Question

A student believes that the average grade on the statistics final examination is 87. A sample of 36 final examinations is taken. The average grade in the sample is 83.96. The sample variance is 144. Compute a 92% confidence interval: (Given t0.08 = -1.43, t0.04 = -1.80 both with 35 degree of freedom; z0.04 = -1.75, z0.08 = -1.41)

Answer 1

We have given that,              
              
Sample mean =83.96      
Sample standard deviation =12  
Sample size =36      
Level of significance=1-0.92=0.08      
Degree of freedom =35      
              
t critical value is (by using t table)=1.803      

Therefore, 92% confidence interval is =( 80.354, 87.566)