Question 1202822
Pharmaceutical companies promote their prescription drugs using television advertising. In a survey of 112 randomly sampled television viewers, 14 indicated that they asked their physician about using a prescription drug they saw advertised on TV.
a. What is the point estimate of the population proportion?

The sample proportion is 14/112 or {{{highlight(0.125)}}}

b. What is the margin of error for a 95% confidence interval estimate? (Round "z-scores" and round your final answers to 2 decimal places.)

Since P(-1.96 < Z < 1.96) = 0.95 where Z is a Standard Normal random variable, then the margin of error is {{{1.96*sqrt(0.125*(1-0.125)/112)}}} = {{{highlight(0.06)}}}
c. Compute the 95% confidence interval for the population proportion. (Round "z-scores" to 2 decimal places and round your answers to 3 decimal places.)
Confidence interval for the proportion mean is between ___ and ___.

We use parts a and b to answer part-c.
lower limit = 0.125 - 0.061 = 0.064
upper limit = 0.125 + 0.061 = 0.186
Answers: between 0.064 and 0.186