How can I solve the Statistics problem below on power and ss?

the difference between means is .25. St. deviation is .67. What sample size is needed for a detectable difference with 90% power and alpha .05?





Also, if n = 30 per group, alpha is .05, power is 90%, what is the minimum detectable difference?

How can I solve the Statistics problem below on power and ss?
1)





First find the rejection region. for a 5% significance two tail test it is ±1.96 standard units about the hypothesized mean difference, 0, in this case.





The power of the test is the probability of being in the rejection region when the null hypothesis is false.





This is much easier to explain when I can draw the images. But oh well.





Now, although the test is two tailed and both tails should be accounted for, the lower tail's contribution to the power is so small i will omit it from the calculations.





the upper rejection region is 1.96 * 0.67 / sqrt(n)


the lower end point for the power is 0.25 - 1.28 * 0.67 / sqrt(n)





set these two equations equal to each other and solve for n.





1.96 * 0.67 / sqrt(n) = 0.25 - 1.28 * 0.67 / sqrt(n)





n = 75.39796





n must be integer valued, always take the ceiling to make sure you have the correct size





n = 76








== -- == -- == -- ==





find the rejection region





0 ± 1.96 * 0.67 / sqrt(30)





Dbar %26lt; -0.2397564 or Dbar %26gt; 0.2397564





the smallest detectable difference is ± 0.2397564

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