Homework SourseIn a random sample of 45 refrigerators the mean repair cost
In a random sample of 45 refrigerators, the mean repair cost was $142.00 and the standard deviation was $19.50. A 90% confidence interval for the population mean repair cost is (137.22, 146.78). change the sample size to n=90. Construct a 90% confidence interval for the population mean repair cost. Which confidence interval is wider?
Solution
a)
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.05
X = sample mean = 143
z(alpha/2) = critical z for the confidence interval = 1.644853627
s = sample standard deviation = 19.5
n = sample size = 45
Thus,
Margin of Error E = 4.78140588
Lower bound = 138.2185941
Upper bound = 147.7814059
Thus, the confidence interval is
( 138.2185941 , 147.7814059 ) [ANSWER]
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b)
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.05
X = sample mean = 143
z(alpha/2) = critical z for the confidence interval = 1.644853627
s = sample standard deviation = 19.5
n = sample size = 90
Thus,
Margin of Error E = 3.380964521
Lower bound = 139.6190355
Upper bound = 146.3809645
Thus, the confidence interval is
( 139.6190355 , 146.3809645 ) [ANSWER]
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The confidence interval of n = 45 is wider. [ANSWER]
