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.

Calculate (), the standard error of the proportion of households spending more than $100 per week on groceries (to 4 decimals).

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 1,800 households (to 4 decimals)?

Answer 1

1)

std error of proportion=σp=√(p*(1-p)/n)=

0.0126

2) probability that the sample proportion will be within +/- 0.02 of the population proportion :

probability =

P(0.13<X<0.17)

=

P(-1.58<Z<1.58)=

0.9429-0.0571=

0.8858

3)

sample size       =n=	1800
std error of proportion=σp=√(p*(1-p)/n)=	0.0084

probability =

P(0.13<X<0.17)

=

P(-2.38<Z<2.38)=

0.9913-0.0087=

0.9826