PSY 302 Stats Midterm #2

1. Which of the following best explains why the t-distribution is used when the population
standard deviation (Οƒ) is unknown?
a. The t-distribution is narrower than the z-distribution to accommodate small
samples.
b. The t-distribution is used only for large sample sizes.
c. The t-distribution always has fixed critical values, regardless of the sample size.
d. The t-distribution has a wider shape to account for the increased variability
and uncertainty associated with estimating Οƒ.
2. What is a defining property of the t-distribution when compared to the z-distribution?
a. It always has critical values of Β±1.96 at Ξ± = 0.05.
b. It has critical values that depend on the sample size (degrees of freedom).
c. It is only used for large samples greater than 30.
d. It assumes the population standard deviation is known.
3. How do degrees of freedom (df) affect the shape of the t-distribution?
a. As the degrees of freedom increase, the t-distribution approaches the shape
of the z-distribution.
b. As the degrees of freedom increase, the t-distribution becomes more spread out.
c. Degrees of freedom have no effect on the shape of the t-distribution.
d. Degrees of freedom make the t-distribution less useful for small samples.
4. What is the assumption of homogeneity of variance in an independent samples t-test?
a. It assumes that the means of the two groups are equal.
b. It assumes that the population mean is known.
c. It assumes that the variance in both groups is approximately equal.
d. It assumes that there is no relationship between the independent and dependent
variables.
5. Why do we use difference scores in a paired samples t-test?
a. To compare the average score of two unrelated groups.
b. To isolate and compare changes within the same individuals, minimizing the
effect of individual differences.
c. To calculate the standard error.
d. To determine the homogeneity of variance between groups.

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6. Why is a paired samples t-test more effective at controlling for individual differences
compared to an independent samples t-test?
a. Because it involves different participants for each condition.
b. Because it measures only one group of individuals.
c. Because it requires matching participants across different conditions.
d. Because it compares each participant to themselves, reducing the variability
caused by individual differences.
7. How does using a within-groups design help in paired samples t-tests?
a. It ensures each group has different participants, increasing variability.
b. It helps control for individual differences, making it a more powerful test.
c. It decreases the sample size required.
d. It introduces more variability, which makes results more generalizable.
8. Why is it problematic to perform multiple t-tests when comparing more than two groups?
a. It increases the risk of making a Type II error.
b. It makes the test less sensitive to differences.
c. It increases the probability of making a Type I error.
d. It leads to a decrease in statistical power.
9. When is it appropriate to use post-hoc tests following an ANOVA?
a. When we suspect that variances are unequal across groups.
b. If the omnibus ANOVA is significant, to identify specific group differences.
c. To check for differences when there are only two conditions.
d. When examining the interaction effect between variables.

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10. Why do larger sample sizes provide more stable or reliable estimates in the t-distribution?
a. Larger sample sizes produce more variability, requiring more conservative
estimates.
b. Larger sample sizes yield more stable estimates, reducing the uncertainty of
population parameter estimates.
c. Larger samples naturally increase the chance of Type I errors.
d. The reliability of estimates remains the same regardless of sample size.
11. What best describes 'between-group variability' in an ANOVA context?
a. It represents the natural variation among individual scores within each group.
b. It explains the variation due to measurement errors within groups.
c. It accounts for the consistency of scores within each treatment condition.
d. It captures the differences between the overall means of the groups.
12. How does matching participants in a paired samples t-test influence the results?
a. It controls for variability between different individuals, increasing power.
b. It introduces more noise into the data, reducing power.
c. It decreases the overall number of comparisons needed, slightly increasing power.
d. It makes comparisons across unrelated individuals, which would lower power.
13. Which metric involves creating a combined estimate of variance from two samples to
represent the variance of the null hypothesis?
a. Pooled Standard deviation
b. Combined Standard error
c. Pooled Variance
d. Combined Sampling variance

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14. What is the mean of the sampling distribution in a paired samples t-test under the null
hypothesis?
a. 𝑴𝑫 = 𝟎
b. 𝑀1 βˆ’ 𝑀2 = 0
c. 𝑀𝐷 β‰  0
d. 𝑀1 βˆ’ 𝑀2 β‰  0
15. What does a significant F-ratio in a one-way ANOVA indicate?
a. All group means are equal.
b. At least one group mean is significantly different from the others.
c. The variances between groups are equal.
d. There is no relationship between the independent and dependent variables.
16. What is the null hypothesis for a one-way ANOVA?
a. There is a significant difference between at least two group means.
b. The variances of the groups are unequal.
c. All group means are equal.
d. The dependent variable has no variation.
17. Which of the following would be an appropriate situation to use a one-way ANOVA?
a. Comparing the test scores of students before and after a training session.
b. Comparing the average heights of plants grown under three different types
of fertilizers.
c. Analyzing the relationship between age and income.
d. Examining the effect of two different diets on weight loss in a matched sample.

