Question

1. A Food Marketing Institute found that 38% of households spend more than $125 a week on groceries. Assume the population proportion is 0.38 and a simple random sample of 56 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.26 and 0.39?
Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations.

2. In a recent year, the Better Business Bureau settled 75% of complaints they received. (Source: USA Today, March 2, 2009) You have been hired by the Bureau to investigate complaints this year involving computer stores. You plan to select a random sample of complaints to estimate the proportion of complaints the Bureau is able to settle. Assume the population proportion of complaints settled for the computer stores is the 0.75, as mentioned above. Suppose your sample size is 229. What is the probability that the sample proportion will be at least 2 percent more than the population proportion?
Note: You should carefully round any z-values you calculate to at least 4 decimal places to match wamap's approach and calculations.

Answer 1

Gisan RE 038 0.39-0.38 o.38xau62 852 0,) __2 < 0,15)- P(2< -1 O,o 322 0.S24 02sx0,25 229 220.