A Food Marketing Institute found that 25% of households spend more than $125 a w

Question

A Food Marketing Institute found that 25% of households spend more than $125 a week on groceries. Assume the population proportion is 0.25 and a simple random sample of 330 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 0.28?

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

Answer =  (Enter your answer as a number accurate to 4 decimal places.)

Explanation / Answer

Calculate the standard deviation and compute z score test statistic.

Standard deviation=Root over (p*(1-p)/n)=0.0238

z=(p-pbar)/sigma=(0.28-0.25)/0.0238=1.2605

Use normal table to find P(z<1.2605)=0.8962 9Ans)

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