Question 1. You decide to carry out a trial on 12 goats to collect coefficients

Question

Question 1. You decide to carry out a trial on 12 goats to collect coefficients of digestibility of dry matter. Suppose you know that =3. You wish to test Ho: = 54 versus H!: 54: (ie. 54 being, say, the known mean parameter for sheep but you didn't know whether the mean for goats would be larger or smaller and so are interested in testing this) Before approving your research protocol, the Animal Use and Care Committee wishes to know if you have sufficient power of test to make sure you don't unnecessarily inconvenience 12 goats. +' a) If you suspect the true mean for goats is in the neighborhood of 57, how much Should you proceed with the trial as power of test did you have for = 0.05? planned then?+' b) To assure a statistical power no less than 90%, how many goats should you have in your trial?-' Question 2. Refer to Question 1 of the previous homework; i.e., you are thinking about conducting such an experiment yourself a) What would be the power of the hypothesis test conducted in part c) of that question if the true mean calcium intake for individuals below the poverty level is 700 mg. - b) How large of a sample size should be considered for the hypothesis test in part c) of that question in the previous homework if the desired power was 90% and the true mean calcium intake for individuals below the poverty level is 700 mg.

Explanation / Answer

Question 1

True Mean = 57

the significance level = 0.05

standard error of the sample mean = /sqrt(n) = 3/ sqrt(12) = 0.866

95% significance level = H +- Z95% /sqrt(n)  = 54 +- 1.96 * 0.866 = (52.30, 55.70)

so We wil have a type II error if we will fail to reject the null hypothesis even if it is false that means we have to find the probability that the given dry matter is under the given confidence interval when true mean is 57.

Pr( Type II error) = Pr(X < 55.70 ; 57; 0.866) =

Z - value = (55.70 - 57)/ 0.866 = -1.50

p value = Pr( Z < -1.50) = 0.0668

Power of the test = 1 - 0.0668 = 0.9332

(b) Here power = 0.90

so Pr(Type II error) = 1 - 0.90 = 0.10

Z - value = -1.28

Let say the sample size = n

so standard error of the mean = /sqrt(n) = 3/ sqrt(n)

-1.28 = (X - 57)/ (3/ n)

-3.84/n + 57 = X

so that value will be at max the out bound of 95% confidence interval of hypothesis mean

-3.84/n + 57 = 54+ 1.96 * 3/n

3 = 9.72/ n

n = 9.72/3

n = 3.24

n = 10.49 or say 11

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