Question

Researchers are studying populations of two squirrels, the eastern gray and the western gray. For the eastern gray squirrel, about 22 percent of the population weighs over 0.5 kilogram (kg). For the western gray squirrel, about 36 percent of the population weighs over 0.5 kg. A random sample of 60 squirrels will be selected from the population of eastern gray squirrels, and a random sample of 120 squirrels will be selected from the population of western gray squirrels. What is the mean of the sampling distribution of the difference in sample proportions (eastern minus western)

Answer 1

Answer:

p₁  -  p₂  = ( -0,28 . 0 )

Step-by-step explanation:

Eastern gray squirrel

Sample size    n₁  = 60

p₁  =  22 %        p₁  = 0,22       then     q₁  = 1 - 0,22      q₁ = 0,88

Western gray squirrel

Sample size  n₂  =  120

p₂ = 36%     p₂ = 0,36     Then    q₂  =  1  - 0,36     q₂  =  0,64

p (s)  = ( p₁ - p₂ ) /√ (p₁*q₁ ) / n₁   +  (p₂*q₂/ n₂)

p(s)  = ( 0,22 -  0,36 ) /√ (0,22*0,88)/60  +  ( 0,36*0,64)/120

p(s)  =  -0,14 /0,071

p(s) = - 1,9718

Then

CI = (  p₁ - p₂ ) ± p(s) *EED

EED =  /√ (p₁*q₁ ) / n₁   +  (p₂*q₂/ n₂)   = 0,071

CI = (  - 0,14  ±  1,9718 * 0,071 )

CI =  ( - 0,14  ±  0,14 )

CI =  ( - 0,28 ;  0 )

we will find with 95% of confidence the mean of the sampling distribution of the difference in sample proportion in the interval  ( -0,28 , 0 )

p₁  -  p₂  = ( -0,28 . 0 )