Determine the margin of error for a confidence interval to estimate the population mean with n=18 and s =14.5 for the confidence levels below.
a) |
808% |
b) |
90% |
c) |
99% |
a) The margin of error for an 80% confidence interval is
(Round to two decimal places as needed.)
(a)
n = Sample Size = 18
s = 14.5
SE= s/
= 14.5/ = 3.4177
ndf = 18- 1 =17
= 0.20
From Table critical values of t = 1.3334
So,
Margin of Error = t X SE
= 1.3337 X 3.4177 = 4.56
(b)
n = Sample Size = 18
s = 14.5
SE= s/
= 14.5/ = 3.4177
ndf = 18- 1 =17
= 0.10
From Table critical values of t = 1.7396
So,
Margin of Error = t X SE
= 1.7396 X 3.4177 = 5.95
(c)
n = Sample Size = 18
s = 14.5
SE= s/
= 14.5/ = 3.4177
ndf = 18- 1 =17
= 0.01
From Table critical values of t = 2.8982
So,
Margin of Error = t X SE
= 2.8982 X 3.4177 = 9.91
Thus, we get:
(a) | 80 % | 4.56 |
(b) | 90 % | 5.95 |
(c) | 99% | 9.91 |
(a) The margin of error for an 80% confidence interval is 4.56
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