An article reports that a sample of 220 juvenile rainbow trout fed a diet fortified with equal amounts of the amino acids taurine and methionine for a period of 70 days had a mean weight gain of 313% with a standard deviation of 25, while 220 fish fed with a control diet had a mean weight gain of 233% with a standard deviation of 19. Units are percent. Find a 99% confidence interval for the difference in weight gain on the two diets. (Round the final answers to two decimal places.)

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step-by-step explanation:

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.