Question

The Belgian Association of Grocery Stores (BAGS) did a survey of their customers and reported that 76% of them read the ingredients listed on the products labels. Assume a sample of 400 customers is selected from the population. Assume that population follows a normal distribution:

a) What is the expected value of the sample proportion?

b) what is the standard deviation of the sample proportion?

c) what is the probability that the sample proportion will be within plus or minus 0.03 of the population proportion?

d)  what is the probability that the sample proportion will be within plus or minus 0.03 of the population proportion if the sample size was increased by 350 customers?

e) what can you conclude from your calculation?

Answer 1

n = 400

p = 0.76

a) = p = 0.76

b) = sqrt(p(1 - p)/n)

        = sqrt(0.76(1 - 0.76)/400)

        = 0.021

c) P(0.73 < < 0.79)

= P((0.73 - )/ < ( - )/ < (0.79 - )/)

= P((0.73 - 0.76)/0.021 < Z < (0.79 - 0.76)/0.021)

= P(-1.43 < Z < 1.43)

= P(Z < 1.43) - P(Z < -1.43)

= 0.9236 - 0.0764

= 0.8472

d) n = 750

   p = 0.76

= p = 0.76

= sqrt(p(1 - p)/n)

     = sqrt(0.76(1 - 0.76)/750)

     = 0.0156

P(0.73 < < 0.79)

= P((0.73 - )/ < ( - )/ < (0.79 - )/)

= P((0.73 - 0.76)/0.0156 < Z < (0.79 - 0.76)/0.0156)

= P(-1.92 < Z < 1.92)

= P(Z < 1.92) - P(Z < -1.92)

= 0.9726 - 0.0274

= 0.9452

e) As the sample size increases, the probability value also increases.