1-20chapter stats

05 to reject the null hypothesis a reject h0 b fail

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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 1-1034 = .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 Chi-Square 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 chi-square distribution is a continuous probability distribution that is skewed to the left. True False 1-1035 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 1-1036 Chapter 01 - An Introduction to Business Statistics 8. One use of the chi-square 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 chi-square 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 chi-square 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 chi-square goodness of fit test, if the value of the chi-square statistic is large enough, we reject the null hypothesis. True False 1-1037 Chapter 01 - An Introduction to Business Statistics 14. The chi-square 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 chi-square 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 chi-square 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 chi-square 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 chi-square test of independence, the chi-square statistic is based on (rc-1) 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 chi-square test of independence, in the alternative hypothesis we state that the two classifications are statistically independent. True False 1-1038 Chapter 01 - An Introduction to Business Statistics 20. When we carry out a chi-square test of independence, the expected frequencies are based on the null hypothesis. True False 21. When we carry out a goodness of fit chi-square 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 chi-square goodne...
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