CFA Level 1 (2009) - 1
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Study Session 3
Cross-Reference to CFA Institute Assigned Reading #11 - Hypothesis Testing
LOS ll.e
With unknown population variance, the t-statistic is used for tests about the
x - fLo ...
mean of a normally distributed population: tn-l = ~. If the population vanance IS
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x -fLo
known, the appropriate test statistic is z= --- for tests about the mean of a alI
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population.
For two independent samples from twO normally distributed populations, the difference in means can be tested with a t-statistic. When the two population variances are assumed to be equal, the denominator is based on the variance of the pooled samples, but when sample variances are assumed to be unequal, the denominator is based on a combination of the two sam pies' variances.
A paired cOllOfJarisom test is concerned with the mean of the differences between the paired observations of two dependent, normall~' distributed samples. At-statistic
d - fld |
1- , where sJ |
Sd |
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is used: t = |
= In' and d is the average difference of the |
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observa tions. |
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LOS 11.f
The test of a hypothesis about the population variance for a normally distributed
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(n _1)s 2 |
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popu IatlOn uses a c |
I-square test statistic: \. - = |
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,were n IS t |
e samp e |
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ao |
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size, s2 is the sample variance, and |
a~ is the hypothesized value for the |
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population variance. Degrees of freedom is n - |
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The test comparing twO variances based on independent samples from twO
normally distributed populations uses an F-disrributed test statistic: F = -s2*-, where Sf IS
s2
the variance of the first sample and s~ is the (smaller) variance of the second sample.
LOS 11.g
Parametric tests. like the t-test, F-tesr, and chi-square tests, make assumptions regarding the distribution of the population from which samples are drawn, while nonparametric tests either do not consider a particular population parameter or have few assumptions about the sampled population.
©2008 Kaplan Schweser |
Page 321 |
Study Session 3
Cross-Reference to CFA Institute Assigned Reading #11 - Hypothesis Testing
CONCEPT CHECKERS
1.Which of the following statements about hypothesis testing is most accurate?
A.A Type II error is rejecting the null when it is actually true.
B.The significance level equals one minus the probability of a Type I error.
e. A two-tailed test with a significance level of 5% has z-critical values of
± 1.96.
2.Which of the following statements about hypothesis testing is feast accurate?
A.The power of test = 1 - P(Type II error).
B.If the computed z-statistic = -2 and the critical z-value = -1. 96, the null hypothesis is rejected.
e. The calculated z-statistic for a test of a sample mean when the population
variance is known is: z = |
X-P'o |
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Use the following data to answer Questions 3 through 7.
Austin Roberts believes that the mean price of houses in the area is greater than $145,000. A random sample of 36 houses in the area has a mean price of $149,750. The population standard deviation is $24,000, and Roberts wants to conduct a hypothesis test at a 1% level of significance.
3.The appropriate alternative hypothesis is:
A.H a : ~ < $145,000.
B.H a : ~ 2:: $145,000.
e.H a : ~ > $145,000.
4.The value of the calculated test statistic is closest to:
A.z = 0.67.
B.z=1.19.
e.z = 4.00.
5.Which of the following most accurately describes the appropriate test structure?
A.Two-tailed test.
B.One-tailed test.
e.Chi-square test.
6.The critical value of the z-statistic is:
A.z = ±1.96.
B.z = +2.33.
e.z = ±2.33.
7.At a 1% level of significance, Roberts should:
A.reject the null hypothesis.
B.fail to reject the null hypothesis.
e. neither reject nor fail to reject the null hypothesis.
Page 322 |
©2008 Kaplan Schweser |
Study Session 3
Cross-Reference to CFA Institute Assigned Reading #11 - Hypothesis Testing
Use the following data to answer Questions 8 through 13.
An analyst is conducting a hypothesis test to determine if the mean time spent on investment research is different from three hours per day. The test is performed at the 5% level of significance and uses a random sample of 64 portfolio managers, where the mean time spent on research is found to be 2.5 hours. The population standard deviation is 1.5 hours.
8.The appropriate null hypothesis for the described test is:
A.H o: [1 = 3 hours.
