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.

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

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

ans.....
(1) 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
(2) 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