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WEEK 4 Problems
Chapters 13 & 15
Chapters:
Packet Pages:
Ch. 13: The Pearson Correlation Coefficient
Ch. 15: Non-parametric Statistical Tests: Chi-Square
1-9
10-17
Refer to the Resource Page on CANVAS for:
StatSheets
Table 6, Critical Values of r
Table 10, Critical Values of Chi-Square
Online calculator/statistical app links & tutorials
Chapter 13
The Pearson Correlation Coefficient
Ch. 13-
Defining Key Terms
Provide definitions for the following key terms:
COEFFFICIENT OF DETERMINATION/EFFECT SIZE (r2):
DIRECT/POSITIVE RELATIONSHIP:
INVERSE/NEGATIVE RELATIONSHIP:
OUTCOME VARIABLE:
OUTLIER:
PARTIAL CORRELATION:
PEARSON CORRELATION COEFFICIENT (and possible values):
PERFECT RELATIONSHIP:
PREDICTOR VARIABLE:
WEEK 4 CH 13 & 15 Problems
Page 1 of 17
Ch. 13- Looking Up Critical Values of r (rcv) in Table 6 & Marking the
Rare & Common Zones
For each of the following studies, look up the critical value of r (Table 6) that separates the rare zone
from the common zone & mark on the sampling distribution of the r ratio. (see textbook pp. 496499). There is also an on-line calculator for finding the critical values.
Marking Rare & Common Zones
The common zone is the middle section
of the distribution, the section centered
around zero where it would be common
to find the r-value for a sample if there
was no relationship between variables.
As the r-value increases (moves farther
away from zero), the results are more
likely to be statistically significant,
meaning a significant relationship.
Example:
A developmental psychologist wanted to determine if there were a relationship between the age at
which children started to walk and their intelligence at age 16. She went to a pediatrician’s office and
randomly selected 10 charts of 16-year-old girls. In the charts, she found the age (in months) at which
each girl started walking and then she gave each girl a standard IQ test. Use alpha=.05 two-tailed test.
Answer:
STEP 1: For N=10 pairs of data, Use Equation 13.1: df = N-2. For this study df = 10-2 = 8.
STEP 2: Use Table 6 (Critical Values of r) to find cutoffs for Rare & Common Zones. Or use
online calculator.
rcv = +/- .632
-.632
WEEK 4 CH 13 & 15 Problems
+.632
Page 2 of 17
Find the critical value of r & mark the common & rare zones on the sampling distribution:
1. For a two-tailed hypothesis test with alpha set at α = .05 and df =7?
2. For a two-tailed hypothesis test with alpha set at α = .05 and df =27?
3. For a two-tailed hypothesis test with alpha set at α = .05 and N =33?
WEEK 4 CH 13 & 15 Problems
Page 3 of 17
Ch. 13-
Reading Research Studies: Pearson Correlation Coefficient
Use the table and excerpt from the research article to answer the questions below. A brief “statistical
guide” summary is included to assist with understanding.
Statistical Guide:
A correlation coefficient indicates the strength and direction of a relationship between two variables.
The most widely used is the Pearson r. When it is positive in value, the relationship is direct (i.e., those
with high scores on one variable tend to have high scores on the other variable and those with low
scores on one variable tend to have low scores on the other). In a direct relationship, the closer r is to
1.00, the stronger the relationship; the closer to 0.00, the weaker. When the value of Pearson r is
negative, the relationship is inverse (i.e., those with high scores on one variable tend to have low scores
on the other one). In an inverse relationship, the closer r is to -1.00, the stronger the relationship; the
closer it is to 0.00, the weaker. For the table below, the lower the probability (p-value), the more
significant the relationship. The p-values are provided in the key below the table.
CORRELATES OF ALCOHOL AND TOBACCO USE
Excerpt from the Research Article
From intact families, 321 adolescents participated in the present study… the age range for all being
12-16 years. There was no significant age difference between the sexes. Participants were asked to
estimate how often (a) they, (b) their mothers, (c) their fathers, and (d) their best friends smoked tobacco
and used alcoholic beverages. The estimations were made on a 5-point scale with anchors of 0: never, 1:
almost never, 2: occasionally, 3: often, 4: very often.
The present study corroborates previous research indicating both parental and peer influence
on tobacco and alcohol consumption among adolescents. The study suggests that the influence of
peers might be greater than that of parents. However, correlations as such do not suggest causal
relationships. There are certainly other social psychological factors that may contribute to the
explanation of or co-vary with adolescent alcohol and tobacco consumption.
