The Food Marketing Institute shows that 15% of households spend more than $100 p

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 600 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)?

Explanation / Answer

1)for n=600

standard error of the proportion=(p(1-p)/n)1/2 =(0.15*(!-0.15)/600)1/2 =0.0146

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

=P(-0.02/0.0146<Z<0.02/0.0146)=P(-1.3720<Z<1.3720)=0.9150-0.0851 =0.8299

2) for n=1800

standard error of the proportion=(p(1-p)/n)1/2 =(0.15*(!-0.15)/1800)1/2 =0.0084

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

=P(-0.02/0.0084<Z<0.02/0.0084)=P(-2.3764<Z<2.3764)=0.9913-0.0087 =0.9825

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.