Solution :
Given that,
Point estimate = sample mean = = 25.8
sample standard deviation = s = 2.8
sample size = n = 87
Degrees of freedom = df = n - 1 = 87 - 1 = 86
a.
At 95% confidence level
= 1 - 95%
=1 - 0.95 =0.05
/2
= 0.025
t/2,df
= t0.025,86 = 1.988
Margin of error = E = t/2,df * (s /n)
= 1.988 * (2.8 / 87)
Margin of error = E = 0.60
The 95% confidence interval estimate of the population mean is,
- E < < + E
25.8 - 0.60 < < 25.8 + 0.60
25.20 < < 26.40
( 25.20 , 26.40 )
The 95% confidence interval is ( 25.20 , 26.40 )
b.
At 90% confidence level
= 1 - 90%
=1 - 0.95 =00.10
/2
= 0.05
t/2,df
= t0.05,86 = 1.663
Margin of error = E = t/2,df * (s /n)
= 1.663 * (2.8 / 87)
Margin of error = E = 0.50
The 90% confidence interval estimate of the population mean is,
- E < < + E
25.8 - 0.50 < < 25.8 + 0.50
25.30 < < 26.30
( 25.30 , 26.30 )
The 90% confidence interval is ( 25.30 , 26.30 )
c.
At 99% confidence level
= 1 - 99%
=1 - 0.99 = 0.01
/2
= 0.005
t/2,df
= t0.005,86 = 2.634
Margin of error = E = t/2,df * (s /n)
= 2.634 * (2.8 / 87)
Margin of error = E = 0.79
The 99% confidence interval estimate of the population mean is,
- E < < + E
25.8 - 0.79 < < 25.8 + 0.79
25.01 < < 26.59
( 25.01 , 26.59 )
The 99% confidence interval is ( 25.01 , 26.59 )
A random sample of 87 observations produced a mean -25.8 and a standard deviations 28 a....
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