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18. What is the main difference between an independent samples t-test and a paired samples
t-test?
a. An independent samples t-test compares means from the same group of
participants at two time points.
b. A paired samples t-test requires more participants than an independent samples t-
test.
c. An independent samples t-test can only be used with categorical data.
d. An independent samples t-test compares means from two different groups,
while a paired samples t-test compares means from the same group at two
time points.
19. In a paired samples t-test, what does it mean if the mean of the difference scores is
significantly different from zero?
a. The two samples are unrelated.
b. There is a significant change or difference between the two related samples.
c. The variances of the samples are equal.
d. The difference scores have no effect on the outcome.
20. Which of the following scenarios would call for a paired samples t-test?
a. Measuring the performance of students before and after a training program.
b. Comparing the exam scores of two different classes.
c. Testing the effect of two different medications on separate groups.
d. Analyzing the relationship between two continuous variables.

Midterm 2 Review
For the following questions you will need to work through all of the steps of hypothesis
testing. On the exam this will all be multiple choice but here I want you to work through it all
without that framework to ensure that you understand the process. You will be asked on
the exam about any of the following components:
β€’ Which test do I use for this scenario?
β€’ What are the assumptions of this test?
β€’ What are the null and alternative hypotheses? (Plain language or in symbols)
β€’ What do you set alpha to?
β€’ What are the critical values for your test?
β€’ Calculate the test statistic (You may have to calculate from scratch, but you may
also be given shortcuts in the form of variance or SS having already been
calculated)
β€’ Do you reject or fail to reject the null hypothesis?
β€’ What is the direction of the effect?
β€’ What is the effect size?
β€’ What is the 95% Confidence Interval?
β€’ What is the interpretation of the findings?

Midterm 2 Review
You are a developmental psychologist interested in how different types of play
environments impact children's problem-solving abilities. You design a study with
three groups of children, each exposed to a different play environment for one hour:
(1) a structured puzzle-based play area, (2) a free-play area with various toys, and (3) a
no-play control group where children read quietly. After the session, you assess each
child's problem-solving ability using a standardized test that yields a score from 0 to
100, where higher scores indicate better problem-solving skills.
You collect data from 15 children, with 5 children in each group. Here are the scores:
β€’ Puzzle-Based Play Group: 85, 78, 90, 88, 84
β€’ Free-Play Group: 72, 75, 80, 70, 77
β€’ No-Play Control Group: 65, 68, 62, 70, 67
Step 1: Pick a Test and check Assumptions
β€’ Test – Between Subjects One-Way ANOVA
β€’ Assumptions
o Independence of cases
o Homogeneity of variances
o Normality
o Random sample
Step 2: List the Null and Alternative Hypotheses
β€’ 𝐻0: There is no difference in the mean problem-solving scores between the three
groups.
o πœ‡π‘π‘’π‘§π‘§π‘™π‘’ = πœ‡π‘“π‘Ÿπ‘’π‘’ = πœ‡π‘›π‘œ
β€’ 𝐻1: There is a difference in the mean problem-solving scores between the three
groups.
o πœ‡π‘π‘’π‘§π‘§π‘™π‘’ β‰  πœ‡π‘“π‘Ÿπ‘’π‘’ β‰  πœ‡π‘›π‘œ
Step 3: Set the Decision Rule
β€’ alpha = 0.05
β€’ Degrees of freedom

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o 𝑑𝑓𝑏𝑒𝑑𝑀𝑒𝑒𝑛 = 2 (k - 1 = 3 - 1 = 2)
o π‘‘π‘“π‘€π‘–π‘‘β„Žπ‘–π‘› = 12 (N - k = 15 - 3 = 12)
β€’ 𝐹𝑐𝑣= 3.885

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Step 4: Calculate the Test Statistic

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Step 5: Decide Whether to Reject or Fail to Reject the Null Hypothesis
β€’ F = 28.29 and the 𝐹𝑐𝑣= 3.885
β€’ Reject the null hypothesis; the probability of obtaining this sample mean difference
when the null hypothesis is true is less than 5% (p < .05)
Step 6: (*For this one we can just interpret at this point – no effect or 95%CI*)
β€’ All you can say with the results that you have at this point is that at least one of the
means is significantly different than at least one other mean.
β€’ You will not be required to calculate a posthoc test but you still need to know what
they are.

