A Food Marketing Institute found that 29% of households spend more than $125 a w

Question

A Food Marketing Institute found that 29% of households spend more than $125 a week on groceries. Assume the population proportion is 0.29 and a simple random sample of 328 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 0.27? Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations AnswerEnter your answer as a number accurate to 4 decimal places.) Licens Points possible: 1 Unlimited attempts. Submit 7 delete

Explanation / Answer

Sol:

n=328

p^=0.27

p=0.29

z=p^-p/sqrt(p(1-p)/n

=0.27-0.29/sqrt(0.29*(1-0.29)/328

=-0.7982

P(Z<-0.7982)

=1-P(Z<0.7982)

=1-0.7881

=0.2119

ANSWER:0.2119

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