## Question 600:

1## Answer:

No answer provided yet.1. The moving average method averages out cyclical and irregular components.

TRUE: The moving average (more or less) reduces cyclical effects. It is often used when reporting average sales when say a company has a seasonal sales cycle (like they get 75% of revenue in 1 quarter).

2. When using Chi Square, the sum of the expected frequencies and the sum of the observed frequencies must be equal.

FALSE: It is not required for them to be equal and in fact, when the difference between the observed and expected values increase, this leads to a higher chi-square value, which leads to rejection of the null-hypothesis.

3. The test statistic chi-square used in a goodness-of-fit test has k-2 degrees of freedom.

False: The goodness of fit Chi-Square test, also called a 1-way Chi-Square test is based on the number of categories or groups being assessed minus 1, where k is the number of groups, so df = k-1.

4. Explained variation equals total variation minus unexplained variation.

TRUE: This question seems a bit out of context, so I'm not sure if this is in reference to a correlation, ANOVA or CHI-Square. As a general rule though it appears like a correct statement and we see this in the ANOVA table: Total variation = Explained + Unexplained so using some simple algebra we see Explained = Unexplained – Total.

5. A Spearman's rank-order coefficient of -0.91 indicates a very weak relationship.

False: You interpret the spearman rho coefficient just like the Pearson r. Cohen 1988 has as the general guide as anything over .5 or -5 as a strong relationship, thus making a .9 or a -.9 as a very strong relationship. Note that with correlations, the strength of the relationship is based on the absolute value. So a -.9 and .9 while both meaning the data have a different relationship (one positive and one inverse), the strength of their association is equal.

6. To employ ANOVA, the data must be at least the ordinal scale.

TRUE: The dependent variable in an ANOVA needs to be quantitative, so nominal values or numbers for categories don't work—the data needs to have at least an order to it.

7. If the computed value of F is 5.01 and the critical value is 2.67, we would conclude that all the population means are equal.

FALSE: The null hypothesis in an ANOVA is that all means are equal. If you obtain a test statistic that is greater than the critical value, then you reject the null hypothesis and conclude at least 1 mean differs from the others.

8. Unlike student's t distribution, there is only one F distribution.

FALSE: The F-distribution is also a family of distributions like the t, which is defined by the numerator degrees of freedom and denominator degrees of freedom.

9. Pearson's coefficient of correlation can be used if the data is nominal scale

FALSE: You'd want to use a contingency table and the phi-correlation coefficient for nominal data.