Homework SourseTwentyfive patients with congestive heart failure were weigh
Twenty-five patients with congestive heart failure were weighed before and after receiving a
new diuretic agent, and the average weight loss (i.e. difference between the two weights) for
this sample was found to be 2 lbs. with a standard deviation of 1.38 lbs.
Assuming the weight loss to be normally distributed find a 90% confidence interval for
the mean weight loss.
If a 90% confidence interval of width 0.75 is desired and the standard deviation is 1.5
lbs., find the number of patients that will be needed.
The following percentiles of the t-distribution with 24 DF might be needed:
t.975(24) = 2.064; t.95(24) = 1.711; t.90(24) = 1.318
The following percentiles of the standard normal distribution might be needed:
z.975 = 1.96; z.95 = 1.645 ; z.90 = 1.282
Solution
A)
Note that
Margin of Error E = t(alpha/2) * s / sqrt(n)
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.05
X = sample mean = 2
t(alpha/2) = critical t for the confidence interval = 1.71088208
s = sample standard deviation = 1.38
n = sample size = 25
df = n - 1 = 24
Thus,
Margin of Error E = 0.472203454
Lower bound = 1.527796546
Upper bound = 2.472203454
Thus, the confidence interval is
( 1.527796546 , 2.472203454 ) [ANSWER]
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b)
Note that
n = z(alpha/2)^2 s^2 / E^2
where
alpha/2 = (1 - confidence level)/2 = 0.05
Using a table/technology,
z(alpha/2) = 1.644853627
Also,
s = sample standard deviation = 1.38
E = margin of error = 0.375
Thus,
n = 36.63955167
Rounding up,
n = 37 [ANSWER]
