Stat 203, Statistics for Social Sciences – Assignment 5, Due Wednesday August 1, 4:30pm.
Questions 2,3, and 5 will be marked fully.
Questions 1,4, and 6 will be marked for completion only.
Student Name _______________________________
Student Number ______________________________________
Question 1: Consider the following crosstab from two samples, one randomly selected from those born
prematurely, and one from those not born prematurely. They were asked if they had chronic oral
infections.
Oral Infection * Born Premature Crosstabulation
Count
Born Premature
No
Oral Infection
Total
Yes
Total
No
57
8
65
Yes
6
44
50
63
52
115
a) What is the probability of a prematurely born person having chronic oral infections?
b) What are the odds of a prematurely born person having chronic oral infections?
c) What are the odds of a non-prematurely born person having chronic oral infections?
d) How many times are large are the odds of chronic oral infection if a child is born
prematurely than if they were born at a normal time. Use your answers from part B and
C to find the odds ratio.
e) How many times are large are the odds of chronic oral infection if a child is born
prematurely than if they were born at a normal time. Use the quick odds ratio formula.
Question 2: Consider the sample of court cases where the gender of the offender and whether or
not the offender was abused as a child. The dataset is A5CrimeAbuse.sav
a) Get the observed frequencies table from SPSS and include it. How many male offenders
were no victims of abuse?
b) Get the odds ratio table from SPSS and include it.
c) Interpret the odds ratio in real world terms.
d) The null hypothesis of a chi-squared test is that being a victim of child abuse and
offender’s gender are independent. What would the frequencies be on average if this were
true?
e) What is the 95% confidence interval of the odds ratio? Can you reject the null hypothesis
a two-tailed alpha=0.05 test with this information?
f) Use SPSS to get the score of the chi-square test. Do we reject the null at alpha = 0.05
with this test? (You can use the sig. here)
Question 3: Consider this survey of students at United States college, which includes their
participation in extra activities at the school, and the region of the US that they’re from. The
data can be found in A5College.sav
a) Build a crosstab using SPSS. Find by hand how many degrees of freedom does this chisquared test have in this situation?
b) State the null hypothesis of the chi-squared test in this situation.
c) Run a chi-squared test in SPSS and test the null hypothesis.
d) Confirm your conclusion from 3C by using the chi-squared value from SPSS and the
Table E on page 523. If you don’t have a textbook, use the Chi-squared table provided in
chisqtable.csv or chisqtable.xls
Question 4: Complete the table
Probability
Odds
0
0.10
0.25
1/3
0.50
0.75
0.99
Question 5: Consider the data set of a sample of families with three different types of father
figures and three difference parenting styles. Found in A5Father.sav
a) Find the degrees of freedom of a chi-squared test by hand.
b) Do the chi-squared test in SPSS. Included the table.
c) Get the observed values from the cross tab. The null hypothesis is that father figure type
and parenting style are independent. Under this assumption how many moderate
stepfathers would we expect?
d) How many authoritarian fathers would we expect?
e) If the chi-squared value were twice as large as they are (from largest deviations from
independence), would you reject the null?
Question 6: Consider the following crosstab of data from the students of a university which
includes the transportation to school and their primary faculty (arts or science).
Arts
Science
Total
Transit
44
23
67
Drives/Rides Car
16
6
22
Drives/Rides Truck
1
2
3
Total
65
31
96
a) What potential problem is there to performing a chi-squared test on this data? (Don’t
actually do the chi-squared test).
b) How could the problem in 6A be fixed? Insert a table of the improved crosstab.
c) Get the odds-ratio of the fixed data set.
Odds Ratio from 2x2 contingency plot: (Quick odds-ratio formula)
Odds Ratio = AD / BC
(AD means A times D)
Expected frequencies formula (as found on page 317 of textbook, examples pp. 317-323)
Expected freq. = (row total) x (column total) / (grand total)
Degrees of freedom for crosstabs (as found on page 316)
Df = (rows – 1) x (columns – 1)
SPSS Tips:
Use
Analyze  Descriptive Statistics  Crosstabs.
For expected frequencies click on the “Cells” button, put a check on “Expected”.
For odds ratio click on the “Statistics” button, and put a check on “Risk”.