SOLUTION: 2. First-semester GPA’s for a random selection of freshmen at a large university are shown below. Estimate the true mean GPA of the frshman class with 99% confidence. Assume stan

Question 1197083: 2. First-semester GPA’s for a random selection of freshmen at a large university are shown below. Estimate the true mean GPA of the frshman class with 99% confidence. Assume standard deviation = 0.62.
1.9 3.2 2.0 2.9 2.7 3.3
2.8 3.0 3.8 2.7 2.0 1.9
2.5 2.7 2.8 3.2 3.0 3.8
3.1 2.7 3.5 3.8 3.9 2.7
2.0 2.8 1.9 4.0 2.2 2.8
2.1 2.4 3.0 3.4 2.9 2.1

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Enter the given data into a spreadsheet.
Let's say the upper left corner is in cell A1
That places the bottom right corner in cell F6

In another blank cell off to the side somewhere, type in =AVERAGE(A1:F6) to find the arithmetic mean of this data set.
Don't forget about the equal sign up front.
The result should be roughly 2.819444
This is the sample mean xbar.

If you wanted to find the value of xbar another way, then add up all the values and divide by 36 since there are 6*6 = 36 items in this list.
The values add up to 101.5 which leads to xbar = 101.5/36 = 2.819444
The command =SUM(A1:F6) will add up the values in cells A1 to F6.

At 99% confidence, the z critical value is roughly z = 2.576
Use a table like this
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
to get that value. Look at the bottom row labeled "Z" and above the 99% confidence level.

An alternative way to calculate this critical value is to type =NORMINV((1-0.99)/2, 0, 1) into the spreadsheet.
The portion (1-0.99)/2 represents the area in each tail at 99% confidence.
The spreadsheet should display -2.5758293035489, but we'll go with the positive version of that number.

----------------------------

We have this information
z = 2.576 = critical value
xbar = 2.819444 = approximate sample mean
sigma = 0.62 = given population standard deviation
n = 36 = sample size

Let's compute the margin of error
E = z*sigma/sqrt(n)
E = 2.576*0.62/sqrt(36)
E = 0.266187

So,
L = lower boundary
L = xbar - E
L = 2.819444 - 0.266187
L = 2.553257
L = 2.55
and
U = upper boundary
U = xbar + E
U = 2.819444 + 0.266187
U = 3.085631
U = 3.09

The 99% confidence interval for the population mean mu is roughly (2.55, 3.09)
This is in the format (L, U)

This can be expressed as 2.55 < mu < 3.09 which is in the format L < mu < U

A third alternative is to say which is in the format

Round each value according to the instructions your teacher provides.

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