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.
1)
for normal distribution z score =(p̂-p)/σ_{p} | |
here population proportion= p= | 0.750 |
sample size =n= | 120 |
std error of proportion=σ_{p}=√(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=σ_{p}=√(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 |
Get Answers For Free
Most questions answered within 1 hours.