In: Math
Solution :
Given that,
= 200,000
s = 18,000.
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
Margin of error = E = t/2,df * (s /
n)
= 1.753 * (18,000 / 16 )
=7888.5
The 90% confidence interval estimate of the population mean is,
- E <
<
+ E
200,000 - 7888.5< < 200,000 + 7888.5
192111.5 < <207888.5
(192111.5, 207888.5 )