Question

1. A random sample of 120 ULS students yields a mean GPA of 2.71 with a sample standard deviation of 0.51. Construct a 95% confidence interval for the true mean GPA of all students at ULS. Find the minimum sample size required to estimate the true mean GPA at ULS to within 0.03 using a 95% confidence interval

2. Suppose we have 3 ULS student candidates running for the position of student representative, A, B, and C. We randomly select 160 ULS students. Of them, 120 vote for candidate A. Find a 95% confidence interval for the true proportion of the population of ULS students who will vote for candidate A. How many students must be surveyed in order to be 95% confident that the sampling percentage is in error by no more than 3 percentage points (or 0.03)?

Answer 1

1)

Standard Deviation ,   σ =    0.51                  
sampling error ,    E =   0.03                  
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   *   0.51   /   0.03   ) ² =   1110.182
                          
                          
So,Sample Size needed=       1111                  

(please try 1133 if above gets wrong)

2)

a)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   120          
Sample Size,   n =    160          
                  
Sample Proportion ,    p̂ = x/n =    0.7500          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.034233          
margin of error , E = Z*SE =    1.960   *   0.03423   =   0.0671
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.75000   -   0.06709   =   0.6829
Interval Upper Limit = p̂ + E =   0.75000   +   0.06709   =   0.817
                  
95%   confidence interval is (   0.683   < p <    0.817   )

b)

sample proportion ,   p̂ = 120/160= 0.75                          
sampling error ,    E =   0.03                          
Confidence Level ,   CL=   0.95                          
                                  
alpha =   1-CL =   0.05                          
Z value =    Zα/2 =    1.960   [excel formula =normsinv(α/2)]                      
                                  
Sample Size,n = (Z / E)² * p̂ * (1-p̂) = (   1.960   /   0.03   ) ² *   0.75   * ( 1 -   0.75   ) =    800.3039
                                  
                                  
so,Sample Size required=       801