Question

1. A fast food company is interested to know how many times residents of a town eat fast food per week. (How many meals per week). The town population is 150,000. The company asked you to survey its residents and find out the answer to the company’s question. You take a random sample of 400 and find that the sample mean is 5.2 with standard deviation of 3.75. You want to generalize your findings in a report to the company. a) Build a confidence interval with 95% certainty. b) How would you report it to the company?

Answer 1

a) We have the following data provided

Sample size n = 400

Mean x_bar = 5.2

Standard deviation SD = 3.75

For 95% confidence, a = 0.05. Now we need to determine Z_a/2 or Z_0.025 = 1.96 (from the normal table)

For 95% confidence interval = x_bar (1.96 * SD) / sqrt (n)

Interval will be between 5.2 + 1.96* 3.75 / sqrt(400) and 5.2 - 1.96* 3.75 / sqrt(400)

Or, between 5.2 + 0.3675 and 5.2 – 0.3675

Or, between 5.56 and 4.83

b) We can inform the company that we can say with 95% confidence that the on an average residents of the town eat fast-food between 4.83 and 5.56 times a week