This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 114. Consider the following partial analysis of variance table from a randomized block design
with 10 blocks and 6 treatments. Test Ho: there is no difference between treatment effects at
A. Reject H0
B. Fail to Reject H0
F = 1.91 and F.05,5,45 = 2.41 AACSB: Analytic
Difficulty: Hard
Learning Objective: 3
Topic: Randomized block design 11034 = .05. Chapter 01  An Introduction to Business Statistics 115. Consider the following partial analysis of variance table from a randomized block design
with 10 blocks and 6 treatments. Test Ho: there is no difference between blocks at
A. Reject H0
B. Fail to Reject H0 = .05. AACSB: Analytic
Difficulty: Hard
Learning Objective: 3
Topic: Randomized block design Chapter 12
ChiSquare Tests
True / False Questions 1. A contingency table summarizes data that has been classified on two dimensions or scales.
True False 2. The actual counts in the cells of a contingency table are referred to as the expected cell
frequencies.
True False 3. The χ2 goodness of fit test requires nominative level of data.
True False 4. The chisquare distribution is a continuous probability distribution that is skewed to the
left.
True False 11035 Chapter 01  An Introduction to Business Statistics 5. Expected cell frequencies for a multinomial distribution are calculated by assuming
statistical dependence.
True False 6. A multinomial probability distribution describes data that is classified into two or more
categories when a multinomial experiment is carried out.
True False 7. The trials of a multinomial probability are assumed to be dependent.
True False 11036 Chapter 01  An Introduction to Business Statistics 8. One use of the chisquare goodness of fit test is to determine if specified multinomial
probabilities in the null hypothesis is correct.
True False 9. In a contingency table, when all the expected frequencies equal the observed frequencies
the calculated χ2 statistic equals zero.
True False 10. In a contingency table, if all of the expected frequencies equal the observed frequencies,
then we can conclude that there is a perfect association between rows and columns.
True False 11. When using chisquare goodness of fit test with multinomial probabilities, the rejection of
the null hypothesis indicates that at least one of the multinomial probabilities is not equal to
the value stated in the null hypothesis.
True False 12. A fastener manufacturing company uses a chisquare goodness of fit test to determine if a
population of all lengths of ¼ inch bolts it manufactures is distributed according to a normal
distribution. If we reject the null hypothesis, it is reasonable to assume that the population
distribution is at least approximately normally distributed.
True False 13. When using the chisquare goodness of fit test, if the value of the chisquare statistic is
large enough, we reject the null hypothesis.
True False 11037 Chapter 01  An Introduction to Business Statistics 14. The chisquare goodness of fit test can only be used to test whether a population has
specified multinomial probabilities or to test if a sample has been selected from a normally
distributed population. It cannot be applied to test if a sample data comes from other
distribution forms such as Poisson.
True False 15. In performing a chisquare test of independence, as the difference between the respective
observed and expected frequencies decrease, the probability of concluding that the row
variable is independent of the column variable decreases.
True False 16. In performing a chisquare goodness of fit test with multinomial probabilities, the smaller
the difference between observed and expected frequencies, the higher the probability of
concluding that the probabilities specified in the null hypothesis is correct.
True False 17. When we carry out a chisquare test of independence, if ri is row total for row i and cj is
the column total for column j, then the estimated expected cell frequency corresponding to
row i and column j equals (ri) (cj)/n.
True False 18. When we carry out a chisquare test of independence, the chisquare statistic is based on
(rc1) degrees of freedom where r and c denote, respectively the number of rows and columns
n the contingency table.
True False 19. When we carry out a chisquare test of independence, in the alternative hypothesis we
state that the two classifications are statistically independent.
True False 11038 Chapter 01  An Introduction to Business Statistics 20. When we carry out a chisquare test of independence, the expected frequencies are based
on the null hypothesis.
True False 21. When we carry out a goodness of fit chisquare test, the expected frequencies are based on
the alternative hypothesis.
True False Multiple Choice Questions 22. The χ2 statistic from a contingency table with 6 rows and five columns will have
A. 30 degrees of freedom
B. 24 degrees of freedom
C. 5 degrees of freedom
D. 20 degrees of freedom
E. 25 degrees of freedom 23. The chisquare goodne...
View
Full
Document
This document was uploaded on 01/20/2014.
 Winter '14

Click to edit the document details