CFA Level 1 (2009) - 1

Study Session 3

Cross-Reference to CFA Institute Assigned Reading #11 - Hypothesis Testing

This decision rule is illustrated in the following figure.

Decision Rule for a Two-Tailed Paired Comparisons Test

(a. =0.05, df =38)

Fail to Reject H o

CoLLect the sampLe and calcuLate the sampLe statistics. Using the sample data provided, the test statistic is computed as follows:

Make a decision regarding the hypotheses. The computed test statistic, 10.2596, is greater than the critical t-value, 2.024-it falls in the rejection region to the right of 2.024 in the previous figure. Thus, we reject the null hypothesis of no difference, concluding that there is a statistically significant difference in betas from before to after deregulation.

Make a decision based on the resuits ofthe test. We have support for the hypothesis that betas are lower as a result of deregulation, providing support for the proposi tion that deregulation resulted in decreased risk.

Keep in mind that we have been describing two distinct hypothesis tests: One about the significance of the difference between the means of two populations and one about the significance of the mean of the differences between pairs of observations. Here are rules for when these tests may be applied:

• The test of the differences in means is used when there are two independent sampLes.

•A test of the significance of the mean of the differences between paired observations is used when the samples are not independent.

Srudy Session 3

Cross-Reference to CFA Institute Assigned Reading #11 - Hypothesis Testing

Professor's Note: The LOS here say "IdentifY the appropriate test statistic and interpret the results... " I can't believe candidates are expected to memorize these formulas (or that you would be a better analyst ifyou did). The CFA exam is

~not known for requiring the use of complicatedformulas from memory. You

~should instead focus on the fact that both ofthese tests involve t-statistics arId depend on the degrees offreedom. Also note that when samples are indeperldcnt,

you can use the difference in means test and when they are dependent, the statistic is the average difference in (paired) observations divided by the standard error of the average difference.

LOS 11.f: Identify the appropriate test statistic and interpret the results for a hypothesis test concerning 1) the variance of a normally distributed

population, and 2) the equality of the variances of two normally distributed populations, based on two independent random samples.

The chi-)'(jllt, n' test is used for hypothesis tests concerning the variance of a normally distributed population. Lerting (Yo:' represent the true population variance and a~ represent the: hypothesized variance. the hypotheses for a two-tailed test of a single population variance are structured as:

The hypothed~s for one-tailed tests arc structured as:

Hypothesis testing of the population variance requires the use of a chi-square distributed test statistic, denoted X2. The chi-square distribution is asymmetrical and approaches the normal distribution in shape as the degrees of freedom increase.

To illustrate the chi-square distriburion. consider a two-tailed test with a 5% level of significance and 30 degrees of freedom. As displayed in Figure 7, the critical chi-square values are 16.791 and 46.979 for the lower and upper bounds, respectively. These values are obtained from a chi-square table, which is used in the same manner as a t-table. A portion of a chi-square table is presented in Figure 8.

Note that the chi-square values in Figure 8 correspond to the probabilities in the right tail of the dimibution. As such. the 16.791 in Figure 7 is from the column headed 0.975 because 95% + 2.5% of the probability is to the right of it. The 46.979 is from the column headed 0.025 because only 2.5<),(, probability is to the right of it. Similarly. at a 5% level of significance with 10 degrees of freedom, Figure 8 shows that the critical chi-square values for a two-tailed test are 3.247 and 20.483.

Study Session 3

Cross-Reference to CFA Institute Assigned Reading #11 - Hypothesis Testing

Figure 7: Decision Rule for a Two-Tailed Chi-Square Test

(a= 0.05, df = 30)

2.5%

o

"(n-1)s2

Similar to other hypothesis tests, the chi-square tcst compares the test statistic, X~-I, to a critical chi-square valuc at a givcn level of significance and n - 1 degrees of freedom. Note that since the ell i-square distribution is bounded below by zero, chi-sq uare values cannot be negative.

Study Session 3

Cross-Reference to CPA Institute Assigned Reading # 11 - Hypothesis Testing

Example: Chi-square test for a single population variance

Historically, High-Return Equity Fund has advertised that its monthly returns have a standard deviation equal to 4%. This was based on estimates from the 1990-1998 period. High-Return wants to verify whether this claim still adequately describes the standard deviation of the fund's returns. High-Return collected monthly returns for the 24-month period between 1998 and 2000 and measured a standard deviation of

monthly returns of 3.8%. Determine if the more recent standard deviation is different from the advertised standard deviation.

