In: Statistics and Probability
A.
If
Upper X overbar
=61,
Upper S=27,
and
n=49,
and assuming that the population is normally distributed, construct a
90 %
confidence interval estimate of the population mean,
B. If
n=200
and
X=40,
construct a
99%
confidence interval estimate of the population proportion.
C.If
n=400
and
X=140,
construct a
99%
confidence interval estimate of the population proportion.
Solution :
Given that,
A )
= 61
s = 27
n = 49
Degrees of freedom = df = n - 1 = 49 - 1 = 48
At 90% confidence level the z is ,
= 1 - 90% = 1 - 0.90 = 0.10
/ 2 = 0.10 / 2 = 0.05
t
/2,df = t0.05,48 =1.677
Margin of error = E = t/2,df
* (s /
n)
= 1.677 * (27 /
49)
= 6.468
Margin of error = 6.468
The 90% confidence interval estimate of the population mean is,
- E <
<
+ E
61- 6.468 <
< 61 + 6.468
54.532 <
< 67.468
(54.532 , 67.468 )
B ) Given that,
n = 200
x = 40
= x / n = 40 / 200 = 0.200
1 -
= 1 - 0.200 = 0.800
At 99% confidence level the z is ,
= 1 - 99% = 1 - 0.99 =
0.01
/ 2 = 0.01 / 2 = 0.005
Z/2
= Z0.005 = 2.576
Margin of error = E = Z
/ 2 *
((
* (1 -
)) / n)
= 2.576 * (((0.200
* 0.800) / 200)
= 0.073
A 99 % confidence interval for population proportion p is ,
- E < P <
+ E
0.200 - 0.073 < p < 0.200 + 0.073
0.127 < p < 0.273
The 99% confidence interval for the population proportion p is : ( 0.127 , 0.200)
C ) Given that,
n = 400
x = 140
= x / n = 140 / 400 = 0.350
1 -
= 1 - 0.350 = 0.650
At 99% confidence level the z is ,
= 1 - 99% = 1 - 0.99 =
0.01
/ 2 = 0.01 / 2 = 0.005
Z/2
= Z0.005 = 2.576
Margin of error = E = Z
/ 2 *
((
* (1 -
)) / n)
= 2.576 * (((0.350
* 0.650) / 400)
= 0.061
A 99 % confidence interval for population proportion p is ,
- E < P <
+ E
0.350 - 0.061 < p < 0.350 + 0.061
0.289 < p < 0.411
The 99% confidence interval for the population proportion p is : ( 0.289, 0.411)