Question

2. Suppose a study followed 1000 women aged 25-45 (when the study began) over 6 years and found that 50 acquired bacteriuria. a) Find a 95% confidence interval for the expected number of bacteriuria cases per 1000 women within 6 years. b) Find a 95% confidence interval for the incidence of bacteriuria per person-year. c) What is a 95% confidence interval for the incidence of bacteriuria per 100 person- years. 3. A study looked at the subgroup of 50-to 55-year-old women with a parental history of myocardial infarction (MI) at age >60. Of the 800 women they followed over 4 years, 4 died from coronary heart disease (CHD). a) What is an exact 95% confidence interval for the probability of dying from CHD in this subgroup? Is normal approximation appropriate here? Why? b) Suppose that the probability of dying from CHD among women in the same age group but without parental history of Ml is 1 in 1500. Can you conclude that the risk of death from CHD among women with parental history of MI is higher than that of women without any parental history of MI?

Answer 1

2)

a)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   50          
Sample Size,   n =    1000          
                  
Sample Proportion ,    p̂ = x/n =    0.0500          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.006892          
margin of error , E = Z*SE =    1.960   *   0.00689   =   0.0135
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.05000   -   0.01351   =   0.0365
Interval Upper Limit = p̂ + E =   0.05000   +   0.01351   =   0.0635
                  
95%   confidence interval is (   36   < p <    64   )

b)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   50          
Sample Size,   n =    1000          
                  
Sample Proportion ,    p̂ = x/n =    0.0500          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.006892          
margin of error , E = Z*SE =    1.960   *   0.00689   =   0.0135
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.05000   -   0.01351   =   0.0365
Interval Upper Limit = p̂ + E =   0.05000   +   0.01351   =   0.0635
                  
95%   confidence interval is (   0.036   < p <    0.064   )

c)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   50          
Sample Size,   n =    1000          
                  
Sample Proportion ,    p̂ = x/n =    0.0500          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.006892          
margin of error , E = Z*SE =    1.960   *   0.00689   =   0.0135
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.05000   -   0.01351   =   0.0365
Interval Upper Limit = p̂ + E =   0.05000   +   0.01351   =   0.0635
                  
95%   confidence interval is (   3.6   < p <    6.4   )

3)

a)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   4          
Sample Size,   n =    800          
                  
Sample Proportion ,    p̂ = x/n =    0.0050          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.002494          
margin of error , E = Z*SE =    1.960   *   0.00249   =   0.0049
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.00500   -   0.00489   =   0.0001
Interval Upper Limit = p̂ + E =   0.00500   +   0.00489   =   0.0099
                  
95%   confidence interval is (   0.000   < p <    0.010   )

Not apprpriate because np <5

b)

Yes we can conclude that

Ho:   p1 - p2 =   0          
Ha:   p1 - p2 >   0          
                  
sample #1   ----->              
first sample size,     n1=   800          
number of successes, sample 1 =     x1=   4          
proportion success of sample 1 , p̂1=   x1/n1=   0.0050          
                  
sample #2   ----->              
second sample size,     n2 =    1500          
number of successes, sample 2 =     x2 =    1          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.0007          
                  
difference in sample proportions, p̂1 - p̂2 =     0.0050   -   0.0007   =   0.0043
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.002173913          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.00204          
Z-statistic = (p̂1 - p̂2)/SE = (   0.004   /   0.0020   ) =   2.13
                  
  
p-value =        0.0168   [excel function =NORMSDIST(-z)]      
decision :    p-value<α,Reject null hypothesis

              

Please let me know in case of any doubt.

Thanks in advance!

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