0 20 using similar reasoning to the above trial and

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= 0 . 20 using similar reasoning to the above. Trial and error – or logs – again results in n = 20. 4. (a) FALSE — The only way one can obtain a neg- ative value of F is if one has made a mistake. Even if all data values are negative, F can never be less than 0. (b) FALSE — Since you are in ‘planning’ mode, the underlying variance should be viewed as known. We know V ar ( ˆ b 1 ) = σ 2 y.x / P ( x i ¯ x ) 2 . With the 10 data points, the sum in the denom- inator is 30; for the 6 data points situation, the sum is 70. Thus, V ar ( ˆ b 1 ) will be smaller for the 6 data point case. (Even if one were concerned with CI width by considering the di²erence in t-values for 4df and 8df, the conclusion would not change.) 5. The random variable of interest is D , the di²er- ence in glucose levels between B and A . The stated condition requires the distribution of D under the given alternative. In this case, D N ( . 5 , 1 . 8). The given condition is P ( ¯ D 0 . 3 | μ D = . 5) = . 90. We know P ( Z ≥ − 1 . 28) = . 90. Thus, 1 . 28 = ( . 3 . 5) / p (1 . 8 /n ). Solving results in n = 8 . 59 and, rounding up, n = 74. Grade Distribution 90-99:20 80-89:19 70-79:37 median = 74 60-69:23 50-59:24 below: 5
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