A study is conducted to asses the extend to which patients who had coronary arte
ID: 3182168 • Letter: A
Question
A study is conducted to asses the extend to which patients who had coronary artery bypass surgery were maintaining their prescribed exercise program The following data reflect the number of times patients reported exercising over the previous month: 14, 11, 8, 6, 5, 3, 6, 13, 12, 8, 1, 4 (i) Compute the mean and standard deviation (ii) Assuming that the data arc approximately normal distributed, compute a 95% confidence interval for the population mean. (iii) Assuming that the data arc approximately normal distributed, we can use the data as a pilot study to decide on a sample size n for a future study to obtain a margin of error of 2 for a 95% confidence interval. Select the appropriate n. (iv) What would be the appropriate n for a 99% confidence interval? (v) Without additional computations conclude which confidence interval is wider: [Circle one:] 90% Confidence Interval 95% Confidence lateral [Circle one:] 80% Confidence Interval, n = 50 80% Confidence Interval, n = 10Explanation / Answer
a.
Mean(x)=7.5833
Standard deviation( sd )=4.1661
b.
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Sample Size(n)=12
Confidence Interval = [ 7.5833 ± t a/2 ( 4.1661/ Sqrt ( 12) ) ]
= [ 7.5833 - 2.201 * (1.203) , 7.5833 + 2.201 * (1.203) ]
= [ 4.936,10.23 ]
c.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 4.1661
ME =2
n = ( 1.96*4.1661/2) ^2
= (8.17/2 ) ^2
= 16.67 ~ 17
d.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.01% LOS is = 2.58 ( From Standard Normal Table )
Standard Deviation ( S.D) = 4.1661
ME =2
n = ( 2.58*4.1661/2) ^2
= (10.75/2 ) ^2
= 28.88 ~ 29
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