Question

15. A 1994 U.S. poll found that 64% of 1000 cigarette smokers who were sampled would quit smoking if the price of cigarettes were increased to $3.00 per pack.

    1. (a) How many individuals in the sample indicated they would quit smoking?

    2. (b) Using a 95% confidence interval estimate the true proportion of American adults who

       would quit smoking if the price of cigarettes was increased to $3.00 per pack. Interpret the interval.

Answer 1

(a) np = 1000 * 0.64 = 640 individuals

(b)

n = 1000    

p = 0.64    

% = 95    

Standard Error, SE = √{p(1 - p)/n} =    √(0.64(1 - 0.64))/1000 = 0.015178933

z- score = 1.959963985    

Width of the confidence interval = z * SE =     1.95996398454005 * 0.0151789327688082 = 0.02975016

Lower Limit of the confidence interval = P - width =     0.64 - 0.029750161550619 = 0.61024984

Upper Limit of the confidence interval = P + width =     0.64 + 0.029750161550619 = 0.66975016

The 95% confidence interval is [61.02%, 66.98%]

Interpretation:     

If samples of size   1000 are repeatedly drawn from the population and the confidence intervals constructed such intervals will contain the true proportion 95% of the time.