Midterm 2 Review
You are a clinical psychologist interested in evaluating the effectiveness of a new
cognitive-behavioral therapy (CBT) intervention designed to reduce anxiety. You recruit
10 participants who suffer from moderate anxiety and measure their anxiety levels
using a standardized anxiety scale (where higher scores indicate greater anxiety)
before and after the 6-week CBT program.
Here are the anxiety scores for each participant before and after the therapy:
β€’ Participant Scores Before Therapy: 25, 28, 30, 27, 32, 26, 29, 31, 33, 30
β€’ Participant Scores After Therapy: 20, 22, 24, 23, 27, 21, 25, 26, 28, 24
Step 1: Pick a Test and Check the Assumptions
β€’ Paired Samples t-Test
β€’ Assumptions
o Independence of cases
o Normality
o Random sample
Step 2: List the Null and Alternative Hypotheses
β€’ 𝐻0: The mean difference in anxiety scores before and after the therapy is zero. (There
is no effect of the CBT intervention on anxiety levels.).
o πœ‡π· = 0
β€’ 𝐻1: The mean difference in anxiety scores before and after the therapy is not zero.
(There is an effect of the CBT intervention on anxiety levels.).
o πœ‡π· β‰  0
Step 3: Set the Decision Rule
β€’ alpha = 0.05
β€’ Degrees of freedom
o 𝑛𝐷 βˆ’ 1 = 9
β€’ 𝑑𝑐𝑣 = 2.262

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Step 4: Calculate the Test Statistic

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Step 5: Decide Whether to Reject or Fail to Reject the Null Hypothesis
β€’ t = -21.86 and the 𝑑𝑐𝑣= 2.262
β€’ Reject the null hypothesis; the probability of obtaining this sample mean difference
when the null hypothesis is true is less than 5% (p < .05)
Step 6: Direction of Effect - Calculate Effect Size and 95%CI
β€’ Scores after therapy were significantly lower than before therapy.
Step 7: Interpret Overall Results
β€’ There is strong evidence to suggest that the CBT intervention significantly reduced
anxiety levels in participants. The therapy appears to be highly effective in reducing
anxiety based on this sample.

Midterm 2 Review
Imagine you are a social psychologist interested in the effects of background music on
concentration. You design an experiment to test whether different types of
background music affect performance on a concentration task. You randomly assign
participants to one of two groups: one group listens to classical music while
completing the task, and the other group listens to pop music. The performance on
the concentration task is measured by the number of correct answers on a series of
logic problems, with higher scores indicating better concentration.
You collect data from 20 participants, with 10 participants in each group. Here are the
scores for each group:
β€’ Classical Music Group: 18, 20, 17, 19, 21, 22, 18, 20, 19, 21
β€’ Pop Music Group: 15, 16, 14, 17, 16, 15, 18, 14, 15, 16
Step 1: Pick a Test and Check the Assumptions
β€’ Independent Samples t-Test
β€’ Assumptions
o Independence of cases
o Homogeneity of variances
o Normality
o Random sample
Step 2: List the Null and Alternative Hypotheses
β€’ 𝐻0: There is no difference in the mean concentration scores between the classical
and pop music groups.
o πœ‡π‘π‘™π‘Žπ‘ π‘ π‘–π‘π‘Žπ‘™ βˆ’ πœ‡π‘π‘œπ‘ = 0
β€’ 𝐻1: There is a difference in the mean concentration scores between the classical
and pop music groups.
o πœ‡π‘π‘™π‘Žπ‘ π‘ π‘–π‘π‘Žπ‘™ βˆ’ πœ‡π‘π‘œπ‘ β‰  0
Step 3: Set the Decision Rule
β€’ alpha = 0.05
β€’ Degrees of freedom
o 𝑁 βˆ’ 2 = 18
β€’ 𝑑𝑐𝑣 = 2.101