B.H o: fl ::; 3 hours. e. H o: ~L ;::: 3 hours.
9.This is a:
A.one-tailed test.
R.rwo-tailed rest.
C.p:lired cOlll11:lrisons rest.
10.Thl' caklll;lrl'd ;:--stalisric is:
A.-2.()-.
B.+0.33. e. +2.67.
1]. The critical z-vallle(s) of the teSl statisric is (are):
A.-1.9G. B. + 1.96.
e.±1.96.
12.The 95% confidence interval for rhe population mean is:
A. 11.00<~L<3.501.
B. {0.54 < ~L < 4.46).
e.(2.13 < 11 < 2.87}.
13.The analysr should most Ilppropriate~y: A. reject the null hypothesis.
B. t:lil to reject the null hypothesis.
e.reach no conclusion because the sample standard deviation was nor given.
14.A study was conducred to determine wherher the standard deviation of monthly maintenance com of a Pepper III aircraft is $300. A sample of 30 Pepper Ills had a mean monthly maintenance cost of $3,025 and a srandard deviarion
of $325. Using a 5% level of significance, which of the following is the most appropriate conclusion regarding the difference between the hypothesized value of the population variance and the sample variance?
A. The population and sample variances are significantly different.
B. The population and sample variances are not significantly different.
e. There are no rests that may be used to test variance differences in small
samples.
©2008 Kaplan Schweser |
Page 323 |
Study Session 3
Cross-Reference to CFA Institute Assigned Reading #11 - Hypothesis Testing
Use the following data to answer Questions 15 through 18.
Two samples were drawn from two normally distributed populations. For the first sample, the mean was $50 and the standard deviation was $5. For the second sample, the mean was $55 and the standard deviation was $6. The first sample consists of 25 observations and the second sample consists of 36 observations. (Note: In the questions below, the subscripts" 1" and "2" indicate the first and second sample, respectively.)
15. |
Consider the hypotheses structured as H o: ~l = $48 versus H a : ~l ~ $48. At a |
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1% level of significance, the null hypothesis: |
A.cannot be rejected. B. should be rejected.
C. cannot be tested using this sample information provided.
16.Using a 5% level of significance and a hypothesis test structure of Ho: of ~ 24 versus H,,: IT? > 24, the null hypothesis:
A. cannot be rejected. R. should be rejected.
C. C.ll1Ilot be tested using thIs sample information provided.
F. Consider the hypotheses structured as Ho: ~ll ~ $48 versus Ha : ~1 > $48. At :l 5°1c) level of significance, the null h~'pothesis:
A.C:lnnot be rejected.
B.should be rejected.
C.cannot be tested using the sample information provided.
18.Using a 5% level of significance for a test of the null of H o: (J 1 = O 2 versus the alternative of H,,: (J I :;z= (J2' tht: null hypothesis:
A.cannot be rejected.
B.should be rejected.
C.cannot be tested using the sample information provided.
19.If the significance level of a test is 0.05 and the probability of a Type II error is 0.15, what is the power of the test?
A.0.850.
B.0.950.
C.0.975.
20.Which of the following statements about the F-distribution and chi-square distribution is least accurate? Both distributions:
A.are asymmetrical.
B.are bound by zero on the left.
C.have means that are less than their standard deviations.
21.The appropriate test statistic for a test of the equality of variances for two normally distributed random variables, based on two independent random samples, is the:
A.t-test.
B.Ftest.
C.X 2 test.
Page 324 |
©2008 Kaplan Schweser |
Study Session 3
Cross-Reference to CFA Institute Assigned Reading #11 - Hypothesis Testing
22.The appropriate test statistic for a test that the variance of a normally distributed population is equal to 13, is the:
A.t-rest.
B.F-test.
C.X2 test.
23.William Adams wants to test whether the mean monthly returns over the last five years are the same for two stocks. If he assumes that the returns distributions are normal and have equal variances, the type of test and test statistic are best described as:
A.paired com parisons test, t-statistic.
B.paired comparisons test, F-statistic.
C.difference in means test, t-statistic.
24.Which of the following assumptions is least likely required for the difference in means test based on two samples?
A.The rwo samples are independent.
