2 this is a two independent sample situation a let y

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2. This is a two-independent sample situation. (a) Let Y A and Y B be the concentrations on A and B respectively and let μ A and μ B be the respective population means. The needed summary statistics are ¯ y A = 6 . 1, s 2 A = 0 . 8733, ¯ y B = 4 . 9, and s 2 B = 0 . 74. Then, due to the balanced data, s 2 p = ( s 2 A + s 2 B ) / 2 = . 8067. The CI for μ A - μ B is ¯ y A - ¯ y B ± t * s 2 p (1 /n A + 1 /n B ). Since s 2 p has 6 df, the appropriate t-value is 1.943. Thus, ( - 0 . 03 < μ A - μ B < 2 . 43). (b) The null hypothesis is written: H 0 : μ A - μ B = 0 . 3. We notice that the value of 0.3 is contained within the CI above. Thus, we can conclude that the p - value > 0 . 10 for the given test. 3. The appropriate formula for the pooled variance is: s 2 p = ( k i =1 ( n i - 1) s 2 i ) / ( N - k ) where k is the number of treatments and N is the total number of observations. You are given the standard de- viations so you must square them. s 2 p = 157 . 02. There are N - k = 46 degrees of freedom asso- ciated with this variance. This can be thought of as the addition of the number of df for each variety (9 + 9 + 6 + 5 + 8 + 9).
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