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A group of doctors published an article in the New England Journal of Medicine. In the article they revealed the findings of a study to determine if there is a difference in survival rates for men who undergo surgery for localized prostate cancer and those who are treated with observation only. The researchers randomly assigned 731 men with localized prostate cancer to radical prostatectomy or observation. During the years of follow-up, 245 of 367 (66.8%) assigned to surgery survived for at least 5 years and 223 of 364 men (61.3%) assigned to observation survived for at least 5 years.

Required:
a. Construct and interpret a 99% confidence interval for the difference in the true proportion of men like these that would survive for at least 5 years after receiving surgery and the proportion that would survive for at least 5 years when only being observed.
b. How does the confidence interval from part a tell you that there is not sufficient evidence at the a=.01 level that there is a difference in the proportion of men with localized prostate cancer who survive at least five years after receiving surgery and the proportion of men with localized prostate cancer who survive at least five years when only being observed. In other words, how would you know that the p-value for a 2 proportion z test would not be significant at the 0.01 level?

Answer :

Answer:

a)  CI =  (  - 0,0352  ;  0,1472 )

b)  The CI  include 0 since that means we   don´t have sufficient evidence at α = 1 % to support differences between the two proportions

Step-by-step explanation:

From sample 1.    ( assigned to surgery)

sample size   n = 367

sample proportion   p₁  =  245/367    p₁  = 66,8 %   or p₁ = 0,668

Then  q₁ = 1 - 0,668   q₁ =  0,332

From sample 2.  ( assigned to observation )

sample size  n₂  = 364

sample proportion   p₂  =  223/364    p₂  = 61,2 %   or p₂ = 0,612

Then  q₂ = 1 - 0,612   q₁ =  0,388

Test Hypothesis is:

Null hypothesis                 H₀        p₂  =  p₁

Alternative Hypothesis    Hₐ        p₂ ≠ p₁

a) CI ( 99 %)     if  CI = 99%     significance level α = 1 %     α = 0,01

α/2 = 0,005  n z-table we look z (c) for  0,005

z(c)  = 2,578

CI    =              ( p₁  -  p₂  ) ± z(α/2) √(p₁*q₁) / n₁] + [ p₂*q₂/n₂

CI =(0,668 -0,612 ) ± 2,578 * √(0,668*0,332) /367 + (0,612*0,388)/364

CI = 0,056 ± 2,578*√6,04 *10⁻⁴ + 6,52 *10⁻⁴

CI = 0,056 ±  2,578*3,54*10⁻²

CI  = ( 0,056 - 0,0912 ;  0,056 + 0,0912 )

CI =  (  - 0,0352  ;  0,1472 )

b) The CI  include 0 since that means we   don´t have sufficient evidence at α = 1 % to support differences between the two proportions

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