The Food Marketing Institute shows that 17% of households spend more than $100 p

Question

The Food Marketing Institute shows that 17% of households spend more than $100 per week on groceries. Assume the population proportion is p = 0.17 and a sample of 900 households will be selected from the population. Use z-table.

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

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

c. 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)?

Explanation / Answer

Ans:

a)standard error=sqrt(0.17*(1-0.17)/900)=0.0125

b)

z=0.02/0.0125=1.597

P(-1.597<=z<=1.597)=P(z<=1.597)-P(z<=-1.597)

=0.9449-0.0551=0.8897

c)

standard error=sqrt(0.17*0.83/1300)=0.0104

z=0.02/0.0104=1.92

P(-1.92<=z<=1.92)=P(z<=1.92)-P(z<=-1.92)

=0.9726-0.0274

=0.9451

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