Question

Overweight Men For a random sample of 55 overweight men, the mean of the number of pounds that they were overweight was 29. The standard deviation of the population is 4.4 pounds. (a) Find the best point estimate of the mean. (b) Find the 95% confidence interval of the mean of these pounds. (c) Find the 99% confidence interval of the mean of these pounds. (d) Which interval is larger? Why?

Answer 1

a)

best point estimate of the mean = 29

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b)

population std dev ,    σ =    4.4000
Sample Size ,   n =    55
Sample Mean,    x̅ =   29.0000

Level of Significance ,    α =    0.05          
'   '   '          
z value=   z α/2=   1.960   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   4.4/√55=   0.5933          
margin of error, E=Z*SE =   1.9600   *   0.5933   =   1.163
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    29.00   -   1.1628   =   27.8372
Interval Upper Limit = x̅ + E =    29.00   -   1.1628   =   30.1628
95%   confidence interval is (   27.84   < µ <   30.16   )
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c)

Level of Significance ,    α =    0.01          
'   '   '          
z value=   z α/2=   2.576   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   4.4/√55=   0.5933          
margin of error, E=Z*SE =   2.5758   *   0.5933   =   1.528
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    29.00   -   1.5282   =   27.4718
Interval Upper Limit = x̅ + E =    29.00   -   1.5282   =   30.5282
99%   confidence interval is (   27.47   < µ <   30.53   )
....................

d)

99% confidence interval is larger

....................

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