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# C here we focus on ˆ y as an estimator se ˆ y e s p

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(c) Here we focus on ˆ Y as an estimator. se ( ˆ Y e ) = s p (1 /n ) + (5 ¯ x ) 2 / ( P ( x i ¯ x ) 2 ) = . 079. Now, t 8 ,. 025 = 2 . 306 and ˆ Y = 3 . 326. Thus, a 95% con±dence interval for ˆ Y e is (3.144, 3.508). Thus, at 95 % con±dence, a plausible range for the mean rate of oxygen consumption is from 3.144 to 3.508 ml/g/hr. Since the value of “3” does not fall within this range, it is not a plau- sible value with 95% con±dence; thus the hy- pothesis that the true rate is 3 is rejected at α = 0 . 05. 3. There are two equivalent approaches. Let m = 4 n be the total number of trials. Due to independence, we can solve directly for m and choose the smallest integer equal to or larger than m that is divisible by 4. P ( atleast 1 success ) = 1 P (0 successes ) = 1 . 98 m . We wish to choose m so that . 98 m = 0 . 20. This can be done by trial and error or, directly, using logs. m ln ( . 98) = ln ( . 20). This results in m = 79 . 66. Rounding up to 80 (and dividing by 4) results in n = 20. Equivalently, let Y be the number of successes out of 4 trials. Then, P ( Y = 0) = . 98 4 = . 92237. Then, with n as the number of ‘sets’ of trials, . 92237 n = 0
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