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

a)here sample proportion =11/114 =0.0965

std error =(p(!-p)/n)1/2 =0.0277

for 90% CI ; z =1.6449

hence  90% confidence interval for the percentage of patients who stop smoking =sample proportion -/+z*Std errror

=0.0510 ; 0.1475 ~ 5.1% to 14.75%

b) above interval gives 90% confidence to contain true  percentage of patients who stop smoking after

advice given by a physician

c) margin of errro =z*std error =0.0455

d) for margin of error is dependent on std error and confidence interval

therefore for margin of error to be cut in half; we need to reduce std error to half and to do that we need to have 4 times the sample size. which is equal to 4*114=456

e)for 300 patients ; std error =(p(!-p)/n)1/2 =0.0170

hence margin of error =z*Std error =0.0280

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