Question 1208012: Suppose that a word-association experiment is conducted using eight people as blocks and making a comparison of reaction times within each person; that is, each person is subjected to both stimuli in a random order. The reaction times (in seconds) for the experiment are as follows.
Person__________Stimulus 1__________Stimulus 2 (IGNORE THE LINES - space)
1__________________2__________________5 (IGNORE THE LINES - space)
2__________________1__________________3 (IGNORE THE LINES - space)
3__________________1__________________4 (IGNORE THE LINES - space)
4__________________2__________________2 (IGNORE THE LINES - space)
5__________________1__________________3 (IGNORE THE LINES - space)
6__________________2__________________3 (IGNORE THE LINES - space)
7__________________3__________________4 (IGNORE THE LINES - space)
8__________________2__________________3 (IGNORE THE LINES - space)
Do the data present sufficient evidence to indicate a difference in mean reaction times for the two stimuli? Test using 𝛼 = 0.05. (Use 𝜇1 − 𝜇2 = 𝜇d.)
a) State the test statistic. (Round your answer to three decimal places.)
t = __________
b) State the rejection region. (If the test is one-tailed, enter NONE for the unused region. Round your answers to three decimal places.)
t > __________
t < ___________
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
Answers:
(a) -4.333
(b) t > 2.365 or t < -2.365
All decimal values mentioned are approximate.
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Part (a)
Given data
| Person | Stimulus 1 | Stimulus 2 | | 1 | 2 | 5 | | 2 | 1 | 3 | | 3 | 1 | 4 | | 4 | 2 | 2 | | 5 | 1 | 3 | | 6 | 2 | 3 | | 7 | 3 | 4 | | 8 | 2 | 3 |
There are n = 8 people which is the sample size.
Subtract the first two columns to form a new column of values.
| Person | Stimulus 1 | Stimulus 2 | Stim1-Stim2 | | 1 | 2 | 5 | -3 | | 2 | 1 | 3 | -2 | | 3 | 1 | 4 | -3 | | 4 | 2 | 2 | 0 | | 5 | 1 | 3 | -2 | | 6 | 2 | 3 | -1 | | 7 | 3 | 4 | -1 | | 8 | 2 | 3 | -1 |
Example: 2-5 = -3 in the first row of this new column.
Add up the values in that new column to get -13.
Divide this sum over the sample size n = 8 to get -13/8 = -1.625 which is the value of xbard
xbar = x with a horizontal bar over top = sample mean
xbar with a subscript "d" indicates we're looking at the sample mean of the differences.
If you want, you can compute the sample standard deviation of the differences by hand, but it's much better to use technology.
Grab your favorite calculator. I used the spreadsheet's calculator since the data is already in spreadsheet form.
You should find that
sd = 1.06066017 approximately
Caution: Do not mix up sample standard deviation with population standard deviation. You'll be using the sample version.
Then,
Test Statistic = (xbard)/( (sd)/sqrt(n) )
Test Statistic = (-1.625)/( (1.06066017)/sqrt(8) )
Test Statistic = -4.333333340605
Test Statistic = -4.333
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Part (b)
We're doing a paired T-test.
𝜇 = greek letter mu = population mean
𝜇1 = population mean of stimulus 1.
𝜇2 = population mean of stimulus 2.
𝜇d = 𝜇1-𝜇2 = population mean of the differences.
Hypotheses
Null: 𝜇d = 0
Alternative: 𝜇d =/= 0
The alternative hypothesis has the "not equal" sign because of the phrasing "Do the data present sufficient evidence to indicate a difference in mean reaction times for the two stimuli?"
If the answer to this question was "no", then we'd stick with the null.
The "not equal" sign in the alternative means we have a two-tailed test.
As mentioned, we're doing a paired T-test.
We'll be using the T distribution.
n = 8 = sample size
df = degrees of freedom
df = n-1
df = 8-1
df = 7
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). Alternatively you can use a stats calculator such as a TI83.
In that table, highlight the df = 7 row and "two tail = 0.05" column.
The 0.05 refers to the alpha value.
This row and column intersect to give us the value 2.365 which is approximate.
It tells us that P(-2.365 < t < 2.365) = 0.95 when df = 7.
i.e. the area under the curve between t = -2.365 and t = 2.365 is roughly 0.95
This applies only when df = 7.
This corresponds to a 95% confidence interval.
The remaining 1-0.95 = 0.05 is the combined area of both tails.
If the test statistic is in the interval -2.365 < t < 2.365, then we fail to reject the null.
Otherwise the test statistic is in the rejection region and we would reject the null.
Therefore the rejection region is:
t > 2.365 or t < -2.365
Extra info:
Test Statistic = -4.333 was found earlier in part (a).
It is outside the interval -2.365 < t < 2.365 and fits with t < -2.365 instead.
We're in the rejection region.
This means we'll reject the null. We'll conclude that there appears to be a difference in the reaction times for the two stimuli.
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