = 0.0668 \) for 'black-sounding' names. To determine if the observed difference is statistically significant, we use hypothesis testing techniques, such as calculating the z-test statistic and comparing against the critical value. This approach allows researchers to make informed decisions about whether the observed differences in proportions are meaningful or could have arisen by random variation.

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

The paper "If It's Hard to Read, It's Hard to Do" (Psychological Science [2008]: \(986-988\) ) described an interesting study of how people perceive the effort required to do certain tasks. Each of 20 students was randomly assigned to one of two groups. One group was given instructions for an exercise routine that were printed in an easy-to-read font (Arial). The other group received the same set of instructions but printed in a font that is considered difficult to read the time (in minutes) they thought it would take to complete the exercise routine. Summary statistics follow. The authors of the paper used these data to carry out a twosample \(t\) test and concluded at the 0.10 significance level that the mean estimated time to complete the exercise routine is significantly lower when the instructions are printed in an easy-to-read font than when printed in a font that is difficult to read. Discuss the appropriateness of using a twosample \(t\) test in this situation.

The article referenced in the previous exercise also described an experiment in which students at Columbia Business School were randomly assigned to one of two groups. Students in one group were shown a coffee mug and asked how much they would pay for that mug. Students in the second group were given a coffee mug identical to the one shown to the first group and asked how much someone would have to pay to buy it from them. It was reported that the mean value assigned to the mug for the second group was significantly higher than the mean value assigned to the same mug for the first group. In the context of this experiment, explain what it means to say that the mean value was significantly higher for the group that was given the mug.

The paper "Short-Term Sleep Loss Decreases Physical Activity Under Free-Living Conditions but Does Not Increase Food Intake Under Time-Deprived Laboratory Conditions in Healthy Men" (American Journal of Clinical Nutrition [2009]: \(1476-1483\) ) describes an experiment in which 30 male volunteers were assigned at random to one of two sleep conditions. Men in the 4 -hour group slept 4 hours per night for two nights. Men in the 8-hour group slept 8 hours per night for two nights. On the day following these two nights, the men recorded food intake. The researchers reported that there was no significant difference in mean calorie intake for the two groups. In the context of this experiment, explain what it means to say that there is no significant difference in the group means. (Hint: See discussion on page 578 )

The article "Dieters Should Use a Bigger Fork" (Food Network Magazine, January/February 2012) described an experiment conducted by researchers at the University of Utah. The article reported that when people were randomly assigned to either eat with a small fork or to eat with a large fork, the mean amount of food consumed was significantly less for the group that ate with the large fork. a. What are the two treatments in this experiment? b. In the context of this experiment, explain what it means to say that the mean amount of food consumed was significantly less for the group that ate with the large fork.

In the paper "Happiness for Sale: Do Experiential Purchases Make Consumers Happier than Material Purchases?" (Journal of Consumer Research [2009]: \(188-197\) ), the authors distinguish between spending money on experiences (such as travel) and spending money on material possessions (such as a car). In an experiment to determine if the type of purchase affects how happy people are after the purchase has been made, 185 college students were randomly assigned to one of two groups. The students in the "experiential" group were asked to recall a time when they spent about \(\$ 300\) on an experience. They rated this purchase on three different happiness scales that were then combined into an overall measure of happiness. The students assigned to the "material" group recalled a time that they spent about \(\$ 300\) on an object and rated this purchase in the same manner. The mean happiness score was 5.75 for the experiential group and 5.27 for the material group. Standard deviations and sample sizes were not given in the paper, but for purposes of this exercise, suppose that they were as follows: \begin{tabular}{|ll|} \hline Experiential & Material \\ \hline\(n_{1}=92\) & \(n_{2}=93\) \\ \(s_{1}=1.2\) & \(s_{2}=1.5\) \\ \hline \end{tabular} Using the following Minitab output, carry out a hypothesis test to determine if these data support the authors' conclusion that, on average, "experiential purchases induced more reported happiness." Use \(\alpha=0.05\) Two-Sample T-Test and Cl Sample \(\begin{array}{rrrrr}\text { ple } & \text { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \\ 1 & 92 & 5.75 & 1.20 & 0.13 \\ 2 & 93 & 5.27 & 1.50 & 0.16\end{array}\) Difference \(=\operatorname{mu}(1)-\operatorname{mu}(2)\) Estimate for difference: 0.480000 \(95 \%\) lower bound for difference: 0.149917 T-Test of difference \(=0(\mathrm{vs}>): \mathrm{T}\) -Value \(=2.40 \mathrm{P}\) -Value \(=\) \(0.009 \mathrm{DF}=175\)

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.

Sign-up for free