Question

1. A researcher in a small town is interested in estimating the true proportion of adults in the town who smoke. 215 adults were randomly selected from the town, and it was found that 36 of them smoke. We would like to construct a 90% confidence interval estimate for the true proportion of adults in the town that smoke. please show work! thanks.
   1. What are the values of, , and  (Round  and  to three decimal places if needed.)

= _______________                 = _______________

= _______________                 = _______________

2. Calculate the margin of error (E) for a 90% confidence interval. (You must show the setup to receive credit. You may round to four decimal places if needed.)

E = _______________

3. Construct the 90% confidence interval for the true proportion of smokers in the town. (Round limits to three decimal places.)

_______________<  < _______________

Answer 1

Solution :

Given that,

a) Point estimate = sample proportion = = x / n = 36 / 215 = 0.167

1 - = 1 - 0.167 = 0.833

n = 215

Z/2 = Z0.05 = 1.645

b) Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 (((0.167 * 0.833) / 215)

= 0.0418

c) A 90% confidence interval for population proportion p is ,

- E < p < + E

0.167 - 0.0418 < p < 0.167 + 0.0418

( 0.125 < p < 0.209 )