In: Statistics and Probability
1. Identify which test the scenario will use: independent samples t test, paired samples t test, chi square test for goodness of fit, chi square test for independence or chi square test for homogeneity.
A. A researcher goes to the parking lot at a large grocery chain and observes whether each person is male or female and whether they return the cart to the correct spot before Chi leaving (yes or no).
B. Amber Sanchez, a statistics student, collected data on the prices of the same items at the Navy commissary on the naval base in Ventura County, California, and a nearby Kmart The items were matched for content, manufacturer, and size and were priced separately
C. A random survey of automobiles parked in the student lot and the staff lot at a large university classified the brands as either domestic or foreign.
D. Surfers and statistics students Rex Robinson and Sandy Hudson collected data on the number of days on which surfers surfed in the last month for 30 random longboard users and 30 random shortboard users. Test the hypothesis that the mean days surfed for all long boarders is larger than the mean days surfed for all short boarders (because longboards can go out in many different surfing conditions)
E. Suppose you have a random sample of students attending a public university in Nevada and want to determine whether the racial distribution of students different from the racial distribution in the state as a whole
F. Compare the weekday and weekend/holiday hours of sleep. Each pair of numbers is from one randomly selected person.
G. Students observe the number of office hours posted for a random sample of tenured and a random sample of untenured professors
H . Based on a random sample of students at a university, you wish to determine if there is an association between whether or not a student is a transfer student and whether he or she belongs to an on-campus club.
A) Answer: chi square test
for independence
Explanation: Since the researcher wants to test the hypothesis
whether the two factors 1) Gender (Male/Female) and 2) Correct Spot
(Yes/No) are associated, chi square test for independence will be
used.
B) Answer: chi square test
for goodness of fit
Explanation: Since the research test the hypothesis whether the
prices of the same items are significantly different among the
different regions such that prices are not from the same population
mean hence the chi square test for goodness of fit will be
used.
C) Answer: chi square test
for homogeneity.
Explanation: In chi square test for homogeneity, we tests whether
the distribution of categorical variable is same for two or more
populations. In this problem we want to test whether the
distribution of cars in student lot and staff lot can be classified
according to the brands (domestic/foreign) hence chi square test
for homogeneity will be used to test the hypothesis.
D) Answer: independent
samples t test
Explanation: Since we are comparing the two independent population
means, independent samples t test will be used.
E) Answer: chi square test
for goodness of fit
Explanation: The chi square test for goodness of fit test whether
the sample data is consistent with the given population.
Here we are comparing whether the racial distribution of students
(sample) different from the racial distribution in the state as a
whole (population), chi square test for goodness of fit will be
used.
F) Answer: paired samples t
test,
Explanation: In paired samples t test, we compares the two set of
observation collected from same subject.
Here weekday and weekend/holiday hours of sleep are being compared
on same subjects hence paired samples t test will be used.
G) Answer: independent
samples t test
Explanation: Since we are comparing the two independent population
means (tenured and untenured professors), independent samples t
test will be used.
H) Answer: chi square test
for independence
Explanation: Since the researcher wants to test the hypothesis
whether the two factors 1) Transfer(Yes/No) and 2) Belongs to an
on-campus club (Yes/No) are associated, chi square test for
independence will be used.