Assume that a sample is used to estimate a population mean μμ.
Find the margin of error M.E. that corresponds to a sample
of size 12 with a mean of 72.1 and a standard deviation of 11.7 at
a confidence level of 80%.
Report ME accurate to one decimal place because the sample
statistics are presented with this accuracy.
M.E. =
Answer should be obtained without any preliminary rounding.
However, the critical value may be rounded to 3 decimal places.
Solution :
Given that,
s = 11.7
n = 12
Degrees of freedom = df = n - 1 = 12 - 1 = 11
At 80% confidence level the t is ,
= 1 - 80% = 1 - 0.80 = 0.20
/ 2 = 0.20 / 2 = 0.10
t /2,df = t0.10,11 = 1.363
Margin of error = E = t/2,df * (s /n)
= 1.363 * (11.7 / 12 )
= 4.6
Margin of error = 4.6
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