Question

In a survey of 1000 US adults, twenty percent say they never exercise. This is the highest level seen in five years.

a) Verify that the sample is large enough to use the normal distribution to find a confidence interval for the proportion of Americans who never exercise. (Are the conditions met?) Use a formula to justify your answer or, if possible, explain your reasoning.

           

b) Construct a 90% confidence interval for the proportion of U.S. adults who never exercise. Show your work. Use three decimal places in your interval.

           

c) Provide an interpretation of your interval in context.

d) Given the results of the study, is it plausible that 1/3 of US adults never exercise? Explain your answer.

e) Suppose you want to estimate the proportion of local adults who never exercise with 95% confidence and a 5% margin of error. Californians tend to exercise more than Americans in other states, so we will use as a conservative estimate of the proportion of adults who never exercise. (Hint: Use p ~ .) How many people should you sample?

           

Answer 1

n = 1000

p = 0.20

(a) n*p = 1000*0.20 = 200 10

n*(1 - p) = 1000*(1 - 0.20) = 800 10

The sample is large enough to use the normal distribution to find a confidence interval for the proportion of Americans who never exercise.

(b) The 90% confidence interval for the proportion of U.S. adults who never exercise.

= (0.1792, 0.2208)

(c) We are 90% confident that the true proportion of U.S. adults who never exercise is between 0.1792 and 0.2208.

(d) Since 1/3 = 0.33 is not in the confidence interval, we cannot say that it is plausible that 1/3 of US adults never exercise.

(e) n = (1.96/0.05)^2*0.5*0.5 = 385