Question

The Federal Election Commission collects information about campaign contributions and disbursements for candidates and politiDetermine the mean and standard deviation of the sample: mean: 568872.5 standard deviation: 829087.11 (Round to the 2nd decimIf we took this same sample 200 times, using a 95% confidence level, how many confidence intervals would you expect to be wro

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Answer #1

1)

Sample Mean,    x̅ = ΣX/n =    568872.5

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   839649.16

2)

point estimate=568872.50

3) option D)

answer: 200*5% = 10

5)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   39          
't value='   tα/2=   2.0227   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   839649.1557   / √   40   =   132760.1884
margin of error , E=t*SE =   2.0227   *   132760.1884   =   268532.8276
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    568872.50   -   268532.827552   =   300339.6724
Interval Upper Limit = x̅ + E =    568872.50   -   268532.827552   =   837405.3276
95%   confidence interval is (   300340 < µ <   837405 )

6)

margin of error , E=t*SE =   2.0227   *   132760.1884   =   268532.83

7) 568872.50 ± 268532.83

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