In: Statistics and Probability
Solution :
Given that,
= 2.8
s =0.37
n = 11
Degrees of freedom = df = n - 1 = 11- 1 = 10
a ) At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
t /2,df = t0.025,10 = 2.228
Margin of error = E = t/2,df * (s /n)
=2.228 * ( 0.37/ 11)
=0.249
The 95% confidence interval estimate of the population mean is,
- E < < + E
2.8 - 0.249 < < 2.8+ 0.249
2.551< < 3.049
( 2.551 , 3.049)