Question 1207874: 1. For a sample of size 20 taken from a normally distributed population with standard deviation equal to 5, a 90% confidence interval for the population mean would require the use of:
a. z = 2.58
b. t = 1.328
c. z = 1.96
d. z = 1.645
e. t = 1.729
Answer by math_tutor2020(3817)   (Show Source):
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Answer: option D
z = 1.645
Explanation
We know the population standard deviation (sigma) so we can use the Z distribution.
We don't need to worry that n = 20 doesn't clear the n > 30 hurdle.
Refer to this table (similar tables should be found in the back of your stats textbook)
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
At the bottom of the table it mentions the various confidence levels. Highlight the 90% column.
Just above it is 1.645 which is the critical z value we want.
It means P(-1.645 < z < 1.645) = 0.90 approximately.
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Want to use a TI83 or TI84?
Press the button labeled "2nd" then the "VARS" key.
Scroll down to invNorm
Here are the inputs- area = (1-0.9)/2 = 0.05
- mean = 0
- standard deviation = 1
This should be pasted into your home screen invNorm(0.05,0,1)
The result will be roughly -1.645 in which we use the positive version of this number.
Note: if you leave off the 0,1 at the end, the calculator will use those as defaults.
This means invNorm(0.05,0,1) is the same as invNorm(0.05)
Want to use a spreadsheet?
The function to use is called normInv
The order of "norm" and "inv" is unfortunately swapped compared to the TI83 version. Luckily the order of the inputs are the exact same as above.
You would type =normInv(0.05,0,1) into the spreadsheet.
Don't forget about the equal sign up front.
Refer to your spreadsheet manual for more information about this function.
If you're curious why I did (1-0.9)/2 = 0.05, it refers to 0.05 being the area of one tail
1-0.9 = 0.1 is the area of both tails combined (since 0.9 is the area of the main body)
So we have to cut that combined tail area in half to get 0.05
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