Question 1196584: Given sample size n = 40, find the t critical value for a 95% confidence interval estimation of population mean. Round to 3 decimal places.
2.023
12.793
2.372
1.280
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Answer: A) 2.023
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Explanation:
We're tasked to find a value of k such that P(-k < T < k) = 0.95 when we have 39 degrees of freedom (n-1 = 40-1 = 39)
The potential values of k are given as multiple choice options.
Using a TI83, we can type in
tcdf(-2.023, 2.023, 39)
and it would display 0.9500333144 as the approximate result
Therefore, P(-2.023 < T < 2.023) = 0.95 approximately
This is why choice A is the answer.
Side Note: The tcdf function is found by hitting the key labeled "2nd" followed by the "VARS" key.
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If we weren't given a list of options, then we'd have to use a calculator like this one
http://www1.psych.purdue.edu/~gfrancis/calculators/inverse_t_dist.shtml
Type 39 for the box labeled "df", which stands for the degrees of freedom.
The area will be 0.95 which is the confidence level
Then click the "between" radio button option and hit "Recalculate"
It should show "Between -2.02269 and 2.02269"
We only focus on the positive version so we focus on 2.02269 which rounds to 2.0223
The nice thing about that calculator is that it also gives a diagram of what the area looks like under the T distribution curve.
There are many free online calculators that can perform the same idea.
Another option is that you could use GeoGebra's inverseT function
So you would type in
InverseTDistribution(39, (1-0.95)/2)
The portion (1-0.95)/2 calculates the area in one tail
This function helps find the value of k such that P(T < k) = A, where A is the area under the curve to the left of k. In this case, A = (1-0.95)/2
Yet another option is to use spreadsheet software like Microsoft Excel, OpenOffice, LibreOffice, or Google Sheets to name a few.
Excel costs money while the other options are free. While Excel does cost money, it's very standard and if you have that program, then stick to it.
The function you'll be using with these spreadsheet programs is called T.INV with the first argument being the area under the curve and the second argument is the df value.
Example: =T.INV( (1-0.95)/2, 39)
Don't forget about the equal sign up front. The dot between the "T" and "INV" is needed.
Some more info about the function can be found here
https://www.excelfunctions.net/excel-t-inv-function.html
Since many of the free spreadsheet options base their function names on excel, the T.INV function should work in all of the options mentioned.
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If you were in an exam setting and unable to use online calculators, or computer software, then you'll have to rely on the method in the first section.
Or you could calculate the t critical value with a custom-built inverseT function as described in this video
https://www.youtube.com/watch?v=5Ft5eZVJtPk
Unfortunately TI83 calculators do not have a native built in inverseT function already good to go.
Luckily the code needed to make such a program isn't too lengthy. Follow the steps provided in the video and feel free to ask any questions.
Another option is to use a table. I don't really recommend this method since calculators/spreadsheets are more commonplace nowadays.
However, if you must use a table, then here is one you could use
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
and here is another similar table (with slightly different entries)
http://www.math.odu.edu/stat130/t-tables.pdf
Such tables are found in the appendix section of your stats textbook.
As you can see, the problem with tables is that they can only fit so much info. The row df = 39 is not present in either table.
So we'll have to go with the closest thing df = 40 or df = 38 (depending which table you use).
For either table, you'll be looking in the column that has "area in two tails = 0.05".
This is because the main body takes up 95% of the area under the curve, so the remaining part in both tails combined is 100% - 95% = 5% = 0.05
For the first table I mentioned, the value 2.021 is in row df = 40 and column "two tails = 0.05". The value 2.021 isn't too far from 2.023
In the second table, the value 2.024 is in row df = 38 and column "two tails = 0.05"
If we didn't know anything about the calculator result, and only knew about the table entries 2.021 and 2.024, then perhaps the true answer is somewhere in between.
It of course isn't the midpoint, but it may be for some cases.
I think your teacher will only expect you to use a table if and only if the df entry is found in the table. Of course it's best to ask your teacher about clarification on the matter.
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