Question

In: Statistics and Probability

Develop a 90% confidence interval for the household income of subscribers. DATA Household Income ($) 58600...

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

Solutions

Expert Solution

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)


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