Test Your Understanding 5

It has been reported (The Boston Globe, July 31, 1997) that as much as 25% of all worker's compensation claims and 40% of all payments are attributable to lower back injuries. In an effort to reduce such injuries, employers have sent workers ``back to school'' to learn to lift heavy objects safely. Suppose that, in order to study the effectiveness of such a program, for one year after the program's completion, you monitor the incidence of back injury for a random sample of 200 workers who have taken the program, and a random sample of 200 workers who have not. Of the 200 workers who completed the course, 18 report at least one lower back injury during this period and of the other group, 21 report at least one lower back injury. Is this convincing evidence for the efficacy of the program? Formulate the scientific and statistical hypotheses, the statistical model, and the standardized test statistic. Obtain the p-value and use it to state your conclusion.

SOLUTION:

• Scientific hypothesis: The program works.
• Statistical model: Two-sample binomial. Let Y₁ denote the number of injuries in the sample of course grads, and Y₂ the number of injuries in the sample of non-grads, and assume that Y₁ and Y₂ are independent with $Y_1\sim b(200,p_1)$ and $Y_2\sim b(200,p_2)$.
• Statistical hypotheses: H₀: p₁-p₂=0, H_a: p1-p2<0.
• p-value: First, compute the standardized test statistic,

  $$Z_0=\frac{\hat{p}_1-\hat{p}_2}{\hat{\sigma}_0(\hat{p}_1-\hat... ...{\hat{p}(1-\hat{p}) \left(\frac{1}{n_1}+\frac{1}{n_2}\right)}},$$

  which under H₀ has an approximate N(0,1) distribution provided the sample sizes are large enough. Here, $\hat{p}=(Y_1+Y_2)/(n_1+n_2)$is the pooled estimate of the common value, p, of p₁ and p₂.

  From the data, we have, y₁=18, n₁-y₁=182, y₂=21, and n₂-y₂=179 are all larger than 10, so the normal approximation is ok. Also, the pooled estimate of p is $\hat{p}=(18+21)/(200+200)=39/400$. The observed value of the test statistic is

  $$z_0^*=\frac{18/200-21/200} {\sqrt{(39/400)(1-39/400)\left(\frac{1}{200}+{1}{200}\right)}}=-0.5057.$$

  Therefore, the p-value is (Z denotes a N(0,1) random variable)

  $$P(Z\leq z_0^*)=P(Z\leq -0.5057)=0.3065.$$

• Conclusion: Do not reject H₀. The data supply insufficient evidence of the efficacy of the program.