WEEK 4 CH 13 & 15 Problems
Page 4 of 17
RESEARCH STUDY QUESTIONS:
1. For boys, which variable correlates most highly with their tobacco habits?
2. For girls, which variable correlates most highly with their alcohol habits?
3. Is the relationship between boys’ alcohol habits and their fathers’ tobacco habits strong?
Explain.
4. Is the relationship between boys’ alcohol habits and their friends’ alcohol habits statistically
significant? If yes, at what probability level?
5. For the question above, should the null hypothesis for the relationship be rejected?
WEEK 4 CH 13 & 15 Problems
Page 5 of 17
Ch. 13-
Computing a Pearson Correlation Coefficient
For the study below, go thru steps of hypothesis testing to conduct a Pearson Correlation
Coefficient r-test & determine effect size (known as “coefficient of determination’, represented by
r2.) Answer questions in the table. (see text pp. 493-518, Ch. 13 StatSheet & online calculators).
EXAMPLE A.
Study: The data in the table are from 10 high school seniors. Each senior took both a Logical
Reasoning Test and a Creativity Test. The investigator wanted to know if there is a
relationship between logical reasoning and creativity.
QUESTIONS:
a.
HYPOTHESES: List the Null & Alternative Hypotheses:
Null Hyp: There is no relationship between Logical Reasoning & Creativity Scores.
Alt Hyp: There is a relationship between Logical Reasoning & Creativity Scores.
b.
SET THE DECISION RULE: Using the decision rule of alpha=.05, find the degrees of freedom
for this study (Equation 13.1), & find the critical values of r (using Table 6 or online
calculator). Mark the common & rare zones on the distribution below.
rCV= -.632
Rare Zone
WEEK 4 CH 13 & 15 Problems
rCV= .632
Remember, the df
formula for this
test is df=N-2.
With N=10 pairs
for this study, that
means the df=8.
Use Table 6 to find
alpha=.05. , df=8,
critical values = +/.632.
Rare Zone
Page 6 of 17
EXAMPLE A (continued…)
c.
CALCULATE THE STATISTIC: What is the calculated or computed r-test statistic for this
problem? (Equations 13.2; & on-line calculator)
The statistic was calculated using Ch. 13 Calculators found on WEEK 7 RESOURCE PAGE.
The Calculated Pearson r statistic = .353, falling short of the critical value of .632 set by the
Decision Rule.
INTERPRET THE RESULTS
(see pp. 506+)
d.
WAS THE NULL HYPOTHESIS REJECTED?: Is the Pearson Correlation result statistically
significant?
No, the Null Hypothesis is not rejected for this problem. The calculated r-value does not
meet the Decision Criteria of .632 needed for significance. The statistic fell in the Common
Zone.
e.
HOW BIG IS THE EFFECT?: Calculate the Effect Size (Coefficient of Determination) using r2
(Equation 13.3; on-line calculator) & interpret:
Since the calculated r=.353 for this problem, the
Effect Size is calculated with Equation 13.3:
r2 = (r) 2 x 100
r2 = (.353) 2 x 100 = 12.46%, medium
f.
CONCLUSION?: What conclusion or specific statement can be made about the relationship
between the pretest and final course grade in statistics?
The calculated r-value was not in the rare zone, meaning it was not statistically significant.
There is not enough evidence to conclude there is a relationship between Logical
Reasoning and Creativity Scores. Possibly a larger sample size might be studied in the
future.
WEEK 4 CH 13 & 15 Problems
Page 7 of 17
Ch. 13-
Computing a Pearson Correlation Coefficient
For the study below, go through the steps of hypothesis testing to conduct a Pearson
Correlation Coefficient r-test & determine effect size (known as the “coefficient of
determination’, represented by r2.) Answer the questions in the table below. (see textbook
pp. 493-518, Ch. 13 StatSheet & online calculators).
Research Study:
The following data represent data from 10 graduate students. Each student took a screening
pretest before starting their statistics class. They each completed the class; their final grades are
shown in the table below. The question being considered is: Can a screening test predict
success in a statistics class? … or Is there a significant relationship between screening test
performance and final grade in the class?
STUDENT
PRETEST
SCORE (x)
FINAL
COURSE
GRADE (y)
1
18
9
2
14
13
3
7
15
4
6
4
5
10
3
6
8
2
7
8
2
8
8
2
9
8
2
10
5
8
QUESTIONS:
a.
HYPOTHESES: List the Null & Alternative Hypotheses:
WEEK 4 CH 13 & 15 Problems
Page 8 of 17
b.
