Question

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 population 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

Answer 1

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 )