Construct the indicated confidence interval for the population mean mu using the t-distribution. Assume the population is normally distributed. c=0.95, x=13.4, s=0.64, n=19
Solution :
Given that,
= 13.4
s =0.64
n =19
Degrees of freedom = df = n - 1 =19 - 1 = 18
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2= 0.05 / 2 = 0.025
t /2,df = t0.025,18 = 2.101 ( using student t table)
Margin of error = E = t/2,df * (s /n)
= 2.101* ( 0.64/ 19)
= 0.3
The 95% confidence interval is,
- E < < + E
13.4-0.3 < < 13.4+0.3
(13.1 , 13.7)
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