Answer:

State the hypothesis. The null hypothesis is that the standard deviation is equal to 4%, and therefore the variance of monthly returns for the population is (0.04)2 = 0.0016. Since High-Return simply wants to test whether the standard deviation has changed, up or down, a two-sided test should be used. The hypothesis test structure takes the form:

Specify the level ofsignificance. Let's use a 5% level of significance, meaning there will be 2.5% probability in each tail of the chi-square distribution.

State the decision rule regarding the hypothesis. With a 24-month sample, there are 23 degrees of freedom. Using the table of chi-square values at the back of this book, for 23 degrees offreedom and probabilities of 0.975 and 0.025, we find two critical values, 11.689 and 38.076. Thus, the decision rule is:

Reject H o ifX2 < 11.689, or X2 > 38.076

Study Session 3

Cross-Reference to CFA Institute Assigned Reading #11 - Hypothesis Testing

This decision rule is iJJ ustrated in the following figure.

Decision Rule for a Two-Tailed Chi-Square Test of a Single Population Variance

(a= 0.05, df = 23)

Fail [0 Reject HII

CoLLect the sampLe and calculate the sampLe statistics. Using the information provided, the test statistic is computed as:

Make a decision regarding the hypothesis. Since the computed test statistic, X2, falls between the twO critical values, we fail to reject the null hypothesis that the variance is equal to

0.0016.

Malu a decision based on the resuLts ofthe test. It can be concluded that the recently measured standard deviation is dose enough to the advertised standard deviation that we cannot say that it is different from 4%, at a 5% level of significance.

Testing the Equality of the Variances of Two Normally Distributed

Populations, Based on Two Independent Random Samples

The hypotheses concerned with the equality of the variances of tWO populations are tested with an F-disrributed test statistic. Hypothesis testing using a test statistic that follows an Fdisuibution is referred to as the Ftest. The F-test is used under the assumption that the populations from which samples are drawn are normally distributed and that the samples are independent.

Study Session 3

Cross-Reference to CFA Institute Assigned Reading #11 - Hypothesis Testing

If we let crf and cr~ represent the variances of normal Population 1 and Population 2,

respectively, the hypotheses for the two-tailed F-test of differences in the variances can be structured as:

and the one-sided test structures can be specified as:

The test statistic for the F-test is the ratio of the sample variances. The F-statistic is computed as:

Note that n 1 - ] and n 2 - ] are the degrees of freedom used to identify the appropriate critical value from the F-table (provided in the Appendix).

An F-distribution is presented in Figure 9. As indicated, the F-distribution is rightskewed and is truncated at zero on the left-hand side. The shape of the F-distribution is determined by two separate degrees o.{freedom, the numerator degrees of freedom, d.f~,

and the denominator degrees of freedom, df..,. Also shown in Figure 9 is that the rejection region is in the right-side tail of the distributron. This will always be the case as long

as the F-statistic is computed with the largest sample variance in the numerator. The labeling of 1 and 2 is arbitrary anyway.

Study Session 3

Cross-Reference to CFA Institute Assigned Reading # 11 - Hypothesis Testing

Figure 9: F-Distribution

numerator dfl = 10, denominator df2 = 10

5%

o

2.98

Fai I to Rejecr H,

Example: F-test for equal variances

Annie Cower is examining the earnings for two different industries. Cower suspects that the earnings of the textile industry are more divergen t than those of the paper industry. To confirm this suspicion, Cower has looked at a sample of 31 textile manufacturers and a sample of 41 paper companies. She measured the sample standard deviation of earnings across the textile industry to be $4.30 and that of the paper industry companies to be $3.80. Determine if the earnings of the textile industry have greater standard deviation than those of the paper industry.

Answer:

State the hypothesis. In this example, we are concerned with whether the variance of the earnings of the textile industry is greater (more divergent) than the variance of the earnings of the paper industry. As such, the test hypotheses can be appropriately structured as:

where:

0"[ variance of earnings for the textile industry

Select the appropriate test statistic. For tests of difference between variances, the appropriate test statistic is: .

2

F=~

s~

Study Session 3

Cross-Reference to CFA Institute Assigned Reading #11 - Hypothesis Testing

Specify the level ofsignificance. Let's conduct our hypothesis test at the 5% level of significance.

