Title 1.09 Parametric and Non-Parametric Tests of Independence.pdf ✅

1. An investor tests the relationship between the monthly returns of two mutual funds with the null hypothesis that the funds' returns are uncorrelated. After analyzing 51 months of data, the investor believes that the distributions deviate from a normal distribution and calculates a Pearson correlation of 0.7894 and a Spearman correlation of 0.7311. If the critical t-value at a 5% significance level is 2.010, the most appropriate decision is to: A. fail to reject the null hypothesis.  B. reject the null hypotheses since the t-statistic is close to 7.50. C. reject the null hypotheses since the t-statistic is close to 9.00. Explanation Hypothesis tests are classified as either parametric or nonparametric. Parametric tests are based on distributional assumptions (eg, normally distributed) whereas nonparametric tests are used when few assumptions about the underlying population can be made. The correlation between two populations (ρ) can be calculated using a parametric measure (ie, the pairwise or Pearson correlation) or a nonparametric measure (eg, the Spearman rank correlation coefficient). Both measures are similar, but while the Pearson correlation assumes that data is normally distributed, the Spearman rank correlation makes no assumptions about the dataset's distribution. Instead, it ranks the observations and calculates the differences between the rank of each pair. In this instance, the Spearman rank correlation is given (0.7311) and it should be used since the investor believes that the distribution is not normally distributed. The hypothesis test based on this nonparametric measure is calculated as: Since the calculated t-statistic is greater than the critical t-value, there is evidence that the population correlation is different from zero and the null hypothesis should be rejected (Choice A). (Choice C) 9.00 is obtained by incorrectly calculating the t-statistic with the Pearson correlation, which is a parametric measure and should not be used since the distribution is believed to deviate from a normal distribution. Things to remember: Nonparametric tests are used when few assumptions about the underlying population can be made. When testing the correlation of a population that is not normally distributed, the test should use nonparametric measures like the Spearman rank correlation coefficient instead of parametric measures (ie, Pearson correlation). Explain parametric and nonparametric tests of the hypothesis that the population correlation coefficient equals zero, and determine whether the hypothesis is rejected at a given level of significance LOS Copyright © UWorld. Copyright CFA Institute. All rights reserved.
2. An analyst selects 190 stocks and organizes them by style and market cap in a contingency table: Small-cap Mid-cap Large-cap Totals Growth 30 20 50 100 Value 45 30 15 90 Total 75 50 65 190 Using marginal frequencies, the expected frequency of large-cap growth stock is closest to: A. 26.3  B. 34.2 C. 50.0 Explanation A contingency table is used to display the frequency distribution for multiple categorical variables simultaneously. Each row displays a unique variable outcome of the first categorical variable, and each column displays a unique variable outcome of the second categorical variable. The contingency table shows: the marginal frequency of each categorical variable, which is the total frequency for each row or column. the joint frequency of two categorical variables, which is the frequency in each cell. The observed values are from a sample (ie, subset of a population) and are not representative of the whole target population. The population's expected joint frequency of the two specified categorical variable outcomes is the product of the individual relative marginal frequency of each categorical variable outcome, in this case, growth stock and large-cap stock: The expected frequency is 34.2. Things to remember: A contingency table summarizes the frequency distribution of multiple categorical variables. The total frequency for a single variable outcome (across one row or column) is called marginal frequency and the number in each cell is the joint frequency of two categorical variable outcomes. The expected joint frequency of the two specified categorical variable outcomes is the product of the individual relative marginal frequencies of each categorical variable outcome. Explain tests of independence based on contingency table data LOS Copyright © UWorld. Copyright CFA Institute. All rights reserved.
3. An analyst studies the relationship between company size and bond rating to predict bond defaults. The following table represents the number of defaults over a 10-year period: Rating Company Size Large Cap Mid Cap Small Cap AAA or A 15 32 51 BBB or BB 24 47 68 High yield 31 55 70 Based on this information, the relative frequency of BBB–BB bond defaults by mid-cap companies, based on all mid-cap company bonds, is closest to: A. 12.0% B. 33.8%  C. 35.1% Explanation Contingency table Rating Company size Large cap Mid cap Small cap Total by rating AAA or A 15 32 51 98 BBB or BB 24 47 68 139 High yield 31 55 70 156 Total by size 70 134 189 393 Contingency tables can be used to find patterns that exist between two variables. Each variable can have one or more categories; the only requirement is that the number of categories be finite. One variable (eg, company size) is placed in rows and the other (eg, bond rating) is placed in columns. Each cell in a contingency table represents a joint frequency between a category of one variable and a category of the other variable. Data in a contingency table can be displayed as a count (ie, absolute frequency) or a percentage of a total (ie, relative frequency). In this question, the variables are bond rating and company size, and each variable has three categories. There are 47 defaults of bonds rated either BBB or BB from mid-cap companies and 134 total bond defaults from mid-cap companies, so the relative frequency is 47 / 134 ≈ 35.1%. When using contingency tables, it is crucial to know precisely which relationships are being compared. For instance, if this question had asked for the frequency of BBB- or BB-rated mid-cap defaults relative to all defaults of BBB—BB bonds, the correct answer would be 47 / 139 ≈ 33.8% (Choice C). If the question had asked for the frequency of mid-cap defaults with those ratings relative to all defaults, the correct answer would be 47 / 393 ≈ 12.0% (Choice A). Things to remember: A contingency table can be used to find patterns that exist between two variables. Each variable must have a finite number of categories. One application of a contingency table is to detect patterns that exist between and among variables. Explain tests of independence based on contingency table data LOS Copyright © UWorld. Copyright CFA Institute. All rights reserved.
4. An investor builds a contingency table for a group of 600 companies based on two variables: inventory valuation method and gross margin. The scaled squared deviation for each combination of these variables is presented below: Gross margin Inventory valuation method Low Average High FIFO 2.92 0.60 0.77 LIFO 1.47 0.34 0.34 Weighted average 1.18 0.21 0.35 At a 5% level of significance and based on the chi-square distribution table, the most appropriate conclusion is that: A. the variables are positively correlated.  B. the variables are independent of each other. C. the chi-square statistic is greater than the critical value. Explanation Contingency tables can be used to find existing patterns between two categorical variables. The rows represent the groups (ie, categories) of the first variable, while the columns represent the groups of the second variable. Each cell represents a joint frequency of a pair of groups. Chi- square (χ ) is a nonparametric test statistic that tests if these variables are independent or related to each other. To obtain the χ , it is necessary to create a contingency table, an expected frequency table, and a scaled squared deviation table. This calculation is extensive, but χ can be calculated adding the cells of the scaled squared deviation table, which is given in this question. The corresponding hypothesis test is: 2 2 2
As the differences are squared (ie, positive), the only rejection region for the chi-square distribution is on the right side. Since the calculated χ is less than the critical value, there is not enough evidence that the variables are not independent, and the null hypothesis should not be rejected (Choice C). (Choice A) Correlation measures the direction and strength of the linear relationship between two variables. Therefore, it does not apply to unordered categorical variables, like the inventory valuation method. Things to remember: The chi-square distributed statistic can be used to test the independence of the variables of a contingency table. The test is performed by calculating expected frequencies and the squared scaled deviation and comparing the statistic with the critical value. Explain tests of independence based on contingency table data LOS Copyright © UWorld. Copyright CFA Institute. All rights reserved. 2