Question

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

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

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: \(p_1 > p_2\).

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 \(p_2 = 167 / 2500 = 0.0668\).

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 \(z = (p_1 – p_2) / SE = (0.1 - 0.0668) / 0.0086 = 3.86\).

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.

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.

Key concepts

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.