Question

The Food Marketing Institute shows that 15% of households spend more than $100 per week on groceries. Assume the population proportion is p = 0.15 and a sample of 600 households will be selected from the population. Use z-table.

1. Calculate (), the standard error of the proportion of households spending more than $100 per week on groceries (to 4 decimals).
     
   
2. What is the probability that the sample proportion will be within +/- 0.03 of the population proportion (to 4 decimals)?
     
   
3. What is the probability that the sample proportion will be within +/- 0.03 of the population proportion for a sample of 1,200 households (to 4 decimals)?
   

Answer 1

Solution

Given that,

p = 0.15

1 - p = 1 - 0.15 = 0.85

n = 600

a)   = p = 0.15

  [p ( 1 - p ) / n] = [(0.15 * 0.85) / 600 ] = 0.0146

b) P( 0.12 < < 0.18)

= P[(0.12 - 0.15) /0.0146 < ( - ) / < (0.18 - 0.15) / 0.0146 ]

= P(-2.05 < z < 2.05)

= P(z < 2.05 ) - P(z < -2.05)

Using z table,   

= 0.9798 - 0.0202

= 0.9596

c) n = 1200

   = p = 0.15

  [p ( 1 - p ) / n] = [(0.15 * 0.85) / 1200 ] = 0.0103

P( 0.12 < < 0.18)

= P[(0.12 - 0.15) /0.0103 < ( - ) / < (0.18 - 0.15) / 0.0103 ]

= P(-2.91 < z < 2.91)

= P(z < 2.91 ) - P(z < -2.91)

Using z table,   

= 0.9982 - 0.0018

= 0.9964