Lost hours after a safety programme

Topic: Confidence intervals · more questions on this topic
‹ Two standard errors Checking equal variances ›

Lost working hours averaged 91 per month over 50 months after a safety programme, with standard deviation 14.1, and 108 per month over the 50 months before, with standard deviation 14.3. Assume two independent normal samples with a common variance.

  1. Compute the pooled standard deviation and a 95% confidence interval for \mu_1-\mu_2. Use t_{98;0.975}=1.984.

  2. Someone reports the result as -17\pm2.84. What coverage does that interval have?

  1. The pooled variance is

s_p^2=\frac{49\times14.1^2+49\times14.3^2}{98}=\frac{14.1^2+14.3^2}{2}=201.65,\qquad s_p=14.20.

The standard error of \overline X_1-\overline X_2 is 14.20\sqrt{1/50+1/50}=14.20/5=2.84. The 95% interval is

-17\pm1.984\times2.84=-17\pm5.64=(-22.6,\ -11.4).

  1. -17\pm2.84 is the estimate plus or minus one standard error. With 98 degrees of freedom the coverage is \mathbb{P}(|T|<1) for T\sim t_{98}, about 0.68. It is a 68% interval, not a 95% one.