CFA EXAM: STATISTICAL CONCEPTS #5

spearman rank correlation coefficient

1) rank from largest to smallest 1. has the largest value and so on if the third and fourth largest values are tied we assign it 3.5 2) calculate the difference between the ranks 3) then use spearman rank equation

30. In which of the following situations would a non-parametric test of a hypothesis most likely be used? a. The sample data are ranked according to magnitude. b. The sample data come from a normally distributed population. c. The test validity depends on many assumptions about the nature of the population.

A non parametric test is used when the data are given in ranks

The sixth step in hypothesis testing is making the statistical decision. .

For our example, because the test statistic z = 3.07 is larger than the rejection point of 1.645, we reject the null hypothesis in favor of the alternative hypothesis that the risk premium on Canadian stocks is positive.

formulation

H0: θ = θ0 versus Ha: θ ≠ θ0 (a "not equal to" alternative hypothesis) H0: θ ≤ θ0 versus Ha: θ > θ0 (a "greater than" alternative hypothesis) H0: θ ≥ θ0 versus Ha: θ < θ0 (a "less than" alternative hypothesis)

When comparing two sample means

H0: μ1 − μ2 = 0 versus Ha: μ1 − μ2 ≠ 0 (the alternative is that μ1 ≠ μ2) H0: μ1 − μ2 ≤ 0 versus Ha: μ1 − μ2 > 0 (the alternative is that μ1 > μ2) H0: μ1 − μ2 ≥ 0 versus Ha: μ1 − μ2 < 0 (the alternative is that μ1 < μ2)

The fifth step in hypothesis testing is collecting the data and calculating the test statistic.

If the mean of the Canadian stock return is 5.3% and the sd is 19.2 then the serror is 1.717% the test statistic is then 5.3/1.717 =z= 3.07

The fourth step in hypothesis testing is stating the decision rule.

When we test the null hypothesis, if we find that the calculated value of the test statistic is as extreme or more extreme than a given value or values determined by the specified level of significance, α, we reject the null hypothesis

13. The value of a test statistic is best described as the basis for deciding whether to: a. reject the null hypothesis. b. accept the null hypothesis. c. reject the alternative hypothesis.

a

14. Which of the following is a Type I error? a. Rejecting a true null hypothesis b. Rejecting a false null hypothesis c. Failing to reject a false null hypothesis

a

24. For a small sample with unknown variance, which of the following tests of a hypothesis concerning the population mean is most appropriate? a. A t-test if the population is normally distributed b. A t-test if the population is non-normally distributed c. A z-test regardless of the normality of the population distribution

a

28. A chi-square test is most appropriate for tests concerning: a. a single variance. b. differences between two population means with variances assumed to be equal. c. differences between two population means with variances assumed to not be equal.

a

10. In the step "stating a decision rule" in testing a hypothesis, which of the following elements must be specified? a. Critical value b. Power of a test c. Value of a test statistic

a critical value is the rejection point for the test

A t distribution is similar to

a normal standard distribution because it has a mean of zero and is symmetrical

Wants to test a hypothesis concerning the value of an underlying or population mean will conduct

a t test which follows a t distribution

11. Which of the following statements is correct with respect to the null hypothesis? a. It is considered to be true unless the sample provides evidence showing it is false. b. It can be stated as "not equal to" provided the alternative hypothesis is stated as "equal to." c. In a two-tailed test, it is rejected when evidence supports equality between the hypothesized value and population parameter.

a. the null is the hypothesis to be tested

A t distribution is defined by

degrees of freedom

T test is helpful to moderate

departures from normality

the only want to decrease both

errors is increasing n

It is different because it has a standard deviation

greater than one where normal has a standard deviation equal to one

For example, economic theory suggests that investors require a positive risk premium on stocks (the risk premium is defined as the expected return on stocks minus the risk-free rate). Following the principle of stating the alternative as the "hoped for" condition, we formulate the following hypotheses:

ho: The population mean risk premium on Canadian stocks is less than or equal to 0. Ha: The population mean risk premium on Canadian stocks is positive.

use z

if have large n

CHI

if is less than this number than reject

If the sample is not actually random

inferences based on a chi square test are false

Any test or procedure either the two characteristics

is a parametric test

Paired comparisons test

is a statistical test for differences in dependent items

the first formulation

is a two sided or two tailed hypothesis

type 2 error

is beta

Nonparametric test

is not concerned with a parameter

Hypothesis testing

is part of statistical inference

30 or more samples is large and 29 or less

is small

T test

is the test of choice

the second and third are

one tailed hypotheses

Hypothesis testing is concerned with

parameters and validity on set of assumptions

The alternative is the statement

that the condition is true and null is not true if evidence shows that we reject the null and accept the alternative then we confirm what we thought was true

if we have a level of .10 we have some evidence if we have a level of .05 then we have strong evidence if we have .01 it is very strong

that the null is false

null hypothesis

the hypothesis to be tested EX) that the population mean risk premium for Canada is less than or equal to zero

When testing two variances against the other

the larger one goes in the numerator

the significance level is the probability of type 1 or rejecting the null when it is true

the power of a test is the probability of correctly rejecting the null when the null is false

P value

the smallest level at which the null can be rejected

As the number of degrees of freedom increases with sample size

the spread decreases and the t distribution approaches the standard normal distribution

