Given Data Set
{
3.1,3.8,3.0,3.2,2.8,2.7,2.5,1.9,2.0,
2.7,3.8,3.0,2.8,3.3,2.7,2.9,2.0,3.2,
1.9,2.8,2.2,4.0,1.9,2.8,2.0,2.7,3.9
}
There are n = 27 values in the data set above (3 rows of 9 each)
Normally since n > 30 is not true, this means we'd have to use a T distribution. However, we're given the value of sigma () so we can use the standard normal distribution.
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Use a stats table to find the z critical value
I'm using this table specifically. Locate the "99%" value in the bottom row that corresponds to confidence level. The value just above that is 2.576
The 99% z critial value is z = 2.576
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Sample mean
Add up all the values to get
3.1+3.8+3.0+3.2+2.8+2.7+2.5+1.9+2.0+2.7+3.8+3.0+2.8+3.3+2.7+2.9+2.0+3.2+1.9+2.8+2.2+4.0+1.9+2.8+2.0+2.7+3.9 = 75.6
Divide this sum by n = 27
75.6/n = 75.6/27 = 2.8
The sample mean is xbar = 2.8
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So we have,
xbar = 2.8 (computed above)
z = 2.576 (approximate; see above)
sigma = 0.62 (given)
n = 27
The lower boundary (L) of the confidence interval is
L = xbar - z*(sigma/sqrt(n))
L = 2.8 - 2.576*(0.62/sqrt(27))
L = 2.49263411269062
L = 2.49
The upper boundary (U) of the confidence interval is
U = xbar + z*(sigma/sqrt(n))
U = 2.8 + 2.576*(0.62/sqrt(27))
U = 3.10736588730938
U = 3.11
Overall the 99% confidence interval is (L, U) = (2.49, 3.11)
We have 99% confidence that the mean (mu) is between 2.49 and 3.11.
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Answer: 2.49 < < 3.11