Suppose that you are interested in determining whether advice given by a physici

Question

Suppose that you are interested in determining whether advice given by a physician during a routine physical examination is effective in encouraging patients to stop smoking. In a study of current smokers, patients were given a brief talk about the hazards of smoking and were encouraged to quit. All patients were given a follow-up exam. In the sample of 114 patients who had received the advice, 11 reported that they had quit smoking. a) Construct a 90% confidence interval for the percentage of patients who stop smoking. b) Interpret your interval. c) What is the margin of error for your confidence interval? d) Suppose you want to cut the margin of error in half, without changing the confidence level, how many patients must you sample? e) What would be the margin of error if you sampled 300 patients? Explain.

Explanation / Answer

p =11/114 = 0.09649

q=1-p

=1-0.09649

= 0.90351

n=114

For 90% confidence, Z/2= 1.645

c) Margin of error,E =Z/2 × [(p×q)/n]

                             E=1.645 x(0.09649*0.90351)/114)

                             = 0.04549

a)

CI = p ± E

=0.09649 ± 0.04549

=0.09649 -0.04549 to 0.09649 + 0.04549

=0.051 to 0.142

b)

It means that we are 90% confident that we population proportion lies in the interval 0.051 to 0.142

d)

E =0.04549/2 = 0.022745

E= Z/2 × [(p×q)/n]

0.022745 =1.645(0.09649*0.90351)/n

0.022745 =0.4857/n

n=0.4857/0.022745

n =21.3544

n= 457

e)

Margin of error,E =Z/2 × [(p×q)/n]

                             E=1.645 x(0.09649*0.90351)/300)

                             =0.02804

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