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Step 4: Calculate the Test Statistic

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Step 5: Decide Whether to Reject or Fail to Reject the Null Hypothesis
β€’ t = 6.09 and the 𝑑𝑐𝑣= 2.101
β€’ Reject the null hypothesis; the probability of obtaining this sample mean difference
when the null hypothesis is true is less than 5% (p < .05)
Step 6: Direction of Effect - Calculate Effect Size and 95%CI
β€’ Classical music concentration scores were significantly higher than pop music
concentration scores.
Step 7: Interpret Overall Results
β€’ There is strong evidence to suggest that the type of background music significantly
affects concentration. Specifically, participants who listened to classical music
performed significantly better on the concentration task compared to those who
listened to pop music. This suggests that classical music may enhance
concentration more effectively than pop music.

Midterm 2 Review
You are a researcher interested in studying the impact of sleep deprivation on reaction
time. Previous research has established that the average reaction time for well-rested
individuals on a computerized task is 300 milliseconds (ms) (ΞΌ = 300 ms). You want to
know if a group of individuals who have been sleep-deprived for 24 hours will have a
different average reaction time.
You conduct a study with 10 participants who have not slept for 24 hours. After
administering the reaction time test, you record the following reaction times (in
milliseconds):
β€’ 320, 310, 305, 295, 315, 325, 330, 290, 300, 310
Step 1: Pick a Test and Check the Assumptions
β€’ Single Sample t-Test
β€’ Assumptions
o Independence of cases
o Normality
o Random sample
Step 2: List the Null and Alternative Hypotheses
β€’ 𝐻0: The mean reaction time of sleep-deprived individuals is no different from well
rested individuals.
o πœ‡π‘ π‘™π‘’π‘’π‘_π‘‘π‘’π‘π‘Ÿπ‘–π‘£π‘’π‘‘ = 300
β€’ 𝐻1: The mean reaction time of sleep-deprived individuals is no different from well
rested individuals.
o πœ‡π‘ π‘™π‘’π‘’π‘_π‘‘π‘’π‘π‘Ÿπ‘–π‘£π‘’π‘‘ β‰  300
Step 3: Set the Decision Rule
β€’ alpha = 0.05
β€’ Degrees of freedom
o 𝑛 βˆ’ 1 = 9
β€’ 𝑑𝑐𝑣 = 2.262

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Step 4: Calculate the Test Statistic
Step 5: Decide Whether to Reject or Fail to Reject the Null Hypothesis
β€’ t = 2.45 and the 𝑑𝑐𝑣= 2.262
β€’ Reject the null hypothesis; the probability of obtaining this sample mean difference
when the null hypothesis is true is less than 5% (p < .05)

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Step 6: Direction of Effect - Calculate Effect Size and 95%CI
β€’ The mean reaction time of sleep deprived individuals is significantly higher than
the mean reaction time of the population of well rested individuals.
Step 7: Interpret Overall Results
β€’ There is statistically significant evidence to suggest that the mean reaction time of
sleep-deprived individuals is different from the established mean of 300 ms.
Specifically, sleep deprivation appears to increase reaction times, indicating a
potential negative impact on performance.

You should focus on the following key concepts related to statistical tests and their applications:

  1. t-distribution vs. z-distribution: Understand when to use the t-distribution, especially when the population standard deviation is unknown, and how it differs from the z-distribution (which assumes a known population standard deviation).

  2. Degrees of Freedom: Recognize how degrees of freedom affect the shape of the t-distribution and the determination of critical values.

  3. Independent vs. Paired Samples t-tests: Know the differences between these tests, including when to use each and the implications for statistical power and individual differences.

  4. Assumptions of t-tests and ANOVA: Familiarize yourself with the assumptions that need to be satisfied for these tests, such as independence of samples, homogeneity of variance, and normality.

  5. Null and Alternative Hypotheses: Understand how to formulate these hypotheses for various tests and the implications for statistical significance.

  6. Type I and Type II Errors: Be aware of the consequences of making errors in hypothesis testing and how they affect the validity of your conclusions.

  7. Effect Size and Confidence Intervals: Learn about measuring the strength of effects and the importance of confidence intervals in interpreting test results.

  8. Post-hoc Tests: Recognize the circumstances that warrant post-hoc testing following ANOVA and how to interpret those results.