Question

in a program designed to help patients stop smoking 198 were given sustained care, 82.8% were no longer smoking after 1 month

among 199 patients given standard care 62.8 no longer smoking after 1 month

A construct 99% Confidence interval estimate of percentage of success using sustained care, interpret finding with conclusion?

B test hypothesis that there no difference between the two type of care in helping patients stop smoking, interpret the results?

Answer 1

Soln

A)

Proportion of success using sustained care (p) = 0.828

n = 198

alpha = 0.01

Z_Critical = 2.58

99%CI = p +/- Z_Critical * {p(1-p)/n}^1/2 = {0.759, 0.897} or {75.9%, 89.7%}

B)

Proportion of success using sustained care (p₁) = 0.828

n₁ = 198

Proportion of success given standard care (p₂) = 0.628

n₂ = 199

alpha = 0.05

Null and Alternate Hypothesis

H₀: µ₁ = µ₂

H_a: µ₁ <> µ₂

p_c = (n₁*p₁ + n₂*p₂)/(n₁+n₂) = 0.73

Test Statistic

t = (p₁-p₂) / (p_c*(1-p_c)/n₁ + p_c*(1-p_c)/n₂ )^1/2 = 4.48

p-value = TDIST(4.48,198+199-2,2) = 9.95068E-06

Result

Since the p-value is less than 0.05, we reject the null hypothesis.

Conclusion

There is difference between the two type of care in helping patients stop smoking,