SET THE DECISION RULE: Using the decision rule of alpha=.05, find the degrees of freedom for
this study (Equation 13.1), & find the critical values of r (using Table 6). Use the critical values
to mark the common & rare zones on the distribution below.
c.
CALCULATE THE STATISTIC: What is the calculated or computed r-test statistic for this
problem? (Equations 13.2; & on-line calculator)
INTERPRET THE RESULTS
(see pp. 506+)
d.
WAS THE NULL HYPOTHESIS REJECTED?: Is the Pearson Correlation result statistically
significant?
e.
HOW BIG IS THE EFFECT?: Calculate the Effect Size (Coefficient of Determination) using r2
(Equation 13.3; on-line calculator) & interpret:
f.
CONCLUSION?: What conclusion or specific statement can be made about the relationship
between the pretest and final course grade in statistics?
WEEK 4 CH 13 & 15 Problems
Page 9 of 17
Chapter 15
Non-parametric Statistical Tests: Chi-Square
Ch. 15-
Defining Key Terms
Provide definitions for the following key terms:
CHI-SQUARE GOODNESS-OF-FIT TEST:
CHI-SQUARE TEST OF INDEPENDENCE:
CONTINGENCY TABLE:
MANN–WHITNEY U TEST:
NONPARAMETRIC TEST:
PARAMETRIC TEST:
SINGLE-SAMPLE TEST:
SPEARMAN RANK-ORDER CORRELATION COEFFICIENT:
WEEK 4 CH 13 & 15 Problems
Page 10 of 17
Ch. 15- Understanding Concepts
(see textbook pp. 567-8, 581-2, 572-4).
Provide a brief answer to the following questions:
QUESTIONS:
1. When would you use a non-parametric test?
ANSWER:
2. What do we mean when we say the non-parametric test is less powerful? Why is it less
powerful?
ANSWER:
3. What parametric test is the Chi-Square Goodness-of-Fit test most similar to and why?
ANSWER:
4. Which parametric test is the Chi-Square Test of Independence most similar to?
ANSWER:
5. What do the calculated expected frequencies tell you?
ANSWER:
WEEK 4 CH 13 & 15 Problems
Page 11 of 17
Ch. 15-
Computing a Chi-Square Goodness-of-Fit Test
For the study below, go through the steps of hypothesis testing to conduct a Chi-Square Goodness-ofFit Test. Answer the questions below. (see text pp. 568-581, Ch. 15 StatSheet & online calculators).
EXAMPLE A.
A researcher investigated whether gender roles are becoming more equal, with men and women
sharing child rearing responsibility, earning equal pay and sharing chores. For the study, a
researcher collected data at a local grocery store in late morning one day and observed who was
shopping with children; noting if the person was a male or a female. For the 42 people observed, 14
were men with children and 28 were women with children. From the data collected, can the
researcher say that men and women, at least as far as this aspect of child rearing goes, are equal?
NOTE: We are using the Chi-Square Goodness-of-Fit Test because comparing an observed value to
an expected value for a nominal level variable. This test is similar to a Single-Sample z or t Test,
since drawing a single sample and comparing the sample data to expected population data.
1.List the Hypotheses:
Null Hyp: Distribution of gender in sample is same as population
Alt Hyp: Distribution of gender in sample differs from population
Note: if men & women shared equally, then half the people with
children in the store would be women, & half men. As half of 42 is
21, we expect that 21 of the people with children would be men &
21 would be women. We compare observed with expected for
this test.
Using standard level of alpha=.05, 2 tailed test. Go to Table 10,
“Critical Values of Chi Square”, for df=k-1 (where k = # of
categories).
2.Set the Decision Rule:
df= 2-1 = 1
10
WEEK 4 CH 13 & 15 Problems
Critical Value= 3.841. If the calculated
the Null Hypothesis.
≥ 3.841, then reject
Page 12 of 17
METHOD 1: Use Equation 15.3 to calculate the Chi-Square value.
Take the observed frequencies of 14 & 28. We already know the
expected frequencies of 21 and 21.
3.Calculate the Test Statistic:
(the calculations are performed for each cell/category & then summed
to obtain the chi-square value).
METHOD 2: Use online Calculator for Chapter 15.
4.Interpret the Results:
Using the decision rule, we are rejecting the null hypothesis
because the calculated test statistic of 4.66 is greater than the
critical value of 3.841.
The observed distribution of gender differs from what would be
expected if men and women split responsibilities equally.
WEEK 4 CH 13 & 15 Problems
Page 13 of 17
Ch. 15-
Computing a Chi-Square Goodness-of-Fit Test
For the study below, go through the steps of hypothesis testing to conduct a Chi-Square
Goodness-of-Fit Test. Answer the questions in the table below. (see textbook pp. 568-581, Ch.
