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T test for the slope t 26 02 3 15 8 26 on df 13 we

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T-test for the slope: t = 26 . 02 / 3 . 15 = 8 . 26 on df= 13. We get a p-value < 2 * . 001. Strong evidence that the slope is > 0. (d) prediction of one new value: 52 . 37+26 . 02 * 19 = 546 . 87 thousands of eggs if the female weighs 19kg, and 52 . 37 + 26 . 02 * 24 = 677 . 0 thousands of eggs at 24kg. Larger. This is because ¯ x = 16 . 1kg so 19kg is closer to the mean female’s weight than 24kg. The SE of the prediction is larger when ( x - ¯ x ) 2 is larger, and here (24 - ¯ x ) 2 is larger than (19 - ¯ x ) 2 . 3. See Ley et al. Nature 444:1022 (2006). (a) Paired t-test, 2 sided. (b) t = 3 . 09, df= 12 - 1 = 11 so the p-value is 2 * . 005 < p < 2 * . 01 i.e. . 01 < p < . 02. We reject the null hypothesis. In obese subjects (such as those sampled for the study), the mean bacteroidetes abundance is higher at week 54 than at week 12 of the diet. 4.(a) We can get SSError as (8 - 1) * 785 2 + · · · +(8 - 1) * 62 2 = 4 , 427 , 213. To get SSCultivar, we can first get the grand mean (8 * 1830+ · · · +8 * 403) / 48 = 635 . 17 then SSCultivar is 8 * (1830 - 635 . 17) 2 + · · · +8 * (403 - 635 . 17) 2 = 13 , 962 , 803. We can also get SSCultivar by subtracting SSError from SSTotal. Source df SS MS F p-value Cultivar 5 13,962,803 2,792,561 26.5 < . 001 Error 42 4,427,213 105,409.8 Total 47 18,390,020 No, this analysis does not provide evidence that the 5 oca cultivars do not all have the same mean. The significant p- value could be due to the rhubarb having a different mean
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