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)