In: Statistics and Probability
Develop a 90% confidence interval for the household income of subscribers. DATA
Household Income ($) |
58600 |
54200 |
38500 |
148600 |
82100 |
64400 |
86200 |
177100 |
68300 |
67900 |
57300 |
83600 |
77500 |
68200 |
61900 |
57600 |
82300 |
64600 |
61100 |
31200 |
92600 |
68300 |
35100 |
85700 |
140300 |
108200 |
61100 |
33900 |
54400 |
61200 |
58000 |
90700 |
95200 |
50500 |
33800 |
147400 |
92600 |
66200 |
45700 |
60500 |
110600 |
60300 |
75700 |
70100 |
42100 |
41700 |
96900 |
65700 |
50200 |
61700 |
44500 |
51900 |
119100 |
49200 |
39000 |
35000 |
104700 |
49300 |
74000 |
57100 |
51400 |
62100 |
103000 |
97900 |
123100 |
322500 |
54800 |
66500 |
33700 |
73600 |
71300 |
74200 |
70000 |
40800 |
72500 |
53300 |
45600 |
73900 |
83600 |
124700 |
101600 |
205900 |
69700 |
95700 |
46100 |
118600 |
65400 |
149300 |
125000 |
39800 |
83500 |
38700 |
102400 |
57700 |
16200 |
43100 |
43700 |
39600 |
127500 |
33500 |
48100 |
52800 |
54800 |
46500 |
60400 |
From the given data the following were calculated
n | 105 |
Sum | 7835500 |
Average | 74623.810 |
SS(Sum of squares) | 1.74735E* 1011 |
Variance = SS/n-1 | 1680141254.579 |
Std Dev=Sqrt(Variance) | 40989.5262 |
_________________________________________________________________
sample mean () = 74623.81, Standard Deviation (s) = 40989.526, n = 105
The Zcritical at = 0.10 = 1.645
Although population standard deviation is not known, n is large and we can assume Z critical values
The Confidence Interval is given by ME, where
The Lower Limit = 74623.810 - 6580.287 = 68043.533
The Upper Limit = 74623.810 + 6580.287 = 81204.087
The Confidence Interval is (68043.533 , 81204.087)
_____________________________________________________
If required to 2 decimal places. then the 90% CI is (68043.53, 81204.09)
If required to 1 decimal places. then the 90% CI is (68043.5, 81204.1)
If required to the nearest dollar. then the 90% CI is (68044, 81204)