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 = .17 and a simple random sample of 800 households will be selected from the population.

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 +/- .02 of the population proportion (to 4 decimals)?

c. What is the probability that the sample proportion will be within +/- .02 of the population proportion for a sample of 1600 households (to 4 decimals)?

Explanation / Answer

a)standard error of the proportion =(p(1-p)/n)1/2 =(0.17*0.83/800)1/2=0.0133

b)probability that the sample proportion will be within +/- .02 of the population proportion

=P(-0.02/0.0133<Z<0.02/0.0133)=P(-1.5060<Z<1.5060)=0.9340-0.0661 =0.8679

c) standard error of the proportion =(p(1-p)/n)1/2 =(0.17*0.83/1600)1/2 =0.0094

probability that the sample proportion will be within +/- .02 of the population proportion

=P(-0.02/0.0094<Z<0.02/0.0094)=P(-2.1297<Z<2.1297)=0.9834-0.0166 =0.9668

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