mong 500 marriage license applications chosen at random in a givenyear, there were 48 in which the woman was at least one year older than the man, and among400 marriage license applications chosen at random six years later, there were 68 in which thewoman was at least one year older than the man. Construct a 99% confidence interval for thedifference between the corresponding true proportions of marriage license applications in whichthe woman was at least one year older than the man. Interpret the CI in the context of theproblem.

Answer:

CI 99% = (  0,022  ;  0,126 )

Step-by-step explanation:

First sample  

n₁  = 500

x₁  = 48

p₁ = x₁ / n₁   =  48 / 500      p₁  =  0,096      p₁  = 9,6 %

Second sample

n₂  = 400

x₂  = 68

p₂ = x₂ / n₂   =  68 / 400      p₂  =  0,17       p₂  = 17 %

CI = 99 %    significance level α = 1 %     α  = 0,01

z(c)  for α = 0,01  is from z- table    z(c) = 2,325

CI  =  ( p₂ - p₁ ) ±  z(c) *√ p*q* ( 1/n₁  +  1 / n₂ )

Where

p₂  -  p₁  =  0,17 - 0,096 = 0,074

p = ( x₁ + x₂ )  / n₁ + n₂

p =  ( 48  +  68  ) /( 500 + 400)

p = 116/ 900     p = 0,1288     and   q = 1 - p     q  = 0,8712

z(c) *√ p*q* ( 1/n₁  +  1 / n₂ ) = 2,325 * √ 0,1288*0,8712 ( 1 / 500 + 1/ 400)

2,235 * 0,02247

z(c) *√ p*q* ( 1/n₁  +  1 / n₂ ) = 0,052

Then

CI 99 %  =  0,074  ±  0,052

CI 99% = (  0,022  ;  0,126 )

The difference between the groups shows that the proportion in the second group was bigger than in the first group.

The CI in the context of the problem is  CI 99% = (  0,022  ;  0,126 )

Given first sample is  

n₁  = 500

x₁  = 48

  [tex]P_{1} =\dfrac{X_{1} }{n_{1} }[/tex]      [tex]P_{1} =\dfrac{48}{500}[/tex]      

p₁  =  0,096      p₁  = 9,6 %

Given second sample

n₂  = 400

x₂  = 68

[tex]P_{2} =\dfrac{X_{2} }{n_{2} }[/tex]      [tex]P_{2} =\dfrac{68}{400}[/tex]        

p₂  =  0,17       p₂  = 17 %

Since given CI = 99 %    so significance level  α = 1 %     α  = 0,01

From Z-Table  z(c)  for α= 0,01  is = 2,325

CI  =   [tex](P_{2} -P_{1}[/tex]   ±    [tex]Z(c)\sqrt[2]{p\times q} (\dfrac{1}{n_{1} } +\dfrac{1}{n_{2} } )[/tex]  

Where

p₂  -  p₁  =  0,17 - 0,096 = 0,074

[tex]P= \dfrac{X_{1} +X_{2} }{n_{1}+n_{2} }[/tex]

[tex]P=\dfrac{48+68}{500+400}[/tex]

[tex]P=\dfrac{116}{900}[/tex]

p = 0,1288     and   q = 1 - p     q  = 0,8712

[tex]Z(c)\sqrt[2]{p\times q} (\dfrac{1}{n_{1} } +\dfrac{1}{n_{2} } )[/tex]  [tex]2325\times\sqrt[2]{0.1288\times 0.8712} (\dfrac{1}{500_{} } +\dfrac{1}{400_{} } )[/tex]

  [tex]Z(c)\sqrt[2]{p\times q} (\dfrac{1}{n_{1} } +\dfrac{1}{n_{2} } )=0.052[/tex]

Then

CI 99 %  =  0,074  ±  0,052

CI 99% = (  0,022  ;  0,126 )

Hence the difference between the groups shows that the proportion in the second group was bigger than in the first group.

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