In: Statistics and Probability
A sample of size n = 16 is made from a normal distribution with
mean μ. It
turns out that the sample mean is x = 23 and the sample standard
deviation is s = 6.
Construct a 90% confidence interval for μ.
Given that,
= 23
s =6
n = 16
Degrees of freedom = df = n - 1 =16 - 1 = 15
At 90% confidence level the t is ,
= 1 - 90% = 1 - 0.90 = 0.1
/ 2 = 0.1 / 2 = 0.05
t /2,df = t0.05,15 =1.753 ( using student t table)
Margin of error = E = t/2,df * (s /n)
= 1.753 * (6 / 16) = 2.6295
The 90% confidence interval estimate of the population mean is,
- E < < + E
23- 2.6295< < 23+ 2.6295
20.3705 < < 25.6295
( 20.3705, 25.6295)