Download t-Test Worksheet Answers

Psych 201: t-Test Worksheet Answers
This worksheet will review the t-Test then give you some problems to work. Refer back
to the previous handout on t-tests using Excel.
Some Review:
Why use the t-Test?
When you want to compare the means of two groups.
When to use One-tailed or Two-tailed tests?
The answer is that it depends on your hypothesis. If your hypothesis states the direction
of the difference or relationship, then you use a one-tailed test. Example null hypothesis
would include: Females will not score significantly higher than males on the SAT. The
null hypothesis (indirectly) predicts the direction of the difference.
A two-tailed test would be used to test this null hypothesis: There will be no significant
difference in SAT scores between males and females.
While it is generally safest to use a two-tailed tests, there are situations where a one-tailed
test seems more appropriate. The bottom line is that it is the choice of the researcher
whether to use one-tailed or two-tailed research questions.
The Formulae:
For the independent-measures t, the sample statistic is the sample mean difference (M1M2). The population parameter is the population mean difference (µ1-µ2). The estimated
standard error for the sample mean difference is computed by combining the errors for
the two sample means. The resulting formula is:
t=
(M 1 − M 2 ) − (µ 1 − µ 2 )
s(M1 − M 2 )
Where the estimated standard error is:
s(M1 − M 2 ) =
s 2p
n1
+
s 2p
n2
The pooled variance in the formula, s 2p , is the weighted mean of the two sample
variances:
T-Test worksheet
Dr. Sullivan
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s 2p =
SS1 + SS 2
df 1 + df 2
This t-statistic has degrees of freedom determined by the sum of the df values for the two
samples:
df = df 1 + df 2 = (n1 − 1) + (n2 − 1)
For hypothesis testing, the null hypothesis normally states that there is no difference
between the two population means:
H 0 : µ1 = µ 2 or µ1 − µ 2 = 0
Appropriate use and interpretation of the t statistic require that the data satisfy the
homogeneity of variance assumption. This assumption stipulates that the two populations
have equal variances. An informal test of this assumption can be made by simply
comparing the two sample variances. If the two sample variances are approximately
equal, the t test is justified.
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Dr. Sullivan
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Problem 1:
Dr. Jittery is exploring the effects of caffeine on college students. Though caffeine is
widely used to defer sleep and increase the available time per day to perform homework
or improve attention in class, too much caffeine may have negative effects. Dr. Jittery
believes fine motor control may be a good measure of the hypothesized negative effect,
and creates a maze-tracing task that will allow her to collect data on how accurately
subjects can move a pointer through the maze. A counter will be kept to measure the
number of times the subject bumps the pointer into a wall of the maze. Her hypothesis is
that too much caffeine will affect the accuracy of the maze tracing.
Null Hypothesis: Caffeine will have no effect on the number of bumps
Alternative Hypothesis: Caffeine will have an effect on the number of bumps.
She selects an alpha level of 0.05 and 8 subjects for each condition (No Espresso vs 6
Espressos). The df (degrees of freedom) are __14___ and the critical t value will be
___2.145_____ (based upon the table in the book).
She runs her subjects and the following data is produced:
No Espresso
4
5
0
4
2
4
1
4
6 Cups of Espresso
8
10
9
8
8
11
10
9
Using Excel, we calculate the t-test assuming equal variances and get the following:
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
Df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
T-Test worksheet
No Espresso
3
3.142857143
8
2.205357143
0
14
-8.248911105
4.78451E-07
1.761310115
9.56903E-07
2.144786681
Dr. Sullivan
6 Cups of
Espresso
9.125
1.267857143
8
3
We see that the calculated t value is -8.25, which exceeds the two tail critical t value of
2.145. Thus, Dr Jittery can REJECT her null hypothesis that caffeine has no effect and
ACCEPT the alternate hypothesis that caffeine does have an effect, decreasing the
accuracy of the tracing task.
Questions to ponder:
What may have Dr. Jittery overlooked in running her experiment?
Had the subject consumed any other caffeinated drinks prior to their
participation?
Had any subjects come from weight training and suffered from fatigue?
Could other factors have influenced her results?
General motor skill differences?
Was the assumption about equal variances correct in performing the t-test?
There is some difference in variance. Don’t know if it is significant, and
should be tested.
How would you determine if the variances are equal?
F-Max test
Fmax =
s 2 (l arg est ) 3.14
=
= 2.47
s 2 ( smallest ) 1.27
The value for F-max with k=2 (the number of samples or groups) and n1=7 is 4.99. Since the our calculated F-max is less than 4.99, this
means our samples can be considered to have equal variances!
T-Test worksheet
Dr. Sullivan
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Problem 2:
Dr. Lunesta is studying the effects of sleep deprivation on a memory recall task. He
hypothesizes that persons deprived of sleep will perform poorly on the task. The task
results are measured in terms of number of errors.
State Dr. Lunesta’s:
Null Hypothesis: H0: Sleep deprivation does not affect performance on a
memory task
Alternate Hypothesis: Ha: Sleep deprivation affects performance on a
memory task
10 subjects will be run in each of two conditions (normal 8 hour sleep in the 24 hours
prior vs. no sleep in the prior 24 hours).
What is the df value? 18
What is the critical t-value, two-tailed for an alpha of 0.05? 2.101
For the following data, use Excel to calculate the t value and report on Dr. Lunesta’s
findings:
Errors with Normal Sleep
4
3
4
6
4
7
2
4
5
6
Errors with No Sleep
6
4
2
8
2
5
2
4
3
4
Calculated value of t? 0.643 (see next page)
Reject or accept the null hypothesis? ACCEPT the Null Hypothesis. There is no
significant difference in mean scores between the two groups.
T-Test worksheet
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t-Test: Two-Sample Assuming Equal Variances
Errors with Normal
Sleep
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean
Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
T-Test worksheet
4.5
2.277777778
10
3.027777778
Errors with No
Sleep
4
3.777777778
10
0
18
0.642529405
0.264314718
1.734063592
0.528629436
2.100922037
Dr. Sullivan
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