Of 1000 randomly selected cases of lung cancer 823 resulted

Of 1000 randomly selected cases of lung cancer, 823 resulted in death.

(a) Test the hypotheses H0 : p = 0.85 vs H1 : p 6= 0.85 with = 0.05. 2

(b) Construct a 95% confidence interval for p.

c) How large a sample would we need if we wanted to build a 95% confidence interval for the death rate from lung cancer that had a margin of error of less than 0.03, assuming that the range of values for p is unknown?

Solution

a)

Formulating the null and alternatuve hypotheses,          
          
Ho:   p   =   0.85
Ha:   p   =/=   0.85
As we see, the hypothesized po =   0.85      
Getting the point estimate of p, p^,          
          
p^ = x / n =    0.823      
          
Getting the standard error of p^, sp,          
          
sp = sqrt[po (1 - po)/n] =    0.01129159      
          
Getting the z statistic,          
          
z = (p^ - po)/sp =    -2.391160191      
          
As this is a    2   tailed test, then, getting the p value,  
          
p =    0.016795223      
significance level =    0.05      

As P < 0.05, we   REJECT THE NULL HYPOTHESIS.      

Thus, there is significant evidence that the proportion of lung cancer cases resulting in death is not 0.85. [CONCLUSION]

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b)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.823          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.012069424          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
Margin of error = z(alpha/2)*sp =    0.023655637          
lower bound = p^ - z(alpha/2) * sp =   0.799344363          
upper bound = p^ + z(alpha/2) * sp =    0.846655637          
              
Thus, the confidence interval is              
              
(   0.799344363   ,   0.846655637   ) [ANSWER]

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c)

Note that      
      
n = z(alpha/2)^2 p (1 - p) / E^2      
      
where      
      
alpha/2 =    0.025  
As there is no previous estimate for p, we set p = 0.5.      
      
Using a table/technology,      
      
z(alpha/2) =    1.959963985  
      
Also,      
      
E =    0.03  
p =    0.5  
      
Thus,      
      
n =    1067.071895  
      
Rounding up,      
      
n =    1068   [ANSWER]