Question

Many individuals over the age of 40 develop intolerance for milk and milk-based products. A

dairy has developed a line of lactose-free products that are more tolerable to such individuals. To
assess the potential market for these products, the dairy commissioned a market research study of individuals over 40 years old in its sales area. A random sample of 250 individuals showed that 86 of them suffer from milk intolerance. Based on the sample results, calculate a 90% confidence
interval for the population proportion that suffers milk intolerance. Interpret this confidence interval.


a) First, show that it is okay to use the 1-proportion z-interval. (Assumptions)

b) calcuate by hand a 90% confidence interval (use the formula)

c) provide an interpretation (conclusion)

d) if the level of confidence was 95% instead of 90% would the resulting interval be narrower or wider? explain. (show your computation before your explanation)

e) if the researchers were interested in the 90% interval with the 3% margin of error what size sample would they require?

f) verify your 90% confidence interval using your calculator. ​

Answer 1

p = 86/250 = 0.344

z_c = 1.65 at 90% confidence.

margin of error, E = sqrt(pq/n)*z_c

= sqrt(0.344*(1-0.344)/250)*1.65

= 0.0496

90% CI = (p -E , p + E)

= (0.344 - 0.0496 , 0.344 + 0.0496)

= (0.2944, 0.3936) (Ans.)

Explanation: