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 193 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.26 and 0.39? 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.)

Answer 1

Solution:

Let

P = Sample proportion (Capital P)

p = Population proportion (Small p)

Here, we have to find P(0.26<P<0.39)

We are given

Population proportion = p = 0.25

Sample size = n = 193

P(0.26<P<0.39) = P(P<0.39) – P(P<0.26)

First find P(P<0.39)

Z = (P – p)/sqrt(pq/n)

q = 1 – p = 1 – 0.25 = 0.75

Z = (0.39 - 0.25)/sqrt(0.25*0.75/193)

Z = 4.491652

P(Z<4.491652) = P(P<0.39) = 0.999996

(by using z-table)

Now, find P(P<0.26)

Z = (P – p)/sqrt(pq/n)

Z = (0.26 - 0.25)/sqrt(0.25*0.75/193)

Z = 0.320832

P(Z<0.320832) = P(P<0.26) = 0.625831

(by using z-table)

P(0.26<P<0.39) = P(P<0.39) – P(P<0.26)

P(0.26<P<0.39) = 0.999996 - 0.625831

P(0.26<P<0.39) = 0.374165

Required probability = 0.3742