Question

A Food Marketing Institute found that 47% of households spend more than $125 a week on groceries. Assume the population proportion is 0.47 and a simple random sample of 89 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is more than than 0.35?

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 = 0.47

1 - p = 1 - 0.47 = 0.53

n = 89

= p = 0.47

=  [p( 1 - p ) / n] = [(0.47 * 0.53) / 89 ] = 0.0529

P( > 0.35) = 1 - P( < 0.35 )

= 1 - P(( - ) / < (0.35 - 0.47) / 0.0529)

= 1 - P(z < -2.2684)

Using z table

= 1 - 0.0117

= 0.9883