Statistics for laboratory scientists

Solutions for the homework problems for lecture 16

  1. Xbar - Ybar = -7.75; SE = 3.55; T = -2.184; k = 11.263

    1. P-value = 2*pt(-2.184,11.263) = 0.051

    2. Conclusion: reasonable, but not terribly strong evidence against H0.

  2. P-value = pt(-2.184,11.263) = 0.025.
    Same conclusion; somewhat stronger evidence in this case.

  3. The sample mean of the differences is 99.83. The sample SD of the differences is 50.50. Thus the estimated standard error of the sample mean of the differences is 50.50/sqrt{6} = 20.62.

    1. The t-statistic for the test is 99.83/20.62 = 4.84. Thus the P-value is 2 * (1 - pt(4.84, 5)) = 4.7 × 10-3.

    2. Thus we conclude that the treatment does have an effect. Note that the 95% confidence interval for the mean effect of the treatment is (47,153).

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Last modified: Sun Feb 26 00:14:13 EST 2006