B.The two populations are normallY distribured.
C.The two populations have equal variances.
25.For a hypothesis test with a probability of a Type II error of 60% and a
probability of a Type I error of 5%, which of the following statements is moN
(/cC1Iltlte?
A.The power of the test is 40%. and there is a 5% probability that the rest statistic will exceed the critical value(sl.
B.There is a 95% probability thar the test statisric will be between the critical values if this is a two-tail test.
C.There is a 5% probability that the null hypothesis will be rejecred when actually true, and the probability of rejecting the null when it is false is
40%.
©2008 Kaplan Schweser |
Page 325 |
Study Session 3
Cross-Reference to CFA Institute Assigned Reading #11 - Hypothesis Testing
COMPREHENSIVE PROBLEMS
1. |
Ralph Rollins, a researcher, believes that the stocks of firms that have appeared |
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in a certain financial newspaper with a positive headline and story return more |
on a risk-adjusted basis. He gathers data on the risk-adjusted returns for these stocks over the six months after they appear on the cover, and data on the riskadjusted returns for an equal-sized sample of firms with characteristics similar to the cover-story firms matched by time period.
A.State the likely null and alternative hypotheses for a test of his belief.
B.Is this a oneor two-tailed test?
C.Describe the steps in testing a hypothesis such as the null you describe.
2.For each of the following hypotheses, describe the appropriate test, idenrif)· the appropriare tes[ sr;ltistic. and explain under what conditions the null hypothesis should he rejcded.
A.A researcher has returns over 52 weeks for an index of natural gas stocks and for an lndcx of oil stocks and wanrs to know if the weekly returns arc equal. Assume that the returns are approximately normallv distributed.
B.A researcher has two independent samples that arc approximately normallv distributed. She wishes to test whether the mean values of the two random variables are equal and assumes that the variances of the populations from which the two samples are drawn are equal. As an additional question here, how should the degrees of freedom be calculated?
c. A researcher wants to determine whether the population I'Jriances of two normally distributed random variables are equal based on two samples of sizes n l and n 2. As an additional question here, how should the degrees of freedom be calculated?
D.A researcher wants to test whether the variance of a normally distributed population is equal to 0.00165. As an additional question here, how should the degrees of freedom be calculated?
Page 326 |
©2008 Kaplan Schweser |
Swdy Session 3
Cross-Reference to CFA Institute Assigned Reading #11 - Hypothesis Testing
ANSWERS - CONCEPT CHECKERS
1.C
2.C
Rej~cting the null when it is acwally rrue is a Type I error. A Type II error is failing ro reject the null hypothesis when it is false. The significance level equals the probability of a Type I error.
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z = --- |
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e varIance. |
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3.C H a: ~ > $145,000.
4. B z= |
149,750 -145,000 |
24,0001 J36 = 1.1875. |
5.B The alternative hypothesis, H a: 11 > $145,000, only allows for values greater than the hypothesized valuc. Thus. this is a one-sided (one-tailcd) tCSl.
6.B For a ont'-t~\iled ;;-test ,It thc 1(l·i, level of significance. thl' critic.l! z-\'aluc is ::(I.lIl = 2.33.
Since the test is one-tailed on the upper end (i.e .. H. : 11 > 14'1,0(0). wc use a rositive
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z-critical value.
7. B The decision rule is ro reject H o if z-computed > z-critical. Since 1.1875 < 2.33, Roberts will fail ro reject the null.
8.A 
9.B This is a two-sided (tailed) test. We want to test if the mean "differs from" 3 hours (i.e.. H,,: p ;;c 3 hours).
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The normally d1srr1 |
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test stallsllC = Z = |
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1.5/,,64 |
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11. |
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At 0./2 =0 0.025, the critical z-values are: ±Zo./2 =0 |
±zO.025 |
=0 ±1.96. |
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12. |
C |
The 95% confidence interval is 12.5 ± |
(1.96)(0.1875)\ = 12.5 ± 0.3675) - [2.1325 < 11· |
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< 2.86751.
