Applied Business Statistics: Hypothesis Testing - Single Sample Part 4

Hypothesis Testing: Single Sample t-test with a Proportion

This section delves into hypothesis testing for a single sample using a t-test specifically for proportions, utilizing a real-world example of a discrimination lawsuit.

Case P: Discriminatory Hiring Lawsuit

• Scenario: Company X is facing a lawsuit for discriminatory hiring practices, alleging a significantly greater number of men than women have been hired, despite an equal pool of qualified candidates. In the last year, only $18$ out of $50$ hires ($36\%$) were women.

• Question: Is there sufficient evidence to conclude that the company is hiring significantly more men than women?

• Variables and Notation:

  • $\pi$ (pi): Represents the true proportion of men hired by the company (population parameter).

  • $\bar{p}$ (p-hat): Represents the sample proportion of men hired from the $50$ observations ($32/50 = 0.64$ or $64\%$). This is the mean of a dichotomous variable (coded $1$ for men, $0$ for women).

• Why Not "Eyeball" the Data?

  • Intuition is poor at considering sample size.

  • Difficulty in discerning significant versus insignificant differences.

  • Legal Context: Decisions have serious implications (e.g., changes in hiring policies, financial compensation). Therefore, rigorous statistical testing is required, not just visual assessment.

Hypothesis Testing Steps

The following steps are applied to the Company X lawsuit scenario:

1. Formulating Hypotheses:

   • Null Hypothesis ($H_0$): $\pi = 0.50$ (The true proportion of men hired is not significantly different from $0.50$, meaning an equal proportion of men and women are hired, indicating no discrimination).

   • This assumption is based on the information that there is an equal number of qualified men and women in the applicant pool.

   • It reflects the legal principle of "innocent until proven guilty", placing the burden of proof on the plaintiff (the person making the accusation of discrimination).

   • Alternative Hypothesis ($H_1$): \pi > 0.50 (The true proportion of men hired is significantly greater than $0.50$, indicating discriminatory hiring practices against women).

   • This is a one-sided test because the lawsuit alleges discrimination specifically against women (meaning more men are hired), not just a difference in either direction.

2. Setting the Significance Level ($\alpha$):

   • The significance level ($\alpha$) is the probability of rejecting the null hypothesis when it is actually true (Type I Error).

   • In legal contexts especially, a commonly accepted level is $\alpha = 0.05$ (or a $5\%$ chance of Type I Error).

   • This means if the p-value is less than or equal to $0.05$, we reject $H_0$.

3. Checking Assumptions:

   • Random Sample: The $50$ hires must represent a random sample of current hiring practices. While not explicitly stated as random, for analysis, we assume representativeness.

   • Independence: Each hiring decision must be independent of others.

   • Sample Size: For a t-test for proportions, we approximate the binomial distribution with a normal distribution. This approximation is generally considered valid if both $n\pi<em>0 \geq 10$ and $n(1-\pi</em>0) \geq 10$.

     • In this case, $n = 50$ and under $H<em>0$, $\pi</em>0 = 0.50$.

     • $n\pi_0 = 50 \times 0.50 = 25$

     • $n(1-\pi_0) = 50 \times (1-0.50) = 25$

     • Since both are $25$, which is greater than $10$, the sample size assumption is met, and a t-test is appropriate.

4. Calculating the Test Statistic:

   • The formula for the z-test statistic for a single proportion is:

     $z = \frac{\bar{p} - \pi<em>0}{\sqrt{\frac{\pi</em>0(1-\pi_0)}{n}}}$

   • Where:

     • $\bar{p}$ is the sample proportion ($0.64$)

     • $\pi_0$ is the hypothesized population proportion under the null hypothesis ($0.50$)

     • $n$ is the sample size ($50$)

   • Substituting the values:

     $z = \frac{0.64 - 0.50}{\sqrt{\frac{0.50(1-0.50)}{50}}} = \frac{0.14}{\sqrt{\frac{0.25}{50}}} = \frac{0.14}{\sqrt{0.005}}$

     $z = \frac{0.14}{0.07071} \approx 1.98$

5. Determining the p-value:

   • The p-value is the probability of observing a test statistic as extreme as, or more extreme than, the one calculated, assuming the null hypothesis is true.

   • Since this is a one-sided test (H_1: \pi > 0.50), we need to find the probability of a z-score greater than or equal to $1.98$

   • Using a standard normal distribution table or statistical software, the p-value for $z \approx 1.98$ is approximately P(Z > 1.98) = 1 - P(Z \leq 1.98) = 1 - 0.9761 = 0.0239.

6. Making a Decision:

   • Compare p-value to $\alpha$: Our calculated p-value is $0.0239$ and our significance level $\alpha$ is $0.05$.

   • Since p{value} (0.0239) < \alpha (0.05), we reject the null hypothesis ($H</em>0$).

7. Formulating a Conclusion:

   • Based on the statistical analysis, there is sufficient evidence (p-value = $0.0239$) to conclude that the true proportion of men hired by Company X is significantly greater than $0.50$.

   • This suggests that the company is hiring significantly more men than women, supporting the claim of discriminatory hiring practices against women.