Statistics for laboratory scientists

Solutions for the homework problems for lecture 13

1. No. The correct statement would be something like, "95% of such confidence intervals would contain the population mean."

2. The confidence interval would be

   xbar ± 1.96 × sigma / sqrt{n}

   with width 2 × 1.96 × sigma / sqrt{n}.

   Thus we need to solve the following for n:

   2 × 1.96 × 1.5 / sqrt{n} = 1

   Thus n = (2 × 1.96 × 1.5)² = 34.6.

   Thus we need to measure at least 35 samples. This is too large! We should get a more precise measuring device.

3. Sample mean = 3.496; sample SD = 0.301. The appropriate t-value (in R, use qt(0.975, 4)) is 2.776.

   Thus the 95% confidence interval is 3.496 ± 2.776 × 0.301 / sqrt{5} = (approx) 3.50 ± 0.37 = (approx) (3.12, 3.87).