Question

Suppose that in the Grand Canyon, a random sample of 23 Anna's hummingbirds ( a species of hummingbird ) was taken, and their weights averaged 4.15 grams(g) with a sample standard deviation of s=0.7. (PLEASE SHOW ALL WORK AND KEEP ALL DIGITS FOR THE ANSWERS.)

(a) Find and interpret 95% and 99% confidence intervals for the population mean weight of Anna's hummingbirds.

(b) Find the sample size you would have to have taken so that the error E for a 92% confidence interval is no more than 0.03 g.

Answer 1

a)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   22          
't value='   tα/2=   2.074   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   0.7000   / √   23   =   0.1460
margin of error , E=t*SE =   2.0739   *   0.1460   =   0.3027
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    4.15   -   0.302703   =   3.8473
Interval Upper Limit = x̅ + E =    4.15   -   0.302703   =   4.4527
95%   confidence interval is (   3.85   < µ <   4.45   )
                  
---------------

Level of Significance ,    α =    0.01          
degree of freedom=   DF=n-1=   22          
't value='   tα/2=   2.819   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   0.7000   / √   23   =   0.1460
margin of error , E=t*SE =   2.8188   *   0.1460   =   0.4114
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    4.15   -   0.411426   =   3.7386
Interval Upper Limit = x̅ + E =    4.15   -   0.411426   =   4.5614
99%   confidence interval is (   3.74   < µ <   4.56   )

===============

b)

Standard Deviation ,   σ =    0.7                  
sampling error ,    E =   0.03                  
Confidence Level ,   CL=   92%                  
                          
alpha =   1-CL =   8%                  
Z value =    Zα/2 =    1.751   [excel formula =normsinv(α/2)]              
                          
Sample Size,n = (Z * σ / E )² = (   1.751   *   0.7   /   0.03   ) ² =   1668.669
                          
                          
So,Sample Size needed=       1669