State the decision rule regarding the hypothesis. Using the sample sizes for the two industries, the critical F-value for our test is found to be 1.74. This value is obtained from the table of the F-distribution at the 5% level of significance with df] = 30 and df2 =40. Thus, if the computed F-statistic is greater than the critical value of 1.74, the null hypothesis is rejected. The decision rule, illustrated in the figure below, can be stated as: .

Reject H o if F > 1.74

Decision Rule for F-Test

(a. = 0.05, dfl = 30, df~ = 40)

Collect the sample and calculate the sample statistics. Using the information provided, the F-statistic can be computed as:

F = s~ = $4.30~ = $18.49 = 1.2805

S2 $3.80 $14.44

~Prof!ssor's Note: Remember to square the standard deviations to get the

~varzances.

Make a decision regarding the hypothesis. Since the calculated F-statistic of 1.2805 is less than the critical F-statistic of 1.74, we fail to reject the null hypothesis.

Make a decision based on the results ofthe test. Based on the results of the hypothesis test, Cower should conclude that the earnings variances of the industries are not statistically significandydifferent from one another at a 5% level of significance. More pointedly, the earnings of the textile industry are not more divergent than those of the paper industry.

Study Session 3

Cross-Reference to CFA Institute Assigned Reading # 11 - Hypothesis Testing

LOS 11.g: Distinguish between parametric and nonparametric tests and describe the situations in which the use of nonparametric tests may be appropriate.

Parametric tests rely on assumptions regarding the distribution of the population and are specific ro population parameters. For example, the z-test relies upon a mean and a standard deviation to define the normal distribution. The z-test also requires that either the sample is large, relying on the central limit theorem to assure a normal sampling distribution, or that the population is normally distributed.

Nonparametric tests either do not consider a particular population parameter or have few assumptions about the population that is sampled. Nonparametric tests are used when there is concern about quantities other than the parameters of a distribution or when the assumptions of parametric tests can't be supported. They are also used when the data are not suitable for parametric tests (e.g., ranked observations). Nonparametric tesrs are often used along with paramerric tests. In this way, the nonparamerric resr is a hackup in case the assumprions underlying the ~lararnerric tesr do nor hold.

One example of a nonparamerric tesr is a tesr using data ranks (e.g., largest. secondlargest, third-largest, etc.) for tWO data sets and examining the correlation of ranks between the two sets. We might use this to resr the correlation between firm size rank and earnings per share rank for a given set of firms.

Another example is a runs rest. If we look at a series of srock price changes (eirher up or down), a runs test would give us the probability thar an observed series of daily price changes (e.g., + + - + - - +) could resulr given that each price change is random.

Study Session 3

Cross-Reference to CFA Institute Assigned Reading #11 - Hypothesis Testing

KEy CONCEPTS

LOS ll.a

The hypothesis testing process requires a statement of a null and an alternative hypothesis, the selection of the appropriate test statistic, specification of the significance level, a decision rule, the calculation of a sample statistic, a decision regarding the hypotheses based on the test, and a decision based on the test results.

The null hypothesis is what the researcher wants to reject. The alternative hypothesis is what the researcher wants to prove, and it is accepted when the null hypothesis is rejected.

A two-tailed test results from a two-sided alternative hypothesis (e.g., H a : fl ~ flo)' A one-tailed test results from a one-sided alternative hypothesis (e.g., H a: fl > flo' or

H,I: P < flo).

The teSl sr;!ristic i, rhe I'Jlue thar a decision abour a hyporhesis \vill be based on.

A Type I error is the rejection of the null hyporhesis when ir is actually true, while a

Type II error is the ~ailure to rejecr the null hypothesis \V'len it is actually false.

The significance levt'l can be in terpreted as the probability that a test sraristic will reject the null hypothesi" by chance when it is actually rrue (i.e .. the probability of a Type I error).

LOS 11.c

Hypothesis resting compares a computed test statisric to a critical value at a stated level of significance, which is the decision rule for the test.

The power of a test is the probability of rejecting the null when it is false. The power of a test = 1 - P(Type II error).

A hypothesis abour a population parameter is rejected when the sample statisric lies outside a confidence interval around the hypothesized value for rhe chosen level of significance.

LOS 11.d

Statistical significance does not necessarily imply economic significance. Even though a test statistic is significant statistically, the size of the gains ro a srrategy ro exploit a sratistically significanr result may be absolutely small or simply nor grear enough to outweigh transactions costs.