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Psych 210 Exam 4 JMU questions with complete solutions graded A+ passed-10.docx, Exams of Nursing

Psych 210 Exam 4 JMU questions with complete solutions graded A+ passed-10.docx

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2024/2025

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Psych 210 Exam 4 JMU questions with complete
solutions graded A+ passed
1.
Analysis
of
Variance
(ANOVA)
tests whether 3 or more groups' means are
different
(is
there
a
significant
difference
among
the
groups)
2.
one-way ANOVA
1 IV with three or more levels (multi-group). between
or within subjects design
3.
two-way ANOVA
two IVs with two or more levels (factorial). between,
within, or mixed subjects design
4.
Why
not
use
multiple
t-tests?
1)
could
take
a
while
2)
the alphas are cumulative across the
experiment,
so the more tests you do, the more
error there is
5.
within-group
variation
(error)
if
the
null
is
true,
any
numerical
differences
will
be
"within" each group since the samples came from
the same population. therefore, any numerical differ-
ences are error
6. between-group variation (sample
means)
if the null is NOT true, then between-group variation
is due to change in the IV
7.
F
F = between-group variance divided by within-
group variance OR MS between divided by MS
within. you want F to be big. if null is true, F
should be 1
8.
sources
of
variance
1)
measurement
error
(DV)
[error]
2)
uncontrolled
variables
[error]
pf3
pf4
pf5
pf8
pf9
pfa
pfd

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1 /

solutions graded A+ passed

  1. Analysis of Variance (ANOVA) tests whether 3 or more groups' means are different (is there a significant difference among the groups)
  2. one-way ANOVA 1 IV with three or more levels (multi-group). between or within subjects design
  3. two-way ANOVA two IVs with two or more levels (factorial). between, within, or mixed subjects design
  4. Why not use multiple t-tests? 1) could take a while
  1. the alphas are cumulative across the experiment, so the more tests you do, the more error there is
  1. within-group variation (error) if the null is true, any numerical differences will be "within" each group since the samples came from the same population. therefore, any numerical differ- ences are error
  2. between-group variation (sample means) if the null is NOT true, then between-group variation is due to change in the IV
  3. F F = between-group variance divided by within- group variance OR MS between divided by MS within. you want F to be big. if null is true, F should be 1
  4. sources of variance 1) measurement error (DV) [error]
  1. uncontrolled variables [error]

2 /

solutions graded A+ passed

  1. participants [error in between-subject, not in with- in]
  2. the IV
  1. total variance for one-way be- tween-subjects ANOVA

4 /

solutions graded A+ passed

  1. identifying ANOVA tables based on the number of parts that variance is broken up into (one-way between has variation in two parts, one-way within has variation in three parts, two-way between in four parts)

5 /

solutions graded A+ passed

  1. determining significance in ANOVA compare Fcalc to Fcrit Fcrit found in F table based on the between and within degrees of freedom Fcalc eFcrit, reject null, p<. conclude there is a significant ditterence somewhere among the groups
  2. Effect Size for one-way between the percent of variance explained. SSbetween is what we can explain because it came from our manipula- tion
  3. Post Hoc Testing tells exactly where the ditterences between groups exist
  4. Issues with Post Hoc Testing Tests vary from conservative (needs a big ditterence to be conservative) to liberal (small ditterences can be significant). Also need to control for experiment- wise error rate
  5. common report of ANOVA findings F(between df, within df, MSE) = Fcalc , p</>.
  6. Tukey's Post Hoc about 65% liberal and controls for error
  7. Mean Square Error (MSE) mean square of the within variation
  8. MS equation for everything SSwhatever divided by dfwhatever (whatever you're trying to find, between, within, etc)
  9. SStotal equation Σ(X-GM)^

7 /

solutions graded A+ passed

  1. SSwithin equation Σ(X-M)^2 (M is of cells in factorial) (THIS IS ERROR)
  2. SStotal full equation Σ(X-GM)^2 = Σ(M-GM)^2 + Σ(X-M)^
  3. SSrows equation Σ(Mrows-GM)^
  4. SScolumns equation Σ(Mcolumns-GM)^
  5. SSinteraction equation Σ[(- X-GM)-(X-Mcell)-(Mrows-GM)-(Mcolumns-GM)]^
  6. dftotal equation N - 1
  7. dfbetween equation one-way be- tween Nlevels - 1
  8. dfwithin equation one-way between df1 +df2 + df3 + etc
  9. dfbetween equation one-way within N - 1
  10. dfconditions equation Nlevels - 1
  11. dferror equation one-way within dfbetween * dfconditions
  12. dfinteraction equation df of IV1*df of IV
  13. dferror (two-way) N-(NlevelsIV1 * NlevelsIV2)
  14. df for IV1 and IV2 two-way between Nlevels - 1
  15. one way between subjects ANOVA table breakdown IV: between groups error: within groups between+within=total dfbetween+dfwithin=dftotal

8 /

solutions graded A+ passed

SS divided by df = MS

10 /

solutions graded A+ passed

  1. cell means means of data within cell
  2. Marginal Means mean of values in rows/columns that are compared to find main ettect. bigger ditterence = more likely significant

11 /

solutions graded A+ passed

  1. Interaction does the ettect of one IV depend on the level of the other? (NOT how one IV attects another). if changes between cells are the same, there is no interaction
  2. two-way ANOVA graphical output not parallel = interaction visualize middle between points, if middle is on same line, there is no main ettect for the variable
  3. Chi-Square Testing non-parametric test (not based on assumptions about population), used when you have nominal data, mostly when DV is normal. also used when IV or DV is ordinal and small samples from non- normal population distributions
  4. Two types of Chi-square analyses goodness of fit and test of independence
  5. goodness of fit use to test if data from ONE nominal variable fits ex- pected values. expected values can assume no pref- erence (flipping coin should have heads 50% of the time) or that sample should fit hypothesized percent- ages (sample class at JMU should be 65% female)
  6. test of independence tests whether TWO nominal variables are indepen- dent of each other. basically asking if two variables are related - like correlation. the expected values are based on totals from each variable
  7. chi squared equation where O is the observed data and E is the

13 /

solutions graded A+ passed

  1. for group of observations, take the ditterence be- tween the observed and expected values
  2. square that ditterence
  3. divide the squared ditterences by the expected value
  4. add all calculated values together
  1. chi-square hypothesis testing steps 1) define null as "no preference" (goodness of fit) or "no relationship" (test of independence)
    1. calculate chi-square and df=Nlevels- 1
    2. compare to critical value - if calc ecrit, reject null
    3. determine what your findings mean