Question

Given the following test scores, find a 95 percent confidence interval for the population mean: 148, 154, 158, 160, 161, 162, 166, 170, 182, 195, 236. Assume population normality.

Answer 1

We have given that,          
          
Sample mean =172  
Sample standard deviation =24.952
Sample size =11  
Level of significance=1-0.95=0.05  
Degree of freedom =10  
          
t critical value is (by using t table)=   2.228  
          
Confidence interval formula is

$\bar{x}\pm t*\frac{S}{\sqrt{n}}$

$=172\pm 2.228*\frac{24.952}{\sqrt{11}}$

=(155.24,188.76)          
          
          
          
Lower confidence limit=   155.24  
          
Upper confidence limit=188.76