Question 1189881: The IRS claims that it takes 90 minutes to prepare a tax form. You believe it takes longer than 90 minutes. You ask 22 tax filers and find the following preparation times:
66, 69, 78, 79, 87, 87, 89, 96, 96, 97, 98, 98, 99, 100, 100, 105, 108, 115, 117, 118, 122, 125
Determine if the preparation time is significantly longer than 90 minutes. Use α = .025.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Hypotheses:
H0: mu = 90
H1: mu > 90
mu = population mean of the tax prep times (in minutes)
This is a right-tailed test due directly to the inequality sign in H1
Given Data Set
{66, 69, 78, 79, 87, 87, 89, 96, 96, 97, 98, 98, 99, 100, 100, 105, 108, 115, 117, 118, 122, 125}
sample size = n = 22
Since we don't know sigma, and n is not larger than 30, this means we'll use the T distribution (instead of the Z distribution).
The T distribution has degrees of freedom = df = n-1 = 22-1 = 21
Add up the values in the data set and divide by n = 22 to compute the sample mean xbar.
You should find that the values add to 2149
So, xbar = 2149/22 = 97.68182 approximately
Based on this xbar value alone, it appears the mean tax prep time is over 90 minutes.
However, this might be due to random chance and bad luck.
Perhaps we could do the experiment again to randomly select 22 other people and get some xbar 90 or smaller?
We'll have to take into account the standard deviation (and subsequently the standard error).
Use your calculator to compute the sample standard deviation of that original set. You should get roughly
s = 16.13746
Make sure you use the sample standard deviation and not the population standard deviation. There's a subtle difference.
This leads to a standard error (SE) of...
SE = s/sqrt(n)
SE = 16.13746/sqrt(22)
SE = 3.44052
which is also approximate
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We found that
mu = 90
xbar = 97.68182
SE = 3.44052
The test statistic is
t = (xbar - mu)/SE
t = (97.68182 - 90)/3.44052
t = 2.23274970062664
t = 2.23
Now to calculate the p-value.
You can use a T Table such as this one (or anything similar found in the back of your stats textbook)
http://www.ttable.org/
Look in the row df = 21
Try to see if 2.23 is in that row. Unfortunately it isn't. But it's between 2.080 and 2.518
Trace back up those respective columns and locate the "one tail" values in bold.
For the 2.080, that one tail value is 0.025
For the 2.518, that one tail value is 0.01
While we can't nail down the p-value exactly, we know that it's somewhere between 0.01 and 0.025
In other words,
0.01 < pvalue < 0.025
The level of significance was alpha = 0.025
Whatever the p-value is, it's definitely smaller than alpha.
We reject the null based on this.
The rule is if pvalue < alpha, then you reject the null.
A good way to remember the rule is to think of the phrase "if the p-value is low, then the null must go".
If you were to use a p-value calculator such as this
https://www.omnicalculator.com/statistics/p-value
then you'd find the p-value is roughly 0.018395 which helps confirm our decision.
Here's a screenshot of a piece of that page to show the inputs and output.
If you were to use a t-value calculator such as this
https://calculators.io/t-value-calculator/
Then you should find the T-value (right-tailed) is roughly 2.0796. Anything larger than this critical cut-off point will be in the rejection region.
Since t = 2.23 is larger than 2.0796, we'll reject the null.
So there are a few approaches we can take to see that we reject the null hypothesis.
We side with the alternative that mu > 90.
It appears that the mean tax prep time is longer than 90 minutes.
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