Question

For a certain wine, the mean pH (a measure of acidity) is supposed to be 3.39 with a known standard deviation of σ = .13. The quality inspector examines 33 bottles at random to test whether the pH is too low, using a left-tailed test at α = 0.01.

(a) What is the power of this test if the true mean is μ = 3.37? (Round your Z-value to 3 decimal places and final answer to 4 decimal places.)

Power             

(b) What is the power of this test if the true mean is μ = 3.35? (Round your Z-value to 3 decimal places and final answer to 4 decimal places.)

Power             

(c) What is the power of this test if the true mean is μ = 3.33? (Round your Z-value to 3 decimal places and final answer to 4 decimal places.)

Power   

Answer 1

A)
Here, n = 33 , xbar = 3.39 and s = 0.13
Critical value is -1.6449
True mean = 3.37

Critical value, c = xbar + (z/t)*s/sqrt(n)
c = 3.39 + -2.3263 * 0.13/sqrt(33)
c = 3.3374

Beta or type II error is the probability of fail to reject the null hypothesis
P(X > 3.3374 | mu = 3.37) = 0.9251
beta = 0.9251
Power of the test is 1 - beta
Power of the test is 0.0749

B)
Power of the Test
Here, n = 33 , xbar = 3.39 and s = 0.13
Critical value is -1.6449
True mean = 3.35

Critical value, c = xbar + (z/t)*s/sqrt(n)
c = 3.39 + -2.3263 * 0.13/sqrt(33)
c = 3.3374

Beta or type II error is the probability of fail to reject the null hypothesis
P(X > 3.3374 | mu = 3.35) = 0.7112
beta = 0.7112
Power of the test is 1 - beta
Power of the test is 0.2888

C)
Here, n = 33 , xbar = 3.39 and s = 0.13
Critical value is -1.6449
True mean = 3.33

Critical value, c = xbar + (z/t)*s/sqrt(n)
c = 3.39 + -2.3263 * 0.13/sqrt(33)
c = 3.3374

Beta or type II error is the probability of fail to reject the null hypothesis
P(X > 3.3374 | mu = 3.33) = 0.3718
beta = 0.3718
Power of the test is 1 - beta
Power of the test is 0.6282