13.A Decision rule: reject H o if zcompured < -1.96 or if zcompurcd > + 1.96. Since -2.67 < -1.96, reject H o'
14.B The wording of the proposition is a little rricky, but the test srrucwre is H o: 0 2 = 3002 versus H,,: 0 2 ;;c 300 2. The appropriate test is a two-tailed chi-square test. The decision
rule is to reject H o if the test statistic is outside the range defined by the critical chisquare values at 0./2 = 0.025 with df =0 29. The test statistic is
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(29}(105,625) |
Xl = |
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= 34.035 |
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90,000 |
The critical chi-square values are 16.047 on the left and 45.722 on the right. Since the X2 falls between these two values, we fail to reject the null hypothesis. This means the population standard deviation is not significantly different than $300.
©2008 Kaplan Schweser |
Page 327 |
Study Session 3
Cross-Reference to CFA Institute Assigned Reading #11 - Hypothesis Testing
15. A A two-tailed t-test is appropriate. The decision rule is to reject H o if the t-statistic is
outside the range defined by ±t |
at ex = 0.01 with df = 24. The t-statistic = t24 |
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x-ILo |
50-48 |
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- r = |
r;:;; = 2.0 . :tt24 |
at (X = 0.0 I = ±2.797; therefore, H o cannot be rejected. |
s/-..{n |
5/-..{25 |
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16. A The chi-square test is used to test hypotheses concerning a single population variance. Since this is a one-tailed test, the decision rule is to reject Ho if X2> the critical chisquare value at (X = 0.05 with df = 24.
,2 _ |
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(24)(25) |
= 25.0 . The right-tail critical chi-square value is 36.415. |
'Xn-l- |
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24 |
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Since X2 = 25 ::; 36.415, H o cannot be rejected.
17. B A one-tailed t-test is appropriate. The decision rule is to reject H o if the computed
I-statistic> I-critical at (\ = 0.05 with df = 24. The computed value of the t-statistic =
X-II" __ 50-4R _ ') |
,._. _ |
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1.'" 11. Since I > I-critical, H u shoulJ bc |
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_.0, and |
1-(;(I(IC,11 - I, |
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,,/·../n |
5/'0/25 |
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rejected.
18.A The Ftest is appropriate to the equality of population variances. The decision rule i" [0 reject H o if the computed test statistic, F, exceeds the critical F-valuc at (~/2. For thc'
information provided, F = s~/s~ = %/25 = |
1.44. At a 0.025 level of significance with |
<II =}'iaJiJ dz = 24, F-critical = 2.18. Since |
F < F-critical (1.44 < 2.18), we fail tll reject |
the null hy pothesis. |
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P"ofessoi''s Note: Many F-tables do not contain numerator df of35. On the exam.
CFA IlIStitute will design problems such that the dfarc contained directly in the table,- that you will be given on the exam. Ifthe tables do not contain the exact dfyou need, interpolate the critical value based on the closest aMilahle df values.
J 9. A The power of a test is I - P(Type 11 error) = I - 0.1 5 = 0.85.
20.C There is no consistent relationship between the mean and standard deviation of the chi-square distribution or F-distribution.
21.B The Ftest is the appropriate test.
22.C A tcst of er 2 = ()'o2is a X2 test.
23.A Since the observations are likely dependent (both related to market returns), a paired comparisons (mean differences) test is appropriate and is based on at-statistic.
24.C When the variances are assumed to be unequal, we just calculate the denominator (standard error) differently and use both sample variances to calculate the t-statistic.
Page 328 |
©2008 Kaplan Schweser |
Study Session ;)
Cross-Reference to CFA Institute Assigned Reading #11 - Hypothesis Testing
25. C A Type I error is rejecting the null hypothesis when it's true. The probability of rejecting a false null is [1 - Prob Type II] = [1 - 0.60] =40%, which is called the power of the test. A and B are not necessarily true, since the null may be false and the probability of rejection unknown.
ANSWERS - COMPREHENSIVE PROBLEMS
1.A. The null hypothesis is typically the one the researcher wants to disprove. In this case, that would be that the mean risk-adjusted return on the cover stocks is less than or equal to the mean risk-adjusted return on the control stocks. The alternative is that the mean risk-adjusted returns on the cover stocks is greater than the mean risk-adjusted return on the control stocks. Rejecting the null will offer statistical support for the proposition the researcher wants to "prove" (the alternative).
