Question

A Food Marketing Institute found that 34% of households spend more than $125 a week on groceries. Assume the population proportion is 0.34 and a simple random sample of 368 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 0.32?

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

Given that,

p = 0.34

1 - p = 1 - 0.34 = 0.66

n = 368

= p = 0.34

=  [p ( 1 - p ) / n] =   [(0.34 * 0.66) / 368 ] = 0.0247

P( < 0.32)

= P[( - ) / < (0.32 - 0.34) / 0.0247]

= P(z < -0.8097)

Using t table

= 0.2091