Question

An article in the Journal of Agricultural Science ("The Use of Residual Maximum Likelihood to Model Grain Quality Characteristics of Wheat with Variety, Climatic and Nitrogen Fertilizer Effects" (1997, Vol. 128, pp. 135-142)] investigated means of wheat grain crude protein content (CP) and Hagberg falling number (HFN) surveyed in the UK. The analysis used a variety of nitrogen fertilizer applications (kg N/ha), temperature (°C), and total monthly rainfall (mm). The data shown below describe temperatures for wheat grown at Harper Adams Agricultural College between 1982 and 1993. The temperatures measured in June were obtained as follows: 12.7 13.1 14.4 12.1 14.2 15.2 16.0 11.5 15.7 14.6 13.6 Assume that the population has a normal distribution and the standard deviation is known to be = 0.6. (a) Construct a 99% two-sided confidence interval on the mean temperature. Round to two decimal places (e.g. 98.76). sus (b) Construct a 95% lower-confidence bound on the mean temperature. Round to two decimal places (e.g. 98.76). <μ (c) Suppose that we wanted to be 95% confident that the error in estimating the mean temperature is less than 2 degrees Celsius. What sample size should be used? n = (d) Suppose that we wanted the total width of the two-sided confidence interval on mean temperature to be 1.5 degrees Celsius at 95% confidence. What sample size should be used? n =

Answer 1

a)

Sample Mean,    x̅ = ΣX/n =    13.9182

Level of Significance ,    α =    0.01          
'   '   '          
z value=   z α/2=   2.5758   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   0.6000   / √   11   =   0.1809
margin of error, E=Z*SE =   2.5758   *   0.1809   =   0.4660
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    13.92   -   0.465985   =   13.4522
Interval Upper Limit = x̅ + E =    13.92   -   0.465985   =   14.3842
99%   confidence interval is (   13.45   < µ <   14.38   )

b)

Level of Significance ,    α =    0.05   
'   '   '          
z value=   z α=   1.6449   [Excel formula =NORMSINV(α) ]      
                  
Standard Error , SE = σ/√n =   0.6000   / √   11   =   0.1809
margin of error, E=Z*SE =   1.6449   *   0.1809   =   0.2976
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    13.92   -   0.297565   =   13.6206

answer: 13.62≤µ

c)

Standard Deviation ,   σ =    0.6                  
sampling error ,    E =   2                  
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   *   0.6   /   2   ) ² =   0.346
                          
                          
So,Sample Size needed=       1                  

d)

Standard Deviation ,   σ =    0.6                  
sampling error ,    E = 1.5/2 = 0.75                  
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   *   0.6   /   0.75   ) ² =   2.459
                          
                          
So,Sample Size needed=       3