If n=31, ¯ x (x-bar)=33, and s=3, find the margin of error at a 90% confidence level Give your answer to two decimal places.
Point estimate = sample mean = = 33
sample standard deviation = s = 3
sample size = n = 31
Degrees of freedom = df = n - 1 = 30
At 90% confidence level the t is ,
t /2,df = t0.05,24 = 1.697
Margin of error = E = t/2,df * (s /n)
= 1.697 * ( 3/ 31)
The Margin of error is 0.91
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