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. Calculate (), the standard error of the proportion of households spending more than $100 per week on groceries (to 4 decimals). 0.0130 What is the probability that the sample proportion will be within +/- 0.03 of the population proportion (to 4 decimals)? What is the probability that the sample proportion will be within +/- 0.03 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

= p = 0.16

=  [p ( 1 - p ) / n] = [(0.16 * 0.84) / 900 ] = 0.0130

a) P( 0.13 < < 0.19 )

= P[(0.13 - 0.16) /0.0130 < ( - ) / < (0.19 - 0.16) / 0.0130 ]

= P(-2.31 < z < 2.31)

= P(z < 2.31) - P(z < -2.31)

Using z table,   

= 0.9896 - 0.0104

= 0.9792

b) n = 1300

= p = 0.16

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

P( 0.13 < < 0.19 )

= P[(0.13 - 0.16) /0.0102 < ( - ) / < (0.19 - 0.16) / 0.0102 ]

= P(-2.94 < z < 2.94)

= P(z < 2.94) - P(z < -2.94)

Using z table,   

= 0.9984 - 0.0016

= 0.9968