Question

Construct the indicated confidence interval for the population mean μ using the t-distribution. Assume the population is normally distributed.

ce=0.99, x =13.9, s=3.0, n=6

The 99% confidence interval using a t-distribution is ____________.(Round to one decimal place as needed.)

Answer 1

Solution:

Mean (x)= 13.9
Standard Deviation (s)= 3
No of sample (n)= 6
Confidence Interval= ??
Step 1: df=(n-1)= (6-1)=5
Step 2: a= (1-ce)/2= (1-0.99)/2= 0.005
Step 3: Find in t-distribution Table with df=5 and a= 0.005
Result is 4.032
Step 4: standard deviation(s)/ n^1/2 = 3/6^1/2 = 1.22
Step 5: 4.032*1.22= 4.9
Step 6: Lower End of Range : Mean - Step 5
Lower End of Range = 13.9-4.9= 9
Upper End of Range = Mean + Step 5
Upper End of Range = 13.9+4.9= 18.8
Confidence Interval: 9<m<18.8