Question

The Grocery Manufacturers of America reported that 60% of consumers read the ingredients listed on a product’s label. Assume the population proportion is p=0.6 and a simple random sample of 380 consumers is selected from the population. Show the sampling distribution of the sample proportion p where p is the proportion of the sampled customers who read the ingredients listed on a product’s label. What is the probability that the sample proportion will be within +.03 of the population proportion? Answer part B for a sample of 650 consumers.

Answer 1

A) = p = 0.6

   = sqrt(p(1 - p)/n)

= sqrt(0.6 * 0.4/380)

= 0.0251

B) = 0.6

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

= sqrt(0.6 * 0.4/650) = ​​​​​0.0192

P(0.57 < < 0.63)

= P((0.57 - )/ < ( - )/ < (0.63 - )/)

= P((0.57 - 0.6)/0.0192 < Z < (0.63 - 0.6)/0.0192)

= P(-1.56 < Z < 1.56)

= P(Z < 1.56) - P(Z < -1.56)

= 0.9406 - 0.0594

= 0.8812