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Amiraneli [1.4K]
3 years ago
13

Suppose we want to estimate the average weight of an adult male in Mexico. We draw a random sample of 2,000 men from a populatio

n of 3,000,000 men and weigh them. We find that the average person in our sample weighs 200 pounds, and the standard deviation of the sample is 30 pounds. Calculate 94%,98%,96% confidence interval
Mathematics
1 answer:
LenKa [72]3 years ago
7 0

Answer:

1.- CI = 94% (  μ₀ -  1,04   <  x   <   μ₀ +  1,04 )

2.- CI = 98 % ( μ₀ - 2,05  <  x   <   μ₀ + 2,05 )

3.- CI = 96 %  (  μ₀ - 1,75  <  x   <   μ₀  + 1,75 )

Step-by-step explanation:

Sample size      n  = 3000000

Sample mean   x  =  200

Standard deviation   s  =  30

From  z-table values of z(c):

CI  94 %        Confidencial level   α = 6 %   α = 0,06   z(c) = 1,55

CI  98 %        Confidencial level   α = 2 %   α = 0,02   z(c) = 2,05

CI  96 %        Confidencial level   α = 4 %   α = 0,04   z(c) =  1,75

MOE = z(c) * σ/√n

1.-MOE =  1,55* 30 / √2000    MOE = 1,04

2.-MOE = 2,05*30/√2000      MOE = 1,38

3.-MOE = 1,75*30/√2000        MOE = 1,17

Then CI

1.-  CI = 94 %     (  μ₀ -  MOE <  x  <   μ₀ -  MOE )

     CI  =  (  μ₀ -  1,04   <  x   <   μ₀ +  1,04 )

2.-CI = 98 %

     CI  = (    μ₀ - 2,05  <  x   <   μ₀ + 2,05 )

3.- CI = 96 %

     CI = (  μ₀ - 1,75  <  x   <   μ₀  + 1,75 )

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