Question

Average Earnings for College Graduates The average earnings of year-round full-time workers with bachelor's degrees or more is dollar 88,641 for men and dollar 58,000 for women - a difference of slightly over dollar 30,000 a year. One hundred of each were randomly sampled, resulting in a sample mean of dollar 90,200 for men, and the population standard deviation is dollar 15,000 ; and a mean of dollar 57,800 for women, and the population standard deviation is dollar 12,800. At the 0.01 level of significance, can it be concluded that the difference in means is not dollar 30,000 ?

Short answer

The difference is not significant at the 0.01 level; fail to reject the null hypothesis.

Step-by-step solution

Define Null and Alternative Hypotheses

The first step is to set up the null hypothesis, denoted as \( H_0 \), and the alternative hypothesis, denoted as \( H_a \). Here, we have two groups: men and women. We want to determine if the difference in sample means differs from the claimed difference of \( \$30,000 \).\\[ H_0: \mu_m - \mu_w = 30,000 \]\\[ H_a: \mu_m - \mu_w eq 30,000 \]

Compute the Sample Mean Difference

Before calculating the test statistic, compute the observed sample mean difference between men's and women's earnings.\\[ \text{Sample Mean Difference} = \overline{X}_m - \overline{X}_w = 90,200 - 57,800 = 32,400 \]

Calculate the Standard Error of the Difference

The standard error (SE) of the difference in sample means when population standard deviations are known is given by the formula:\\[ SE = \sqrt{\frac{\sigma_m^2}{n_m} + \frac{\sigma_w^2}{n_w}} \] \Substituting the given values, \( \sigma_m = 15,000 \), \( \sigma_w = 12,800 \), \( n_m = 100 \), and \( n_w = 100 \):\\[ SE = \sqrt{\frac{15,000^2}{100} + \frac{12,800^2}{100}} = \sqrt{2,250,000 + 1,638,400} = \sqrt{3,888,400} \approx 1,972.24 \]

Calculate the Z-Statistic

The Z-statistic is calculated using the formula:\\[ Z = \frac{(\overline{X}_m - \overline{X}_w) - D}{SE} \] \Where \( D \) is the hypothesized difference in population means (30,000). Substituting our values, we get:\\[ Z = \frac{32,400 - 30,000}{1,972.24} \approx 1.217 \]

Determine Critical Value and Make Decision

For a two-tailed test at the 0.01 significance level, the critical Z-values are approximately \( \pm 2.576 \). Compare the calculated Z-statistic to these critical values. \Since \( 1.217 \) is less than \( 2.576 \), we fail to reject the null hypothesis.

Key concepts

Understanding Null Hypothesis

In hypothesis testing, the null hypothesis is a foundational concept. It is a statement that there is no effect or no difference, and it serves as a starting point for statistical analysis. In this exercise related to earnings differences, the null hypothesis (\( H_0 \)) suggests that the difference in mean earnings between men and women is exactly $30,000. The null hypothesis is constructed with the assumption that any observed difference is due to chance rather than a true difference in means. By formulating this hypothesis, statisticians aim to disprove it with sufficient evidence to make a conclusion. If the null hypothesis is not rejected, it implies that we do not have strong enough evidence to state a significant difference exists.

Exploring the Alternative Hypothesis

The alternative hypothesis is the statement that contradicts the null hypothesis. In statistical testing, it represents the presence of an effect or difference. For this exercise, the alternative hypothesis (\( H_a \)) is that the mean earnings difference between men and women is not $30,000, implying it could be more or less. This hypothesis is crucial because it defines what we're trying to prove. It usually reflects the suspicion or expectation that led to the analysis. When evidence suggests the null hypothesis might not hold true, the alternative hypothesis becomes the more likely explanation for observed differences. Unlike the null hypothesis, the alternative does not assume the absence of effect. Instead, it argues for some form of relationship or disparity, requiring strong evidence to validate this argument and potentially reject the null hypothesis.

Z-Statistic in Hypothesis Testing

The Z-statistic is a key part of hypothesis testing, especially when testing differences between means. It measures how far away our observed sample mean difference is from the hypothesized population mean difference under the null hypothesis. This 'distance' is measured in terms of standard errors.To calculate the Z-statistic, you use the formula:\[ Z = \frac{(\overline{X}_m - \overline{X}_w) - D}{SE} \]where \( D \) is the hypothesized difference (in this case, $30,000), and SE is the standard error of the difference.Analyzing the Z-statistic helps us decide whether our observed results are statistically significant. If the Z-statistic falls within a certain range, determined by critical values (e.g., \(\pm 2.576\) for a 0.01 significance level), it suggests that the null hypothesis is likely true. If it falls outside this range, the evidence would suggest rejecting the null hypothesis, supporting the alternative hypothesis.

Role of Standard Error

Standard error (SE) in statistics represents the variability of a sample statistic from the population parameter it estimates. It is central to understanding the difference between sample means because it accounts for the variability across samples that might occur by random chance.In this exercise, the formula for calculating the standard error of the difference in sample means is:\[ SE = \sqrt{\frac{\sigma_m^2}{n_m} + \frac{\sigma_w^2}{n_w}} \]Where \( \sigma_m \) and \( \sigma_w \) are the population standard deviations, and \( n_m \) and \( n_w \) are the sample sizes for men and women, respectively.The calculated standard error gives us a sense of how much the sample mean difference might vary due to sampling variability. A smaller standard error suggests that the sample mean difference is closer to the true population mean difference, while a larger standard error implies more variability and less certainty about the sample estimates. Understanding standard error helps contextualize the calculated Z-statistic and assess the reliability of the test results.