If n=10, ¯ x (x-bar)=32, and s=10, find the margin of error at a 90% confidence level Give your answer to two decimal places.
solution
Given that,
s =10
n = 10
Degrees of freedom = df = n - 1 =10 - 1 =9
At 90% confidence level the t is ,
= 1 - 90% = 1 - 0.90 = 0.1
/ 2 = 0.1 / 2 = 0.05
t /2,df = t0.05,9 = 1.833 ( using student t table)
Margin of error = E = t/2,df * (s /n)
= 1.833 * (10 / 10)
= 5.80
Margin of error = E =5.80
Get Answers For Free
Most questions answered within 1 hours.