Question

A Food Marketing Institute found that 35% of households spend more than $125 a week on groceries. Assume the population proportion is 0.35 and a simple random sample of 143 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.32 and 0.46?

Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations.

Answer 1

Population Percentage of households spending more than $125 a week = p = 35%      
       
Given p = 0.35         Population Proportion    
       
n = 143             Sample Size    
Let p̂ be the sample proportion      
To find       
P(0.32 < p̂ < 0.46)      
Converting to z-scores      
= P(z1 < Z < z2)      
= P(Z < z2) - P(Z < z1)      
where      
  
z1 = -0.7521      
z2 = 2.7578      
P(0.32 < p̂ < 0.46) = P(Z < 2.7578) - P(Z < -0.7521)      
We use Excel function NORM.S.DIST to find the probability      
                 = NORM.S.DIST*(2.7578, TRUE) - NORM.S.DIST*(-0.7521, TRUE)      
                = 0.7711      
       
P(sample proportion of households spending
more than $125 a week is between 0.32 and 0.46) = 0.7711