In a 5-year follow-up of bilateral stimulation of the subthalamic nucleus among

Question

In a 5-year follow-up of bilateral stimulation of the subthalamic nucleus among 49 patients with advanced Parkinson's disease, investigators assessed changes in symptoms as measured by the Unified Parkinson's Disease Rating Scale (range = 0 to 108, with higher values denoting more symptoms). Assume this measure follows a normal distribution. The mean score at baseline was 55.7. The standard deviation of change from baseline to 5 years was 15.3 1. How much power does this study have to detect a mean difference of 5 units in the symptom scale 2. What minimu|n sample size would be needed to detect a mean change of 5 units with 80% power 3. The above study had no control group. Assuming that the same standard deviation of change would if a two-sided test is used with a = 0.05? if a two-sided test is used with a = 0.05? occur among controls and the mean change among controls0, how many participants would be necessary to detect a mean difference of 5 units of change between those receiving stimulation and a control group of equal size with 80% power if a two-si ed test is used with a = 0.05?

Explanation / Answer

Solution:

1) We have n = 49, = 5, = 15.3
Power = (-z 1 – /2 + /2/n)
Power = (-z 1 – 0.05/2 + 5/(15.3)2/49)
Power = (-z0.975 + 5/(15.3)2/49)
Power = (-1.96+ 2.2876)
Power = (0.326)
Power = 0.628

2) n = (2/k) (z1 – /2 + z1- )2/²
n = (15.32/1) (z0.975+ z0.20)2/52
n = (15.3)2(1.96 + 0.84)2/25
n = 73.41 ~ 74

3) n = (²/k) (z1 – /2 + z1- )2/²
n = (15.32/1) (z0.975+ z0.20)2/52
n = (15.3)2(1.96 + 0.84)2/25
n = 73.41 ~ 74 participants
Both of them require to calculate the sample size or number of participants with the mean difference of 5 units with same power of 80% and level of significance = 0.05.

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