Question

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 120. What is the probability that the sample proportion will be within 3 percentage points of the population proportion?

Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations.

A Food Marketing Institute found that 26% of households spend more than $125 a week on groceries. Assume the population proportion is 0.26 and a simple random sample of 366 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is less than 0.24?

Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations.

Answer 1

1)

for normal distribution z score =(p̂-p)/σp
here population proportion=     p=	0.750
sample size       =n=	120
std error of proportion=σp=√(p*(1-p)/n)=	0.0395

probability that the sample proportion will be within 3 percentage points of the population proportion :

probability =

P(0.72<X<0.78)

=

P(-0.76<Z<0.76)=

0.7761-0.2239=

0.5521

( please try 0.5522 if this comes wrong and revert)

2)

here population proportion=     p=	0.260
sample size       =n=	366
std error of proportion=σp=√(p*(1-p)/n)=	0.0229

probability that the sample proportion of households spending more than $125 a week is less than 0.24 :

probability =

P(X<0.24)

=

P(Z<-0.87)=

0.1915