Question

The Food Marketing Institute shows that 15% of households spend more than \$100 per week on...

 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)?

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

3)

 sample size       =n= 1800 std error of proportion=σp=√(p*(1-p)/n)= 0.0084
 probability = P(0.13