Question

Find the power for a difference in means of 2.79 using 95% and 99% confidence intervals.

[11.540 8.203 8.214 13.165 11.451 13.015 11.060 10.488 8.849 8.271]

[4.708 8.013 9.886 7.026 6.051 5.546 7.914 9.951 9.880 7.381]

Answer 1

In order to calculate the Power, first we have to calculate the mean and standard deviation for the two groups.

Mean for Group 1 =

Mean for Group 2 = (4.708+...+7.381)/10 = 7.63

Sample Std.dev for Group 1 = =3.199

Sample std.dev for Group 2 = 1.87

Next we have to construct the confidence interval for 95% and 99%

Group 1:

For 95% confidence interval: = 9.82 1.833(3.199)

For 99% confidence interval : 9.823.25(3.199)

Similarly, confidence interval for group can be constructed.

To calculate Power:

Power of a test is that is probability of correctly rejecting the null when it is false

Find the P-value for group 1: Null hypothesis   vs

The P value for the two groups are

Assumed mean = sample mean + 2.79

Let the U = ((mean - std.error) - assumed mean) / (std.dev/sqrt(n))

V= ((mean + std.error) - assumed mean) / (std.dev/sqrt(n))

Then, P = CDF (U) - CDF (V) which gives the p-value as follows

95% - (0.188,0.0991)

99% - (0.522, 0.046)

Hence the Power for difference in mean is: 1-Pvalue

For Group 1 : 95% - 81.2 , 99% - 47.8

For Group 2: 95% - 99.1 , 99% - 95.4