Suppose that we are planning an experiment to test hypotheses about the mean of

Question

Suppose that we are planning an experiment to test hypotheses about the mean of a population that is known to be normal with standard deviation 3. We wish to test the null hypothesis Ho : 0 vs. the alternative HA : HA > 0. We intend to use a likelihood ratio test with significance level a = 0.05 (a) For which values of X, the sample mean, will we reject the null hypothesis. Express your answer as a function of sample size, n. (b) Suppose that we plan to obtain a sample of size 36. The researcher = 0.3. Calcu- late the power of the test to reject the null hypothesis under this thinks that if the alternative is true then perhaps particular alternative hypothesis. (c) Continuing from (b), the scientist who is planning the experiment wishes to have power at least 80%. (Thus your calculation in (b) shows that more than 36 observations are required.) Find approximately the smallest sample size at which this power can be achieved, against the specific alternative 0.3 Hint: repeat the calculation performed in (a) at different sample sizes using trial and error: you may wish to use R to speed up this calculation

Explanation / Answer

Here Population standard deviation = 3

Hypothesis are

H0 : 0 = 0

Ha : 0 > 0

significance level = 0.05

(a) sample mean x above which we reject the null hypothesis.

x = 0 + Z0.05 (/n) = 1.645 * 3/ n = 4.935/n

so we will reject the null hypothesis if sample mean is greater than 4.935/n

(b) Here sample size = 36

so sample mean above which we will reject the null hypothesis = 4.935/ 36 = 0.8225

Here, true mean = 0.3

standard error of sample mean se0 = /n = 3/ 36 = 3/6 = 0.5

Pr (Type II error) = Pr(x < 0.8225; 0.3 ; 0.5)

Z = (0.8225 - 0.3)/ 0.5 = 1.045

Pr (Type II error) = Pr(Z < 1.045) = 0.852

Power = 1 - Pr (Type II error) = 1- 0.852 = 0.148

(c) Here power will be atleast 80%

POwer of the test > 0.80

Probability of type II error < 0.20

standard error of samplemean = 3/n

Pr(x < 4.935/n ; 0.30 ; 3/n) < 0.20

Z - value for p - value = 0.20

Z = -0.842

-0.842 = (4.935/n - 0.30)/ (3/n)

-2.526/n = 4.935/n - 0.30

0.30 = 7.461/n

n = 618.52

n = 619

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