1. A chi-square test for independence with 8 degrees of freedom results in a test statistic of 18.21. Using the chi-square table, the most accurate statement that can be made about the p-value for this test is that: (Points: 2) A) p-value < 0.01 B) 0.025 > p-value > 0.01 C) 0.05 > p-value > 0.025 D) 0.10 > p-value > 0.05
Answer: B) 0.025 > p-value > 0.01
Hint: The p-value corresponding to 20.09 and 17.535 are respectively 0.01 and 0.025.
2. 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: (Points: 2) A) 11.1433 B) 9.3484 C) 7.8147 D) 9.4877
Answer: C) 7.8147
Hint: From Chi-square table with (k-r-1)=(6-2-1) = 3 d.f, where k is the number of classes and r is the number of parameters estimated
3. A test for independence is applied to a contingency table with 3 rows and 5 columns for two nominal variables. The degrees of freedom for this test will be: (Points: 2) A) 20 B) 12 C) 8 D) 15
Answer: C) 8
Hint: (r-1)(c-1) =(3-1)(5-1) = 8 df, where r and c are the number of rows and columns.
4. What is the critical value of the Chi-Square test at the .05 level for the following table?
1 2 3 Total
1 20 36 45 101
2 80 114 205 399
Total 100 150 250 500 (Points: 2) A) 5.991 B) 2.436 C) 7.245 D) 3.447
Answer: A) 5.991
Hint: From Chi-Square table with (r-1)(c-1) =(2-1)(3-1) = 2 df, where r and c are the number of rows and columns.
5. To use the chi-square distribution table to select a table value requires the following: (Points: 2) A) numerator degrees of freedom. B) level of significance. C) degrees of freedom. D) B and C above.
Answer: D) B and C above
6. A regression analysis between sales (in $1000) and advertising (in $) resulted in the following least squares line: y = 80,000 + 5x. This implies that an: (Points: 2) A) increase of $1 in advertising is expected to result in an increase of $5 in sales B) increase $5 in advertising is expected to result in an increase of $5,000 in sales C) increase of $1 in advertising is expected to result in an increase of $80,005 in sales D) increase of $1 in advertising is expected to result in an increase of $5,000 in sales
Answer: D) increase of $1 in advertising is expected to result in an increase of $5,000 in sales
7. Which of the following table values would be appropriate for a 95% confidence interval for the mean of y from a simple linear regression problem if the sample size is 7? (Points: 2) A) 1.895 B) 2.015 C) 2.365 D) 2.571
Answer: D) 2.571
Hint: From Student’s t table with (n-2)=(7-2) =5 df.
8. In publishing the results of some research work, the following values of the coefficient of determination were listed. Which one would appear to be incorrect? (Points: 2) A) 0.99 B) 0.47 C) -0.64 D) 0.00
Answer: C) -0.64
Hint: Coefficient of determination is always lies between 0 and 1.
9. Given the least squares regression line = -2.88 + 1.77x, and a coefficient of determination of 0.81, the coefficient of correlation is: (Points: 2) A) -0.88 B) +0.88 C) +0.90 D) –0.90
Answer: C) +0.90
Hint: The coefficient of correlation is the square root of coefficient of determination, with sign is same as the coefficient of x in the regression equation.
10. In a regression problem the following pairs of (x, y) are given: (2,1), (2,-1), (2,0), (2,-2) and (2,2). That indicates that the: (Points: 2) A) coefficient of correlation is –1 B) coefficient of correlation is 0 C) coefficient of correlation is 1 D) coefficient of determination is between –1 and 1
Answer: B) coefficient of correlation is 0