Example

Comparing Wal-Mart’s with Other Firms’ Insurance Coverage

Recall the 2003 press release by the AFL-CIO:

• Wal-Mart exemplifies the harmful trend among America’s large employers to shirk health insurance responsibilities at the cost of their workers and community….Fewer than half of Wal-Mart workers are insured under the company plan – just 46 percent. This rate is dramatically lower than the 66 percent of workers at large private firms who are insured under their companies’ plans, according to a new Commonwealth Fund study released today.

This press release claims that there is a 20% difference in the proportion of workers with insurance when we compare Wal-Mart to other large private firms. In hypothesis testing for two population proportions, we cannot test a claim about a specific difference between two population proportions. Instead, we test a claim that the proportion of Wal-Mart workers with health insurance is less than the proportion of workers at large private firms with health insurance.

Suppose we select a random sample of 50 Wal-Mart workers and find 23 have health insurance. Suppose also that a random sample of 70 workers of large private firms had 43 with health insurance.

For this test, we choose a 5% level of significance (α = 0.05).

Step 1: State the hypotheses.

Let p₁ and p₂ represent the proportions of workers with health insurance among Wal-Mart and large private company employees respectively.

The null hypothesis is a claim of “no difference”: H₀: p₁ − p₂ = 0. The alternative hypothesis states that the population proportion is lower for Wal-Mart employees: p₁ < p₂. The difference is less than zero, so it is negative: H_a: p₁ − p₂ < 0.

Step 2: Collect the data.

Of 50 Wal-Mart workers, 23 have health insurance. Of 70 workers from large private firms, 33 have health insurance. From the data, we can calculate the difference in sample proportions.

[latex]{\stackrel{ˆ}{p}}_{1}-{\stackrel{ˆ}{p}}_{2}=\frac{23}{50}-\frac{43}{70}\approx-0.154[/latex]

Step 3: Assess the evidence.

Check the Normality Criteria

Determine if a normal model is a good fit for the sampling distribution. Verify that there are at least 10 successes and failures in each sample. Here, a success is an employee with health insurance. In the Wal-Mart sample, there are 23 successes and 50 − 23 = 27 failures. In the large private firms sample, there are 43 successes and 27 failures. Each of these is at least 10, so we can use the normal model.

Compute the Test Statistic (only if the normal model is a good fit)

The test statistic requires the standard error. To compute the standard error, we first compute the pooled proportion.

[latex]\stackrel{ˆ}{p}\text{}=\text{}\frac{{x}_{1}\text{}+\text{}{x}_{2}}{{n}_{1}\text{}+\text{}{n}_{2}}\text{}=\text{}\frac{23\text{}+\text{}43}{50\text{}+\text{}70}\text{}=\text{}\frac{66}{120}\text{}=\text{}0.55[/latex]

We use the pooled proportion to estimate the standard error.

[latex]\sqrt{\frac{\stackrel{ˆ}{p}(1-\stackrel{ˆ}{p})}{{n}_{1}}+\frac{\stackrel{ˆ}{p}(1-\stackrel{ˆ}{p})}{{n}_{2}}}\text{}=\text{}\sqrt{\frac{0.55(0.45)}{50}+\frac{0.55(0.45)}{70}}\text{}\approx \text{}0.092[/latex]

Recall the difference in sample proportions from the data.

[latex]{\stackrel{ˆ}{p}}_{1}-{\stackrel{ˆ}{p}}_{2}\text{}=\text{}\frac{23}{50}-\frac{43}{70}\text{}\approx \text{}-0.154[/latex]

We use the z-score to determine how many standard errors −0.154 is from the mean of 0.

[latex]Z\text{}=\text{}\frac{({\stackrel{ˆ}{p}}_{1}-{\stackrel{ˆ}{p}}_{2})\text{}-\text{}({p}_{1}-{p}_{2})}{\sqrt{\frac{\stackrel{ˆ}{p}\text{}(1-\stackrel{ˆ}{p})\text{}}{{n}_{1}}+\frac{\stackrel{ˆ}{p}\text{}(1-\stackrel{ˆ}{p})}{{n}_{2}}}}\text{}\approx \text{}\frac{-0.154-0}{0.092}\text{}\approx \text{}-1.67[/latex]

Note: A z-score of −1.67 tells us that the observed difference of [latex]{\stackrel{ˆ}{p}}_{1}-{\stackrel{ˆ}{p}}_{2}[/latex] = −0.154 is 1.67 standard errors below the assumed difference of zero. Does this suggest that the observed difference is statistically significant? Since we stated a significance level of 5%, we need to find the P-value and compare it to 0.05.

Identify the P-Value

We use a simulation. We want the probability that the difference in sample proportions is less than −0.154. This corresponds to the probability that Z is less than −1.67. So we use the area to the left of the Z-test statistic. The P-value is about 0.047. If you like symbols, we can write this in mathematical notation.

[latex]P\text{}(\text{}({\stackrel{ˆ}{p}}_{1}-{\stackrel{ˆ}{p}}_{2})\text{}<-0.154\text{})\text{}=\text{}P\text{}(Z\text{}<-1.67)\text{}\approx \text{}0.047[/latex]

The P-value is small, about 4.7%. It means that if there is no difference in the population proportions, there is about a 4.7% chance that random samples will have a difference less than −0.154. The difference we observed in our samples, then, is fairly unlikely. We do not think this difference is due to chance. We see that the P-value is less than 5%, so we conclude that the difference we observed is statistically significant.

Step 4: State a conclusion.

Use the P-Value to Make a Decision

A P-value less than the significance level means we reject the null hypothesis. So we support the alternative hypothesis, p₁ − p₂ < 0, or more simply, p₁ < p₂. The given sample data support the claim that the proportion of Wal-Mart workers with health insurance is lower than the proportion of workers for large private companies.