Question

Is Gender Bias Influenced by Faculty Gender? Exercise 6.215 describes a study in which science faculty members are asked to recommend a salary for a lab manager applicant. All the faculty members received the same application, with half randomly given a male name and half randomly given a female name. In Exercise \(6.215,\) we see that the applications with female names received a significantly lower recommended salary. Does gender of the evaluator make a difference? In particular, considering only the 64 applications with female names, is the mean recommended salary different depending on the gender of the evaluating faculty member? The 32 male faculty gave a mean starting salary of \(\$ 27,111\) with a standard deviation of \(\$ 6948\) while the 32 female faculty gave a mean starting salary of \(\$ 25,000\) with a standard deviation of \(\$ 7966 .\) Show all details of the test.

Short answer

To conclude whether gender of the faculty influences the suggested starting salary for female applicants, first calculate the t-statistic using given means, sample sizes, and standard deviations. Compare this t-score to the critical value for a 95% confidence level and get the p-value. If the magnitude of t is smaller than critical value and p-value greater than 0.05, then there is no significant difference in the starting salaries suggested by male and female faculty for female candidates.

Step-by-step solution

Stating the Hypotheses

The null hypothesis \(H_0\) is that there is no difference in the mean recommended salary given by male and female faculty members. The alternative hypothesis \(H_1\) is that there is a difference. Mathematically, the hypotheses can be stated as follows: \(H_0: \mu_{male} = \mu_{female}\), \(H_1: \mu_{male} \neq \mu_{female}\)

Calculating the Test Statistic

We use the formula for the test statistic in an independent two-sample t-test when population standard deviations are assumed unknown but equal: \(t = \frac{{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\). Here, \(\bar{x}_1\) and \(\bar{x}_2\) are sample means, \(s_1\) and \(s_2\) are sample standard deviations and \(n_1\) and \(n_2\) are sample sizes.

Calculating the test statistic values

Substitute the given values into formula: \(t = \frac{{(\$ 27,111-\$ 25,000) - 0}}{\sqrt{\frac{6948^2}{32} + \frac{7966^2}{32}}}\), which on solving, gives us \(t\approx1.15\).

Determining the Critical Value and P-Value

We need to compare the test statistic (\(t\)) to critical values from the t-distribution table with degree of freedom \((n_1 + n_2 - 2)= (32+32-2)=62\). For a two-tailed test at a 0.05 significance level, the critical value is approximately 2.00. To get the p-value, lookup the t-score in the t-distribution table. Alternatively, software or calculators could be used to get the p-value from t-score.

Conclusion

If absolute value of \(t\) is less than 2.000 (critical value), and p-value is > 0.05, we do not reject the null hypothesis. If not, we reject the null hypothesis. If we fail to reject the null hypothesis, this implies that gender of evaluator does not significantly impact the mean starting salary suggested for female applicants. The specific conclusion will depend on the calculated p-value.

Key concepts

Null Hypothesis

The null hypothesis is a fundamental concept in statistics, serving as a starting point in hypothesis testing. It is a formal statement that assumes there is no effect or no difference between two or more groups. In the context of gender bias in faculty salary recommendations, the null hypothesis (\(H_0\)) posits that the gender of evaluators does not influence the salary recommended for a lab manager applicant with a female name. Mathematically, it is expressed as the equality of the mean recommended salaries given by male and female faculty members: \(H_0: \mu_{male} = \mu_{female}\). By assuming no disparity, it sets the benchmark against which the actual observations will be tested. Rejecting the null hypothesis implies that there is a statistically significant difference, meaning the data collected provides enough evidence to support that the gender of the evaluator has an actual effect on salary recommendations.

Alternative Hypothesis

Conversely, the alternative hypothesis (\(H_1\)) represents what a researcher is trying to demonstrate or is suspecting to be the reality. It suggests there is an effect or a difference, which, in our case, is that the mean recommended salary for a lab manager applicant does depend on whether a male or female faculty member is evaluating. The alternative hypothesis challenges the status quo established by the null hypothesis and is mathematically articulated as inequality: \(H_1: \mu_{male} eq \mu_{female}\). If the hypothesis testing leads to the rejection of the null hypothesis, the alternative hypothesis is considered to have enough support based on the data provided.

Two-Sample T-Test

A two-sample t-test is a statistical technique used to determine whether there is a significant difference between the means of two independent samples. It is particularly useful when the sample sizes are small and the population standard deviations are unknown and assumed to be equal. In our exercise, we use this test to compare the mean salaries recommended by male and female faculties. The test statistic is calculated using the formula: \(t = \frac{{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\). Here, \(\bar{x}_1\) and \(\bar{x}_2\) represent the sample means, \(s_1\) and \(s_2\) are the sample standard deviations, and \(n_1\) and \(n_2\) are the sample sizes for male and female evaluators, respectively. This statistic helps us determine whether the observed data falls outside the range of what we would expect under the null hypothesis.

P-Value

The p-value is a critical concept in statistical hypothesis testing, which provides the probability of obtaining the observed results, or more extreme, assuming the null hypothesis is true. It quantifies the strength of the evidence against the null hypothesis. A small p-value (typically ≤ 0.05) indicates strong evidence against the null hypothesis, so you would reject it in favor of the alternative hypothesis. In evaluating gender bias, we calculate the p-value based on the test statistic (\(t\)) from the two-sample t-test. If the p-value is larger than the chosen significance level (e.g., 0.05), we fail to reject the null hypothesis, meaning we do not have sufficient evidence to say that the evaluator's gender affects salary recommendations.

T-Distribution

The t-distribution, also known as Student's t-distribution, is a probability distribution that arises when estimating the mean of a normally distributed population in situations where the sample size is small and the population standard deviation is unknown. It plays a central role in the two-sample t-test. The spread of the t-distribution varies by the number of degrees of freedom—the more data points, the more the t-distribution approaches the normal distribution. In our gender bias study, we compare the calculated t-value to the critical t-values from a t-distribution table with 62 degrees of freedom (\(n_1 + n_2 - 2 = (32+32-2) = 62\)) to determine if our results are statistically significant. The closer our t-value is to 0, the less likely it is that there is a significant difference between the groups, aligning with the null hypothesis.