15 StatSheet & online calculators).
STUDY:
Vallone studied attitudes about smoking on a college campus. She hypothesized that 60% of
undergraduates would want a smoke-free campus. She surveyed 200 undergraduates and found
that 90 preferred a smoke-free campus.
Hint: create a table to organize the data.
Observed Frequency
Expected Frequency
Smoke
Free
90
100
Non-Smoke
Free
110
100
N=200
QUESTIONS:
a.
LIST THE HYPOTHESES:
b. SET THE DECISION RULE: Use the decision rule of alpha=.05, 2-tailed test. Use Table 10 to find
the critical value.
c.
CALCULATE THE TEST STATISTIC:
d. INTERPRET THE RESULTS:
WEEK 4 CH 13 & 15 Problems
Page 14 of 17
Ch. 15-
Computing a Chi-Square Test of Independence
For the study below, go through the steps of hypothesis testing to conduct a Chi-Square Test
of Independence. Answer the questions in the table below. (see textbook pp. 581-597, Ch.
15 StatSheet & online calculators).
NOTE: We are using the Chi-Square Test of Independence to determine whether two groups differ
on a categorical dependent variable.
EXAMPLE A.
The researcher in the previous example was intrigued so did a larger study to see if a difference
existed in gender roles between small towns that had colleges and small towns without colleges.
The researcher drew random samples of both kinds of towns, went to grocery stores during the
day, and for those people shopping with kids, noted the gender of the adult. The results are in the
table below.
Shopping with Kids
Observed
Male
College Towns
130
Non-College Towns
100
230
1. List the Hypotheses:
H0= There is no relationship
between the variables.
H1= There is a relationship
between the variables.
Female
370
400
770
500
500
Null Hyp: The two variables are independent of each other (not
related).
Alt Hyp: The two variables are not independent of each other
(are related).
Using standard level of alpha=.05, 2 tailed test. Use df formula
from Equation 15.4: df= (R-1) x (C-1).
df = (2-1) (2-1) = 1
Go to Table 10, “Critical Values of Chi Square”, for df=1 &
alpha=.05
Critical Value= 3.841. If the calculated X2 ≥ 3.841, then reject the
Null Hypothesis.
WEEK 4 CH 13 & 15 Problems
Page 15 of 17
3. Calculate the Test Statistic:
METHOD 1: Use Eq. 15.3 & 15.5 to calculate expected frequencies
& Chi-Square value.
X2 = (130-115)2+ (370-385)2+ (100-115)2+ (400-385)2= 5.08
115
385
115
385
METHOD 2: Use online Calculator for Chapter 15.
Using the decision rule, we are rejecting the null hypothesis
because the calculated test statistic of 5.08 is greater than the
critical value of 3.841.
4. Interpret the Results:
A difference exists between the two populations in terms of the
percentages of men who take children shopping: It is more
Using the
rule, we
arethan
rejecting
the null hypothesis
common
in decision
college towns
(26%)
in noncollege
towns (20%).
Usingthe
thecalculated
decision rule,
we are rejecting
null than the
because
test statistic
of 5.08 isthe
greater
hypothesis
critical
value of because
3.841. the calculated test statistic of
5.08 is greater than the critical value of 3.841.
A difference exists between the two populations in terms of the
A difference
exists
between
the two
populations
percentages
of men
who
take children
shopping:
It is more
in
terms
of
the
percentages
of
men
who
take towns (20%).
common in college towns (26%) than in noncollege
children shopping: It is more common in college
towns (26%) than in noncollege towns (20%).
WEEK 4 CH 13 & 15 Problems
Page 16 of 17
Ch. 15-
Computing a Chi-Square Test of Independence
For the study below, go through the steps of hypothesis testing to conduct a Chi-Square Test of
Independence. Answer the questions in the table below. (see textbook pp. 581-597, Ch. 15
StatSheet & online calculators).
STUDY:
Harris & Sanborn were interested in gender differences in movie preference. They surveyed 50
men & 50 women & asked them whether they had seen a horror movie in the last month. Of the
men, 35 had seen a horror movie. Of the women, 20 had seen a horror movie.
Hint: create tables to organize the data.
QUESTIONS:
a.
LIST THE HYPOTHESES:
b. SET THE DECISION RULE: Use the decision rule of alpha=.05, 2-tailed test. Use Table 10 to find
the critical value.
c.
CALCULATE THE TEST STATISTIC:
d. INTERPRET THE RESULTS:
WEEK 4 CH 13 & 15 Problems
Page 17 of 17

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