Question

The Food Marketing Institute shows that 16% of households spend more than $100 per week on groceries. Assume the population proportion is p = 0.16 and a sample of 900 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 +/- 0.02 of the population proportion (to 4 decimals)?
   
   
3. What is the probability that the sample proportion will be within +/- 0.02 of the population proportion for a sample of 1,300 households (to 4 decimals)?

Answer 1

Solution

Given that,

p = 0.16

1 - p = 1 - 0.16 = 0.84

n = 900

a)   = p = 0.16

=  [p ( 1 - p ) / n] = [(0.16 * 0.84) / 900 ] = 0.0122

b) 0.16  ± 0.02 = 0.14, 0.18

P( 0.14 <   < 0.18 )

= P[(0.14 - 0.16) / 0.0122 < ( - ) / < (0.18 - 0.16) / 0.0122 ]

= P( -1.64 < z < 1.64)

= P(z < 1.64) - P(z < -1.64)

Using z table,   

= 0.9495 - 0.0505

= 0.8990

c) n = 1300

   = p = 0.16

=  [p ( 1 - p ) / n] = [(0.16 * 0.84) / 1300 ] = 0.0102

0.16  ± 0.02 = 0.14, 0.18

P( 0.14 <  ​​​​​​​ < 0.18 )

= P[(0.14 - 0.16) / 0.0102 < ( - ) / < (0.18 - 0.16) / 0.0102 ]

= P( -1.96 < z < 1.96)

= P(z < 1.96) - P(z < -1.96)

Using z table,   

= 0.9750 - 0.0250

= 0.9500