Inferential Statistics Exercise
For each problem, state a. which statistical test should be used to analyze the data
b. whether the test is a one-tailed or a two-tailed test
c. the null and alternative hypotheses
d. what decision should be made regarding the null and alternative hypotheses
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1.
A social psychologist was interested in determining if viewing political advertisements increases
viewers’ knowledge about the issues presented. Advertisements were selected, and a test of the
information contained was constructed (higher score indicated more knowledge). One hundred
participants took this test—50 who did not view the advertisements and 50 who did view the
advertisements.
a.
independent samples t-test
b.
one-tailed test
c. H0: population of viewers ≤ population of non-viewers
H1: population of viewers > population of non-viewers
The mean test score of the group that did not view the advertisements was 46, and the mean test score
of the group that viewed the advertisements was 46.
d.
retain null, reject alternative
It turns out that many of the participants were political science majors, and most played an active role
in the most recent election. When tested again, the mean scores for the group that did not watch the
advertisement was 40, and the mean score for the group that did watch the advertisement was 46.
Based on these results, the previous decision was a Type I error OR Type II error.
2.
An environmental psychologist developed program to decrease littering on the campuses of seven
state universities. It was implemented in August of 2004, and the amount of litter collected (in pounds)
was measured in July of 2004 and then again in July of 2005.
a.
related/paired samples t-test
b.
one-tailed test
c. H0: population before ≤ population after
H1: population before > population after
The results showed that there was an average of 25 pounds less litter was collected in July of 2005.
d.
reject null, accept alternative
When this study was made public, another state university system decided to implement the program,
in hopes of replicating the results. This system, however, did not find a difference after implementing
the program. It turned out that the original system had many additional trash containers on campus
that affected the results, and the second university system did not have as many.
Based on these results, the previous decision was a Type I error OR Type II error.
3.
A clinical psychologist was asked to determine whether there was a clear difference in the length of
stay of patients with different categories of disorders in a mental hospital. The three diagnosis
categories were affective disorders, cognitive disorders, and drug- related conditions.
a.
ANOVA
b.
two-tailed test
c. H0: population of affective disorders = population of cognitive disorders =
population of drug-related conditions
H1: There will be a difference between the three categories of mental disorders.
The results showed that the mean lengths of time (in weeks) patients with affective disorders, cognitive
disorders, and drug-related conditions stayed in the hospital were six weeks, ten weeks, and ten
weeks, respectively. The result of the statistical test (α = .05) was F(2, 9) = 9.60, p < .01.
d.
4.
reject null, accept alternative
A cultural psychologist is interested in studying romantic relationships in other cultures and knows
that in India, college seniors report falling in love an average of 2.2 times during their college years. A
sample of 50 seniors, originally from India but who have spent their college years in the United States
were asked how many times they had fallen in love during college. It was believed that these college
seniors would be different from the population of college seniors in India.
a.
one sample t-test
b.
two-tailed test
c. H0: population in that country = population in United States
H1: population in that country ≠ population in United States
The results showed that students who attended college in the United States fell in love an average of
3.4 times, and the result of the statistical test (α = .05) was t(49) = 1.69, p > .10.
d.
retain null, reject alternative