B.This would be a one-tailed test since the alternative is "greater than."
C. The steps in a hypothesis tcst are as follows: State the hypothesis.
Select tht· 'Ippropriall" lest statistic.
Decide on [he appropriate level or significance. Determine the decision ruJe.
Collect the data.
Calculate the sample statistics.
iVlake a decision bas~d on t he decision rule for the test. Make decisions or inferences based on the results.
A.Since these two returns likely exhibit significant correlation and are therefore not
independent, a paired comparisons test is appropriate. Differences between the returns on the two indices each week will be used. The standard deviation of the differences
is used to construct a t-test of the hypothesis that the mean weekly difference is significantly different from (not equal to) zero. Reject if the I-statistic is greater/less than the positive/negative critical value.
B. This is a test of a difference in means and is a t-test. The test statistic is the difference in means over a standard deviation calculated from the pooled variances of the twO samples. Reject if the (-statistic is greater/less than the positive/ negative critical value. When the variances arc assumed to be equal for a difference in means test, we can use the variance of the pooled samples, and the degrees of freedom are simply n 1 + n 2 - 2 (total number of observations in both samples minus two).
C. The test statistic is the ratio of the larger sample variance (0 the smaller sample variance. This statistic follows an F-distribution with n l - 1 and n 2 - 1 degrees of freedom. Reject equality if the test statistic exceeds the upper critical value.
D. The test of whether the population variance is equal (0 a particular value is done with a test statistic with (n - 1) times the sample variance in the numerator and the
hypothesized variance (0.00165 here) in the denominator |
(n _1)s2 |
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The test statistic follows a Chi-square distribution. Reject the null of a population variance equal to 0.00165 if the test statistic is greater than the upper critical value or less than the lower critical value. The degrees of freedom are simply n - 1.
©2008 Kaplan Schweser |
Page 329 |
The following is a review of the Quantitative Methods: Application principles designed to address the learning outcome statements set forth by CFA Institute@. This topic is also covered in:
TECHNICAL ANALYSIS
Study Session 3
EXAM Focus
This topic review inrroduces the "story" that underlies technical analysis, and you should understand how this differs from
fundamenral analysis. You should learn
what the technical indicator names mean.
Confusion regarding |
which indicators are |
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contrarian indicators |
and which are smart |
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money |
indicators is normal. |
I suggest |
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vou try |
[0 remem ber which are |
rhe smart |
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money indicators because there are only three of them; then you will know that the others are conrrarian indicators. The real distinction here is whose actions are driving the indicator. For smart money indicators, the "smart" people driving the indicator values are bond traders (confidence index and TED spread) and investors buying on margin (margin debtl.
LOS 12.a: Explain the underlying assumptions of technical analysis.
Underlying all of rechnical analysis are the following assumptions:
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Values. and thus prices, are determined by supply and demand. |
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Supply and demand are driven by both rational and irrational behavior. |
•Security prices moye in trends that persist for long periods.
•While the causes of changes in supply and demand are difficult to determine, the actual shifts in supply and demand can be observed in market price behavior.
The major challenge to technical analysis is the efficienr markets hypothesis (EMH). Followers of the EMH believe that all available information associated with both fundamenral and technical analysis is impounded in current security prices. EMH followers argue that technical trading rules require too much subjective interpretation and that decision variables change oyer time.
Fundamenral analysts believe that a security's price is determined by the supply and demand for the underlying security based on its economic fundamentals, such as expected return and risk. Fundamenralists believe they can forecast value changes by analyzing earnings and other publicly available data.
The difference between fundamenral analysis and technical analysis is the assumption about the speed at which new information is impounded inro prices. Technicians believe the reaction is slow, while fundamentalists believe prices adjust quickly. In addition, efficienr market hypothesis analysts feel the price adjustmenr happens almost instanraneously.
Fundamenralists, through their research, look for changes in the basis of value, which eventually leads to changes in the supply and demand for the stock. Technicians look for evidence of changes in supply and demand through market signals and indicators.
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©2008 Kaplan Schweser |
