Question 6.16: Shipments of coffee beans are checked for moisture content. ...
Shipments of coffee beans are checked for moisture content. High moisture content indicates possible water contamination, leading to rejection of the shipment. Let μ represent the mean moisture content (in percent by weight) in a shipment. Five moisture measurements will be made on beans chosen at random from the shipment. A test of the hypothesis H_0: μ ≤ 10 versus H_1: μ > 10 will be made at the 5% level, using the Student’s t test. What is the power of the test if the true moisture content is 12% and the standard deviation is σ = 1.5%?
The following computer output (from MINITAB) presents the solution:
Power and Sample Size
1- Sample t Test
Testing mean = null ( versus > null )
Calculating power for mean = null + difference
Alpha = 0.05 Assumed standard deviation = 1.5
Sample
Difference Size Power
2 5 0.786485
The power depends only on the difference between the true mean and the null mean, which is 12 – 10 = 2, and not on the means themselves. The power is 0.786. Note that the output specifies that this is the power for a one-tailed test.