3. Confidence interval for the population mean of paired differences Aa Aa Toure

Question

3. Confidence interval for the population mean of paired differences Aa Aa Tourette's syndrome is an inherited neuropsychiatric disorder with onset in childhood. It is characterized by the presence of multiple physical or motor tics and at least one vocal tic; these tics characteristically wax and wane. Tourette's used to be most commonly associated with the exclamation of obscene words or socially inappropriate and derogatory remarks. However, this symptom is present in only a small minority of people with Tourette's. Donald Gilbert, director of the Tourette's Syndrome Clinic at Cincinnati Children's Hospital Medical Center, and his colleagues tested a dopamine agonist to determine whether it would decrease tic severity. Suppose Gilbert and his team use a sample of n-40 patients with Tourette's and test them on the Yale Global Tic Severity Scale before and after 2 weeks of treatment with the dopamine agonist. The mean of the sample differences is a =-26.6, and the standard deviation of the sample differences is sd = 21.2. You will use these sample results to estimate how much success the dopamine agonist would have for the entire population. First, calculate the standard error. The standard error is s.e.(a) = Next, compute the sample estimate. The sample estimate is a | Select a Distribution Distributions 0123

Explanation / Answer

SE (dbar) = 21.2/sqrt(40) = 3.352

dbar = 26.6

The 95% confidence interval is (dbar - t(alpha/2) SE, dbar + t(alpha/2) SE)
= (26.6 - 2.0227*3.352, 26.6 + 2.0227*3.352) = (19.82, 33.38)

This means that you are 95% confidence that

n=80 the SE(dbar) = 21.2/sqrt(80) = 2.3702
The new 95% confidence interval is (dbar - t(alpha/2) SE, dbar + t(alpha/2) SE)
= (26.6 - 2.0227*2.3702, 26.6 + 2.0227*2.3702) = (21.8058,31.3942)

The new confidence interval is less width than the original one, because that new sample size is greater than the original sample size

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