Question 1082200: First semester GPAs for a random selection of freshmen at a large university are shown. Estimate the true mean GPA of the freshmen class with 99% confidence. Assume o=0.62. Round intermediate and final answers to two decimal places. Assume the population is normally distributed.
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
__________ < u < _________
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! 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
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