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Problem set 1 solutions (ABC)

# Problem set 1 solutions (ABC) - Problem set 1 solutions A...

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Unformatted text preview: Problem set 1 solutions A: t-test practice 1. (a) H : μ x ≥30 H A : μ x <30 (b) 1-tailed 1-sample t-test (c) t = -0.4657, df = 12, P = 0.3249 (d) At all levels of alpha, p> α so we fail to reject the null hypothesis and conclude that this species does not maintain a coelomic calcium concentration less than that of their environment. 2. (a) H : μ x =24.3 H A : μ x ≠24.3 (b) 2-tailed, 1-sample t-test (c) t = 2.7128, df = 24, P = 0.01215 (d) For α = 0.1 and 0.05, p< α so we reject the null hypothesis and conclude that the crabs maintain a body temperature different than air. For α = 0.01 and 0.001, p> α so we fail to reject the null hypothesis and conclude that the crabs maintain a body temperature the same as the air. 3. (a) H : μ x =29.5 H A : μ x ≠29.5 (b) 2-tailed, 1-sample t-test (c) t = -2.2859, df = 13, P = 0.03968 (d) For α =0.1 and 0.05, p< α so we reject the null hypothesis and conclude that average human menstrual cycle does not equal a lunar month. For α =0.01 and 0.001, p> α so we fail to reject the null hypothesis and conclude that the average human menstrual cycle equals a lunar month. 4. (a) H : μ x ≤45 H A : μ x > 45 (b)1-tailed, 1-sample t-test (c) t = 0.3665, df = 7, P = 0.3624 (d) For α =0.1, 0.05, 0.01, and 0.001, p> α so we fail to reject the null hypothesis and conclude that the mean dissolving time is less than or equal to 45 seconds. 5. (a) H : μ x ≥45 H A : μ x < 45 (b) 1-tailed, 1-sample t-test (c) t = 0.3665, df = 7, P = 0.6376 (d) For α =0.1, 0.05, 0.01, and 0.001, p> α so we fail to reject the null hypothesis and conclude that the mean dissolving time greater than or equal to 45 seconds. [Combining questions 4 and 5, if we use a 2-tailed, 1-sample t-test with H : μ x =45 and H A : μ x ≠45, (t = 0.3665, df = 7, P = 0.7248). p> α so we fail to reject the null hypothesis and conclude that the mean dissolving time is 45 seconds.] 6. (a) H : μ Β = μ G H A : μ Β ≠ μ G (b) Welch 2-tailed, two-sample t-test (c) t = -2.54, df = 10.7, P = 0.0277 (d) For α =0.1 and 0.05, p< α so we reject the null hypothesis and conclude that the drugs differ in their effectiveness (mean blood-clotting times are not the same). For α =0.01 and 0.001, p> α so we fail to reject the null hypothesis and conclude that there is no difference in the mean effectiveness of the drugs (the mean blood-clotting times are the same). Note that if we had assumed equal variances then P=0.031, which doesn’t change our conclusions. The variances differ by a factor of two, which is relatively large, but because of the small sample size we can’t reject the null hypothesis that they are the same (use Statistics -> Variances -> Two variances F-test). Nonetheless, the Welch test makes fewer assumptions, so is preferable....
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Problem set 1 solutions (ABC) - Problem set 1 solutions A...

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