Question

for youl review Ul Question with last attempt is displayed A Food Marketing Institute found that 25% of households spend more than S 125 a week on groceries. Assume the population proportion is 0.25 and a simple random sample of 211 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 r88Enter your answer as a number accurate to 4 decimal places.) BANG 6 7 8 A/
The right answer needed please.

Answer 1

Solution:

Given in the question

Population proportion =0.25

Sample size = 211

P(Xbar<0.27)

Z = (sample proportion- population proportion)/sqrt(p*(1-p)/n))

Z = (0.27-0.25)/sqrt(0.25*(1-0.25)/211))

Z = (0.02/sqrt(0.0008886)

Z =0.02/0.0298

Z = 0.6711

From Z table we found p-value

P(Xbar<0.27)= 0.7486