Question

A Food Marketing Institute found that 45% of households spend more than $125 a week on groceries. Assume the population proportion is 0.45 and a simple random sample of 113 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 0.43?

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 ,

p = 0.45 (population proportion)

n = 113 .. (sample size)

Let be the sample proportion.

The sampling distribution of is approximately normal with

mean = =  p = 0.45

SD =   =     

=

=  0.04680027229

Now ,

P(sample proportion is more than 0.43)

= P[ > 0.43]

=

=  P(Z > (0.43 - 0.45)/0.04680027229)

= P(Z > -0.4273)

= 1 - P(Z < -0.4273)

= 1 - 0.3346

= 0.6654 ...use z table

Answer = 0.6654