Question
A simple random sample with N=52 provided a sample mean of 21.0 and a sample standard...
A simple random sample with N=52 provided a sample mean of 21.0 and a sample standard deviation of 4.4.
a. Develop a 90% confidence interval for the population mean (to 1 decimal).
(________, _________)
b. Develop a 95% confidence interval for the population mean (to 1 decimal).
(________, _________)
c. Develop a 99% confidence interval for the population mean (to 1 decimal).
(________, _________)
d. What happens to the margin of error and the confidence interval as the confidence level is increased?
Homework Answers
Answer #1
a)
90% confidence interval for 
is
- t * S / sqrt(n) <
<
+ t * S / sqrt(n)
21 - 1.675 * 4.4 / sqrt(52) <
< 21 + 1.675 * 4.4 / sqrt(52)
20.0 <
< 22.0
90% CI is ( 20.0 , 22.0)
b)
95% confidence interval for
is
- t * S / sqrt(n) <
<
+ t * S / sqrt(n)
21 - 2.008 * 4.4 / sqrt(52) <
< 21 + 2.008 * 4.4 / sqrt(52)
19.8 <
< 22.2
95% CI is (19.8 , 22.2)
c)
99% confidence interval for
is
- t * S / sqrt(n) <
<
+ t * S / sqrt(n)
21 - 2.676 * 4.4 / sqrt(52) <
< 21 + 2.676 * 4.4 / sqrt(52)
19.4 <
< 22.6
99% CI is ( 19.4 , 22.6)
d)
As confidence level increases, margin of error and the width of confidence interval both increases.
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