The random sample shown below was selected from a normal distribution.
3,6,8,3,8,8
Complete parts a and b.
a. Construct a 99% confidence interval for the population mean μ.
(Round to two decimal places as needed.)
Solution:
Given that,
x | x2 |
3 | 9 |
6 | 36 |
8 | 64 |
3 | 9 |
8 | 64 |
8 | 64 |
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a ) The sample mean is
Mean
=
x / n)
= ( 3+6+8+3+8+8 / 6 )
= 36 / 6
= 6
Mean
= 6
The sample standard is S
S =
(
x2 ) - ((
x )2 / n ) / 1 -n )
=
( 246 ( -(36 )2 / 6 ) / 5
=
30 / 5
=6
= 2.45
The sample standard is S = 2.45
b )
= 6
s = 2.45
n = 6
Degrees of freedom = df = n - 1 = 6- 1 = 5
At 99% confidence level the z is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
t
/2,df = t0.005,5 =4.031
Margin of error = E = t/2,df
* (s /
n)
= 4.031* (2.45/
6 )
= 4.03
Margin of error = 1.00
The 99% confidence interval estimate of the population mean is,
- E <
<
+ E
6 - 4.03<
< 6 + 4.03
1.97<
< 10.03
(1.97, 10.03)
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