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 800 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 1,400 households (to 4 decimals)?

Answer 1

(1) the proportion of success p = 0.15 so q= 1-p = 1 - 0.15 = 0.85 and sample size n = 800

standard error of the proportion of households spending = sqrt [ p x q / n ]

= squrt [ 0.15*0.85/800] = 0.0126

(2) What is the probability that the sample proportion will be within +/- 0.02 of the population proportion (to 4 decimals)

sample proportion will be within +/- 0.02 means 0.15 + or - 0.02

or we need to estimate z(0.17) and z(0.13)

z(0.17)= difference / standard error = (0.17-0.15) / squrt [ 0.15*0.85/800] = 1.5873

z(0.13) = - 1.5873

(3) What is the probability that the sample proportion will be within +/- 0.02 of the population proportion for a sample of 1,400 households (to 4 decimals)

the new sample size is 1400

standard error of the proportion of households spending = sqrt [ p x q / n ]

= sqrt [ 0.15*0.85/1400] = 0.0095

sample proportion will be within +/- 0.02 means 0.15 + or - 0.02

or we need to estimate z(0.17) and z(0.13)

z(0.17)= difference / standard error = (0.17-0.15) / squrt [ 0.15*0.85/1400] = 2.1053

z(0.13) = - 2.1053