Statistics for laboratory scientists

Solutions for the homework problems for lecture 14

1. The pooled estimate of the population SD is sqrt{ [10.67² × 9 + 9.30² × 4] / 13 } = 10.27.

   The estimated SE of the difference between the sample means is 10.27 × sqrt{ 1/10 + 1/5 } = 5.62

   The 97.5 percentile of the t distribution with 13 degrees of freedom is 2.16. (In R, use qt(0.975,13).)

   Thus the 95% confidence interval is (103-67) ± 2.16 × 5.62 = 36 ± 12 = (24, 48).

   In R, if x is the data for the sample from strain A and y is the data for the sample from strain B, type t.test(x,y,var.equal=TRUE) to get the above confidence interval.

2. 1. The 97.5 percentile of the t distribution with 5 degrees of freedom = qt(0.975, 5) = 2.57.

      Thus the 95% confidence interval for the population mean is 100.8 ± 2.57 × 8.28 / sqrt{6} = 100.8 ± 8.7 = (92.1, 109.5).

   2. The 95% confidence interval for the population SD is (8.28 × sqrt{5/12.8}, 8.28 × sqrt{5/0.831}) = (5.2, 20.3).

3. The difference between the sample means is 96.58-92.33=4.25. Our estimate of the standard error of the difference between the sample means is sqrt{29.09²/12 + 12.17²/9} = 9.33.

   The "degrees of freedom" parameter we use for the t-distribution is [29.09²/12 + 12.17²/9]² / {[29.09²/12]²/11 + [12.17²/9]²/8} = 15.56.

   The 97.5 percentile of the t distribution with 15.56 degrees of freedom = qt(0.975, 15.56) = 2.12.

   Finally, the 95% confidence interval for the difference between the population means is 4.25 ± 2.12 × 9.33 = 4.3 ± 19.8 = (-15.6, 24.1).