Question

A store surveys customers to see if they are satisfied with the service they receive. Samples of 75 surveys are taken. One in six people are unsatisfied. What is the variance of the mean of the sampling distribution of sample proportions for the number of unsatisfied customers? What is the variance for satisfied customers?

Answer 1

Consider

P₁ : Population propertion of unsatisfied customers = 1/6.

P₂ : Population propertion of satisfied customers = 5/6.

n= number of customers in sample = 75.

X : Number of unsatisfied customers in the sample.

The probability distribution of number of unsatisfied customers (X) is binomial with P₁ =1/6 and n= 75.

i.e. X ~ Bin ( n = 75 , P₁ = 1/6)

E(X) = n * P₁ = 75 * 1/6 = 12.5

Var(X) = n * P₁ * ( 1-P₁) = 75 * 1/6 *5/6 = 10.4167.

let p₁ be the sample propertion of unsatisfied customers.

p₁ = X /n

E(p₁) = E ( X/n) = 1/n E(X) = 1/n * n*P₁ = P₁ =1/6.

Var(p₁) = Var(X/n) =( 1/n²) * ( n* P₁ * (1-P₁)) = ( P₁ * (1-P₁) ) / n = ( 1/6 * 5/6 ) / 75 = 0.0019.

The variance sample propertion of unsatisfied customer is 0.0019.

The sampling distribution of sample propertion of unsatisfied customers is Normal.

Y : Number of satisfied customers in the sample.

The probability distribution of number of satisfied customers (Y) is binomial with P₂ =5/6 and n= 75.

i.e. Y ~ Bin ( n = 75 , P2 = 5/6)

E(Y) = n * P₂ = 75 * 5/6 = 62.5

Var(Y) = n * P2* ( 1-P₂) = 75 * 5/6 *1/6 = 10.4167.

let p₂ be the sample propertion of satisfied customers.

p₂ =Y /n

E(p₂) = E (Y/n) = 1/n E(Y) = 1/n * n*P₂ = P₂ =5/6.

Var(p₂) = Var(Y/n) =( 1/n²) * ( n* P₂ * (1-P₂)) = ( P₂ * (1-P₂) ) / n = (5/6 *1/6 ) / 75 = 0.0019

The variance sample propertion of satisfied customer is 0.0019.

The sampling distribution of sample propertion of satisfied customers is Normal.