Question 1207926: To test the effect of alcohol in increasing the reaction time to respond to a given stimulus, the reaction times of seven people were measured. After consuming 89 mL of 40% alcohol, the reaction time for each of the seven people was measured again. Do the following data indicate that the mean reaction time after consuming alcohol was greater than the mean reaction time before consuming alcohol? Use πΌ = 0.05. (Use πbefore β πafter = πd.Round your answers to three decimal places.)
DATA:-
Person___1___2___3___4___5____6_____7 (IGNORE THE LINES -represents space)
Before___5___6___5___5___3____6_____1 (IGNORE THE LINES -represents space)
After____7___8___3___6___3____4_____6 (IGNORE THE LINES -represents space)
a. Test statistic: t = __________
b. Rejection region: If the test is one-tailed, enter NONE for the unused region.
t > _______
t < ________
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Answers:
(a) -0.915
(b) t > NONE and t < -1.943
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Explanation for part (a)
Here is the original data given to us.
| Person | 1 | 2 | 3 | 4 | 5 | 6 | 7 | | Before | 5 | 6 | 5 | 5 | 3 | 6 | 1 | | After | 7 | 8 | 3 | 6 | 3 | 4 | 6 |
There are n = 7 pairs of values, i.e. there are n = 7 people.
Subtract the before and after in the order before-after.
Place the results in a new row.
| Person | 1 | 2 | 3 | 4 | 5 | 6 | 7 | | Before | 5 | 6 | 5 | 5 | 3 | 6 | 1 | | After | 7 | 8 | 3 | 6 | 3 | 4 | 6 | | Before - After | -2 | -2 | 2 | -1 | 0 | 2 | -5 |
Add up the values in the bottom row. Divide that sum by n = 7 to find the sample mean of these differences is roughly xbard = -0.85714286
xbar refers to x having a horizontal bar over top.
xbar is the sample mean.
The subscript d represents the xbar of these differences.
Use technology (I recommend a spreadsheet) to find the sample standard deviation of the differences.
sd = 2.4784788 approximately
t = test statistic
t = ( xbard )/( (sd)/sqrt(n) )
t = (-0.85714286)/( 2.4784788/sqrt(7) )
t = -0.915 approximately which is the answer to part (a)
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Explanation for part (b)
Whenever you have before-and-after type of problems like this, it means you'll be doing a paired T test.
The null hypothesis is πbefore = πafter
Which is the same as saying πbefore β πafter = 0
I'll replace "before" with the subscript 1.
I'll replace "after" with the subscript 2.
So we have π1 = π2 aka π1 β π2 = 0 as the null.
The instructions ask this question
"Do the following data indicate that the mean reaction time after consuming alcohol was greater than the mean reaction time before consuming alcohol?"
It boils down to
"is before < after?"
i.e.
"is π1 < π2 true?"
This would form the alternate hypothesis.
Here are both hypotheses
Null: π1 = π2
Alternate: π1 < π2
The inequality sign in the alternate hypothesis gives the direction of the test.
We have a left-tailed test.
This means we ignore the "t > " portion of part (b).
In other words, we'll have t > NONE as part of the answer.
n = number of pairs of values = number of people
n = 7
df = degrees of freedom
df = n-1
df = 7-1
df = 6
I'll now use this T table
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
A similar table should be found in your textbook somewhere. Likely in the back (appendix section).
In that table, highlight the df = 6 row and "one tail = 0.05" column.
The 0.05 refers to the alpha value.
This row and column intersect to give us the value 1.943 which is approximate.
It tells us that
P(T > 1.943) = 0.05 approximately when df = 6
By symmetry we can say
P(T < -1.943) = 0.05 approximately when df = 6
Therefore the rejection region is t < -1.943
Extra info:
Recall the test statistic, from part (a), was found to be roughly -0.915
It is not in the rejection region, so we fail to reject the null.
We conclude that πbefore = πafter appears to be the case.
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