Question

3. The FDA's web site provides some data on mercury content of fish. Based on a sample of 90 Pacific croaker white fish (Genyonemus lineatus), a sample mean and sample standard deviation were computed as 0.278 and 0.077 ppm (parts per million), respectively.
(a) Construct a 90% confidence interval for the mean mercury content of all croaker white fish.
(b) Construct a 95% confidence interval for the mean mercury content of all croaker white fish.
(c) Construct a 99% confidence interval for the mean mercury content of all croaker white fish.
(d) If the population standard deviation is known to be 0.069 ppm, construct a 95% confidence interval for the mean mercury content of all croaker white fish.
(e) If the margin of error is to be reduced to 0.018 ppm, what is the required sample size? Assume the population standard deviation is 0.069 ppm, and use the confidence level of 99%.
(f) Suppose researchers reported a 95% confidence interval for the mean mercury content as 0.266 to 0.290 ppm. What was the sample size used? Show your calculations, and assume the population standard deviation is 0.069 ppm.

Answer 1

a)

sample std dev ,    s =    0.0770
Sample Size ,   n =    90
Sample Mean,    x̅ =   0.2780
      
Level of Significance ,    α =    0.1          
degree of freedom=   DF=n-1=   89          
't value='   tα/2=   1.6622   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   0.077   / √   90   =   0.0081
margin of error , E=t*SE =   1.6622   *   0.008   =   0.013
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    0.28   -   0.013   =   0.2645
Interval Upper Limit = x̅ + E =    0.28   -   0.013   =   0.2915
90%   confidence interval is (   0.26   < µ <   0.29   )

b)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   89          
't value='   tα/2=   1.9870   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   0.077   / √   90   =   0.0081
margin of error , E=t*SE =   1.9870   *   0.008   =   0.016
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    0.28   -   0.016   =   0.2619
Interval Upper Limit = x̅ + E =    0.28   -   0.016   =   0.2941
95%   confidence interval is (   0.26   < µ <   0.29   )

c)

Level of Significance ,    α =    0.01          
degree of freedom=   DF=n-1=   89          
't value='   tα/2=   2.6322   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   0.077   / √   90   =   0.0081
margin of error , E=t*SE =   2.6322   *   0.008   =   0.021
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    0.28   -   0.021   =   0.2566
Interval Upper Limit = x̅ + E =    0.28   -   0.021   =   0.2994
99%   confidence interval is (   0.26   < µ <   0.30   )

d)

Level of Significance ,    α =    0.05          
'   '   '          
z value=   z α/2=   1.9600   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   0.069   / √   90   =   0.0073
margin of error, E=Z*SE =   1.9600   *   0.007   =   0.014
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    0.28   -   0.014   =   0.2637
Interval Upper Limit = x̅ + E =    0.28   -   0.014   =   0.2923
95%   confidence interval is (   0.26   < µ <   0.29   )

e)

Standard Deviation ,   σ =    0.0690                  
sampling error ,    E =   0.018                  
Confidence Level ,   CL=   99%                  
                          
alpha =   1-CL =   1%                  
Z value =    Zα/2 =    2.576   [excel formula =normsinv(α/2)]              
                          
Sample Size,n = (Z * σ / E )² = (   2.576   *   0.069   /   0.018   ) ² =   97.5
                          
                          
So,Sample Size needed=       98                  

f)

confidence interval is                  
lower limit =    0.266              
upper limit=   0.29              
margin of error = (upper limit-lower limit)/2= (   0.29   -   0.266   ) / 2 =   0.012

Standard Deviation ,   σ =    0.0690                  
sampling error ,    E =   0.012                  
Confidence Level ,   CL=   95%                  
                          
alpha =   1-CL =   5%                  
Z value =    Zα/2 =    1.960   [excel formula =normsinv(α/2)]              
                          
Sample Size,n = (Z * σ / E )² = (   1.960   *   0.069   /   0.012   ) ² =   127.0
                          
                          
So,Sample Size needed=       128