PS390 Statistical Reasoning in Psychology Assignment 04

PS390 Statistical Reasoning in Psychology Assignment 04 includes questions related to frequency table, histogram, hypothesis testing, effect size, and APA format reporting. The assignment covers topics such as sleep deprivation, short-term memory, concern for farm workers, and driving speed. The solutions are provided with detailed explanations and calculations.

ASSIGNMENT 04
PS390 Statistical Reasoning in Psychology
Student Name:
Instructor Name:
Course Number:
20th August 2018
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1. (25 points) A prison psychologist recorded the number of rule infractions for 15 prison
inmates over a six-month period to be 5, 4, 2, 4, 3, 5, 2, 0, 4, 4, 5, 5, 3, 4, and 3.
a. Make a frequency table.
Answer
Class Frequency
0 1
2 2
3 3
4 5
5 4
b. Make a histogram based on the frequency table.
Answer
c. Describe in words the shape of the histogram.
Answer
The shape of the histogram shows that the data is not normally distributed but is
rather skewed to the left (longer tail to the left)
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2. (25 points) Identify and solve this problem by hand.
The head of public safety notices that the average driving speed at a particular
intersection averages μ = 35 mph with a standard deviation of σ = 7.5 mph. After a school
speed limit sign of 20 mph is placed at the intersection, the first 40 cars travel past at an
average speed of 32 mph. Using the .01 significance level, was there a significant change
in driving speed?
a. Use the five steps of hypothesis testing (report results in APA format).
Answer
Step 1: State the null hypothesis and the alternate hypothesis
The hypothesis to be tested in this case is;
H0 : μ=35
H A : μ ≠35
Step 2: Select the appropriate test statistic and level of significance
The test statistic for this problem would be the z-score
Z= x−μ
σ / √n
The level of significance is 0.01 (i.e. α = 0.01)
Step 3: State the decision rules
The critical z value is 2.576. This means that the null hypothesis is rejected if the
computed z-score value is greater than the critical z-value. Otherwise we fail to reject
the null hypothesis.
Step 4: Compute the appropriate test statistic and make the decision
Z= x−μ
σ / √n = 32−35
7.5 / √40 =−2.52982
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|Z|=|−2.52982|=2.52982
The computed Z score (absolute value) is 2.52982; this value is less than the critical
z-value of 2.576. This means that we do not reject the null hypothesis.
Step 5: Interpret the decision
By not rejecting the null hypothesis we conclude that there is no significant change in
driving speed at 1% level of significance.
b. Sketch the distributions involved.
c. Figure the confidence limits for the 99% confidence interval.
Answer
The confidence interval (C.I) is given as;
C . I : x ± ME
ME=
zα
2
σ
√ n = 2.576∗7.5
√ 40 =3.05476
x=32
C . I : x ± ME → 32 ±3.05476
Lower limit: 32−3.05476=28.94524
Upper limit: 32+3.05476=35.05476
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3. (25 points) You will have to use a combination of StatCrunch and hand calculations (e.g.,
effect sizes) to solve the following problems.
A social psychologist gave a questionnaire about concern for farm workers to seven
participants before and after they attended a film about union organization of farm
workers. The results are shown below with high scores meaning high concern. Using
the .05 significance level, do these results support the hypothesis that the film affected
concern for the lives of farm workers?
Scores on the Concern Measure
Participant Before After
A 17 20
B 7 4
C 10 11
D 13 15
E 8 5
F 9 8
G 11 14
a. Use the five steps of hypothesis testing.
Answer
Step 1: State the null hypothesis and the alternate hypothesis
The hypothesis to be tested in this case is;
H0 : μB =μA
H A : μB ≠ μA
Where μB=Mean score for before
μA =Mean score for after
Step 2: Select the appropriate test statistic and level of significance
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The test statistic for this problem would be the paired t-test
t= x−μ
σ / √ n
The level of significance is 0.05 (i.e. α = 0.05)
Step 3: State the decision rules
The critical t value is computed based on 6 degrees of freedom at 5% level of
significance. The critical t-value is 2.446912. The null hypothesis is rejected if the
computed t-value is greater than the critical t-value. Otherwise we fail to reject the
null hypothesis.
Step 4: Compute the appropriate test statistic and make the decision
Using Statcrunch we obtained the t-test as given below;
t-Test: Paired Two Sample for Means
Before After
Mean 10.71429 11
Variance 11.57143 33.33333
Observations 7 7
Pearson Correlation 0.967433
Hypothesized Mean
Difference 0
df 6
t Stat -0.28768
P(T<=t) one-tail 0.391636
t Critical one-tail 1.94318
P(T<=t) two-tail 0.783271
t Critical two-tail 2.446912
The computed t-value (absolute value) is 0.28768; this value is less than the critical t-
value of 2.4469. This means that we do not reject the null hypothesis.
