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Unformatted text preview: Problem set 1 solutions A: ttest practice 1. (a) H : x 30 H A : x <30 (b) 1tailed 1sample ttest (c) t = 0.4657, df = 12, P = 0.3249 (d) At all levels of alpha, p> 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 : x =24.3 H A : x 24.3 (b) 2tailed, 1sample ttest (c) t = 2.7128, df = 24, P = 0.01215 (d) For = 0.1 and 0.05, p< so we reject the null hypothesis and conclude that the crabs maintain a body temperature different than air. For = 0.01 and 0.001, p> 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 : x =29.5 H A : x 29.5 (b) 2tailed, 1sample ttest (c) t = 2.2859, df = 13, P = 0.03968 (d) For =0.1 and 0.05, p< so we reject the null hypothesis and conclude that average human menstrual cycle does not equal a lunar month. For =0.01 and 0.001, p> so we fail to reject the null hypothesis and conclude that the average human menstrual cycle equals a lunar month. 4. (a) H : x 45 H A : x > 45 (b)1tailed, 1sample ttest (c) t = 0.3665, df = 7, P = 0.3624 (d) For =0.1, 0.05, 0.01, and 0.001, p> 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 : x 45 H A : x < 45 (b) 1tailed, 1sample ttest (c) t = 0.3665, df = 7, P = 0.6376 (d) For =0.1, 0.05, 0.01, and 0.001, p> 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 2tailed, 1sample ttest with H : x =45 and H A : x 45, (t = 0.3665, df = 7, P = 0.7248). p> so we fail to reject the null hypothesis and conclude that the mean dissolving time is 45 seconds.] 6. (a) H : = G H A : G (b) Welch 2tailed, twosample ttest (c) t = 2.54, df = 10.7, P = 0.0277 (d) For =0.1 and 0.05, p< so we reject the null hypothesis and conclude that the drugs differ in their effectiveness (mean bloodclotting times are not the same). For =0.01 and 0.001, p> so we fail to reject the null hypothesis and conclude that there is no difference in the mean effectiveness of the drugs (the mean bloodclotting times are the same). Note that if we had assumed equal variances then P=0.031, which doesnt change our conclusions. The variances differ by a factor of two, which is relatively large, but because of the small sample size we cant reject the null hypothesis that they are the same (use Statistics > Variances > Two variances Ftest). Nonetheless, the Welch test makes fewer assumptions, so is preferable....
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This note was uploaded on 08/06/2008 for the course ESM 206 taught by Professor Kendall,berkley during the Spring '08 term at UCSB.
 Spring '08
 KENDALL,BERKLEY
 Environmental Science

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