lab4_exercises_new

Lab 4 Exercises A. Use QUANTO (available on the virtual desktop) for the two following questions 1. Enter the settings described in Section F of the Lab4 Notes. What is the required N for each of the following changes? (for each below, make only the indicated change to the base settings described in Section F. Revert back to the Section F settings before addressing the next question) a. Increase the desired power to 90% N=____373______ b. Increase the treatment difference to 15 N=____122______ c. Increase the standard deviation to 50 N=____781______ d. Decrease the Type I error to 0.01 N=____415______ e. Assume a 1-sided alternative hypothesis N=____219______ Relative to the sample size N=279 from our base model settings, why do each of the above N’s make sense based on the model parameter that we are changing (the explanation for a. is given as an example)? a. To have a greater chance of detecting a difference, we need larger N b. Increase in treatment difference among N is smaller because shows the strength of effect size. c. Increase in mean of sample and SD increases N d. Type 1 error decreased N increases and margin for error is increased and significance level is decreased. e. One sided hypothesis has N smaller is only testing one side of the sample 2. Revert back to the Section F settings. What is the power to detect the desired treatment difference for the following sample sizes? N=100 Power =____0.39_______ N=200 Power =____0.66_______ N=400 Power =____0.919______

3. Again, revert back to the Section F settings. In the Outcome dialog, click the R- squared radio button. Power will now be computed as a function of the R- squared between Y and treatment. Set R-squared to 3% (0.03). What is the required sample size? N=258 What is the required N for each of the following changes? a. Increase the standard deviation to 50 N=___258_________ b. Change the proportion treated to 30% N=___116_________ What do you notice about the effect of each of these changes on required N when you have constrained the value of R-squared? I noticed that when we constrained the value of R-squared increased N also increase. As power increased to detect the desired treatment difference N also increased. However, when proportion treated increased N decreased, there was no change when standard deviation increased. Why is this so? (Hint: recall the formula for computing a t-statistic based on a correlation coefficient) I think it’s because the greater the proportion the lower the t-statistic mean difference is between the N of variables is so small. Therefore, the correlation coefficient relationship strength is decreased between the two variables

zbeta =probit(power); N =(( 4 *pibar)*( 1 -pibar)*(zalpha + zbeta)** 2 )/(delta** 2 ); proc print data =a; run ; 3) How does your estimate in 1 change if we reduce the power to 80%? ___1538.38___ data a; pi= 0.03 ; pi2= 0.01 ; delta=pi-pi2; pibar=(pi+pi2)/ 2 ; alpha = 0.05 ; sides = 2 ; power = 0.80 ; zalpha =probit( 1 -alpha/sides); zbeta =probit(power); N =(( 4 *pibar)*( 1 -pibar)*(zalpha + zbeta)** 2 )/(delta** 2 ); proc print data =a; run ; 4) How does your estimate in 1 change if we require a stricter significance level of 0.01? ___2916.36____ data a; pi= 0.03 ; pi2= 0.01 ; delta=pi-pi2; pibar=(pi+pi2)/ 2 ; alpha = 0.01 ; sides = 2 ; power = 0.90 ; zalpha =probit( 1 -alpha/sides); zbeta =probit(power); N =(( 4 *pibar)*( 1 -pibar)*(zalpha + zbeta)** 2 )/(delta** 2 ); proc print data =a; run ; 5) How does your estimate in 1 change if we can assume a 1-sided alternative? ___1678.51____ data a; pi= 0.03 ; pi2= 0.01 ; delta=pi-pi2; pibar=(pi+pi2)/ 2 ; alpha = 0.05 ; sides = 1 ; power = 0.90 ; zalpha =probit( 1 -alpha/sides); zbeta =probit(power); N =(( 4 *pibar)*( 1 -pibar)*(zalpha + zbeta)** 2 )/(delta** 2 ); proc print data =a; run ;