Step 5: Interpret the decision
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By not rejecting the null hypothesis we conclude that these results do not support the
hypothesis that the film affected concern for the lives of farm workers at 5% level of
significance.
b. Figure the effect size and find the approximate power of this study.
Answer
Cohe n' sd= M A −MB
σ Pooled
M A =11, MB =10.71429
σ Pooled= √ σ A
2 + σB
2
2 = √ 11.57143+33.3333
2 = √ 44.90473
2 = √ 22.452365=4.73839
Cohe n' sd= M A −MB
σ Pooled
=11−10.71429
4.73839 =0.60298
c. Report results in APA format.
Answer
A paired-samples t-test was conducted to compare Scores on the Concern Measure
In before and after attending a film about union organization of farm workers
conditions. There was no significant difference in the concern scores for before (M =
10.71, SD = 3.40) and after (M = 11.00, SD = 5.77) conditions; t(6)=-0.288, p =
0.7833. These results suggest that watching film about union organization of farm
workers does not have a significant influence on concern for the lives of farm
workers.
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4. (25 points) You will have to use a combination of StatCrunch and hand calculations (e.g.,
effect sizes) to solve the following problems.
A team of cognitive psychologists studying the effects of sleep deprivation on short-term
memory decay had eight participants stay in a sleep lab for two days. Four participants
were randomly assigned to a condition in which they were not permitted to sleep during
that period, while the other four participants were allowed to sleep when they wanted to.
At the end of the two days, the participants completed a short-term memory task that
yielded the results in the table that follows. Using the .05 significance level, did sleep
deprivation reduce short-term memory?
Mean Number of Letters Remembered
Sleep Deprived Normal Sleep
7 9
8 8
7 11
9 7
a. Create the appropriate graph for this problem.
Answer
b. Use the five steps of hypothesis testing.
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Answer
Step 1: State the null hypothesis and the alternate hypothesis
The hypothesis to be tested in this case is;
H0 : μD =μN
H A : μD ≠ μN
Where μD=Mean number of letters remembered for deprived
μN=Mean number of letters remembered for normal
Step 2: Select the appropriate test statistic and level of significance
The test statistic for this problem would be the an independent samples t-test
t= x−μ
σ / √ n
The level of significance is 0.05 (i.e. α = 0.05)
Step 3: State the decision rules
The critical t value is computed based on 6 degrees of freedom at 5% level of
significance. The critical t-value is 2.446912. The null hypothesis is rejected if the
computed t-value is greater than the critical t-value. Otherwise we fail to reject the
null hypothesis.
Step 4: Compute the appropriate test statistic and make the decision
Using Statcrunch we obtained the t-test as given below;
t-Test: Two-Sample Assuming Equal Variances
Sleep
Deprived
Normal
Sleep
Mean 7.75 8.75
Variance 0.916667 2.916667
Observations 4 4
Pooled Variance 1.916667
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Hypothesized Mean
Difference 0
df 6
t Stat -1.02151
P(T<=t) one-tail 0.173211
t Critical one-tail 1.94318
P(T<=t) two-tail 0.346422
t Critical two-tail 2.446912
The computed t-value (absolute value) is 1.0215; this value is less than the critical t-
value of 2.4469. This means that we do not reject the null hypothesis.
Step 5: Interpret the decision
By not rejecting the null hypothesis we conclude that sleep deprivation did not
significantly reduce short-term memory at 5% level of significance.
c. Figure the effect size.
Answer
Cohe n' sd= M N −M D
σ Pooled
M N =8.75 , M D =7.75
σ Pooled = √ σ A
2 + σB
2
2 = √ 2.91666667+0.91666667
2 = √ 3.833333
2 = √ 1.91666667=1.384437
Cohe n' sd= M N −M D
σ Pooled
=8.75−7.75
1.384437 =0.7223
d. Report results in APA format.
Answer
An independent samples t-test was performed to compare the average participant’s short-
term memory based on whether the participant was deprived sleep or had a normal sleep.
Results showed that the average words remembered by the participants who were
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deprived sleep (M = 7.75, SD = 0.96, N = 4) was not significant different with the
average words remembered by the participants who had normal sleep (M = 8.75, SD =
1.71, N = 4), t (6) = -1.022, p > .05, two-tailed. The difference of 1.00 showed an
insignificant difference. Essentially results showed that sleep deprivation did not
significantly reduce short-term memory at 5% level of significance.
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