Question
A simple random sample with n=56 provided a sample mean of 23.5 and a sample standard...
A simple random sample with n=56 provided a sample mean of 23.5 and a sample standard deviation of 4.5.
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
Solution :
a.
t
/2,df = 1.673
Margin of error = E = t/2,df
* (s /
n)
= 1.673 * (4.5 /
56)
Margin of error = E = 1.0
The 90% confidence interval estimate of the population mean is,
- E < 
<
+ E
23.5 - 1.0 <
< 23.5 + 1.0
22.5 <
< 24.5
A 90% confidence interval for the population mean : (22.5 , 24.5)
b.
t
/2,df = 2.004
Margin of error = E = t/2,df
* (s /n)
= 2.004 * (4.5 /
56)
Margin of error = E = 1.2
The 95% confidence interval estimate of the population mean is,
- E <
<
+ E
23.5 - 1.2 <
< 23.5 + 1.2
22.3 <
< 24.7
A 95% confidence interval for the population mean : (22.3,24.7)
c.
t
/2,df = 2.668
Margin of error = E = t/2,df
* (s /n)
= 2.668 * (4.5 /
56)
Margin of error = E = 1.6
The 99% confidence interval estimate of the population mean is,
- E <
<
+ E
23.5 - 1.6 <
< 23.5 + 1.6
21.9 <
< 25.1
A 99% confidence interval for the population mean :(21.9 , 25.1)
d.
Margin of error and the confidence interval also increased
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