Question

A Food Marketing Institute found that 28% of households spend more than $125 a week on groceries. Assume the population proportion is 0.28 and a simple random sample of 93 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.25 and 0.44? Answer - (Enter your answer as a number accurate to 4 decimal places.) Question Help: D Post to forum Submit Question

Answer 1

p = 0.28

n = 93

$\mu_p=p=0.28$

$\sigma_p=\sqrt{\frac{p*(1-p)}{n}}=\sqrt{\frac{0.28*(1-0.28)}{93}}=0.0466$

$P(0.25<\widehat{p}<0.44)=P(\frac{0.25-\mu_p}{\sigma_p}<\frac{\widehat{p}-\mu_p}{\sigma_p}<\frac{0.44-\mu_p}{\sigma_p})$

                                   

0.25 – 0.28 = PG 0.0466 << 0.44 -0.28 0.0466

                                    = P(-0.64 < Z < 3.44)

                                    = P(Z < 3.44) - P(Z < -0.64)

                                    = 0.9997 - 0.2611

                                    = 0.7386