SME_8e_Ch_15_Section_4 - CHAPTER 15 SECTION 4: CHI-SQUARED...

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CHAPTER 15 SECTION 4: CHI-SQUARED TESTS (OPTIONAL) MULTIPLE CHOICE 163. The number of degrees of freedom in testing for normality is the: a. number of intervals used to test the hypothesis minus one. b. number of parameters estimated minus one. c. number of intervals used to test the hypothesis minus one minus the number of parameters estimated. d. None of these choices. ANS: C PTS: 1 REF: SECTION 15.4 (OPTIONAL) 164. The number of degrees of freedom in a chi-squared test for normality, where the number of standardized intervals is 5 and there are 2 population parameters to be estimated from the data, is equal to: a. 5 b. 4 c. 3 d. 2 ANS: D PTS: 1 REF: SECTION 15.4 (OPTIONAL) 165. In a goodness-of-fit test, the null hypothesis states that the data came from a normally distributed population. The researcher estimated the population mean and population standard deviation from a sample of 500 observations. In addition, the researcher used 6 standardized intervals to test for normality. Using a 5% level of significance, the critical value for this test is: a. 11.1433 b. 9.3484 c. 7.8147 d. 9.4877 ANS: C PTS: 1 REF: SECTION 15.4 (OPTIONAL) TRUE/FALSE 166. The null hypothesis states that the sample data came from a normally distributed population. The researcher calculates the sample mean and the sample standard deviation from the data. The data arrangement consisted of five categories. Using α = 0.05, the appropriate critical value for this chi-squared test for normality is 5.99147. ANS: T PTS: 1 REF: SECTION 15.4 (OPTIONAL) 167. The number of degrees of freedom associated with the chi-squared test for normality is the number of intervals used minus the number of parameters estimated from the data. ANS: F
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SME_8e_Ch_15_Section_4 - CHAPTER 15 SECTION 4: CHI-SQUARED...

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