We have identified a test statistic to test the null hypothesis. What probability distribution does it follow? We will encounter four distributions for test statistics in this reading:

the t-distribution (for a t-test); the standard normal or z-distribution (for a z-test); the chi-square (χ2) distribution (for a chi-square test); and the F-distribution (for an F-test).

level of significance of .05 for a test means

there is a 5% probability of rejecting a true null

alternative hypothesis

this is accepted when the null is rejected EX) the mean risk premium for Canada is greater than zero

we decrease the probability of type 1 error by increasing the probability of

type 2 error

if n is greater than 30

use

If the population has a variance and is large then

use the z test

In the first step

we always state two hypotheses: the null and alternative hypothesis

when we believe that the population under consideration meaningfully departs from assumptions

we can employ a spearman rank correlation coefficient - a measure of correlation based on ranked data

when we use nonparametric tests

we convert the data into ranks

If the p-value is less than our specified level of significance,

we reject the null hypothesis

For our risk premium example, the population parameter of interest is the population mean risk premium, μRP.

we test H0: μRP ≤ μ0 versus Ha: μRP > μ0.

In test with the variance of a single population

we use the chi square test

We use these in three situations

when data does not meet assumptions data are given in ranks hypothesis we are addressing does not concern a parameter

Difference between t and z tests is smaller

when you have a large sample

three significance levels when testing

.10, .05, and .01

You are analyzing Sendar Equity Fund, a midcap growth fund that has been in existence for 24 months. During this period, it has achieved a mean monthly return of 1.50 percent with a sample standard deviation of monthly returns of 3.60 percent. Given its level of systematic (market) risk and according to a pricing model, this mutual fund was expected to have earned a 1.10 percent mean monthly return during that time period. Assuming returns are normally distributed, are the actual results consistent with an underlying or population mean monthly return of 1.10 percent? 1. Formulate null and alternative hypotheses consistent with the verbal description of the research goal. 2. Identify the test statistic for conducting a test of the hypotheses in Part 1. 3. Identify the rejection point or points for the hypothesis tested in Part 1 at the 0.10 level of significance. 4. Determine whether the null hypothesis is rejected or not rejected at the 0.10 level of significance. (Use the tables in the back of this book.)

1. H0: μ = 1.10 Ha: μ ≠ 1.10. 2. T Test 3. This is a two tailed test. Significance = .05 at 23 DoF Variable = 1.714 4. (1.50-1.10) / 3.6 / SQR24 = .54

FashionDesigns has generally maintained an average of 45 days in receivables. Because it would be too costly to analyze all of the company's receivables frequently, the controller's office uses sampling to track customers' payment rates. A random sample of 50 accounts shows a mean number of days in receivables of 49 with a standard deviation of 8 days. 1. Formulate null and alternative hypotheses consistent with determining whether the evidence supports the suspected condition that customer payments have slowed. 2. Identify the test statistic for conducting a test of the hypotheses in Part 1. 3. Identify the rejection point or points for the hypothesis tested in Part 1 at the 0.05 and 0.01 levels of significance. 4. Determine whether the null hypothesis is rejected or n

1. H0: μ ≤ 45 versus Ha: μ > 45. 2. Because the population variance is not known, we use a t-test with 50 − 1 = 49 degrees of freedom. 3. The rejection point is found across the row for degrees of freedom of 49. To find the one-tailed rejection point for a 0.05 significance level, we use the 0.05 column: The value is 1.677. To find the one-tailed rejection point for a 0.01 level of significance, we use the 0.01 column: The value is 2.405. 4. Test statistic = 3.536 Null is rejected at both levels. There is high confidence that there is a slowdown in customer payments.

The realized mean monthly return on the S&P 500 Index in the first half of the 2000s appears to have been substantially different than the mean return in the second half of the 2000s. Was the difference statistically significant? The data, shown in Table 3, indicate that assuming equal population variances for returns in the two decades is not unreasonable. 1. Formulate null and alternative hypotheses consistent with a two-sided hypothesis test. 2. Identify the test statistic for conducting a test of the hypotheses in Part 1. 3. Identify the rejection point or points for the hypothesis tested in Part 1 at the 0.10, 0.05, and 0.01 levels of significance. 4. Determine whether the null hypothesis is rejected or not rejected at the 0.10, 0.05, and 0.01 levels of significance.

1. H0: μ1 - μ2 = 0 versus Ha: μ1 - μ2 ≠ 0 2. 60 + 60 − 2 = 118 degrees of freedom. 3. In the tables (Appendix B), the closest number of degrees of freedom to 118 is 120. For a two-sided test, the rejection points are ±1.658, ±1.980, and ±2.617 for, respectively, the 0.10, 0.05, and 0.01 levels for df = 120. To summarize, at the 0.10 level, we will reject the null if t < −1.658 or t > 1.658; at the 0.05 level, we will reject the null if t < −1.980 or t > 1.980; and at the 0.01 level, we will reject the null if t < −2.617 or t > 2.617. 4. Use sp^2 to get = 21.86 then plug into test statistic = -.27 SO DO not reject any of them

You analyze the monthly returns on both ETFs from August 2008 to July 2013 and prepare the following summary table. STRATEGY Worldwide MEAN RETURN 0.61% STANDARD DEVIATION 5.43% STRATEGY US MEAN RETURN 0.39% STANDARD DEVIATION 6.50% STRATEGY Difference MEAN RETURN 0.22% STANDARD DEVIATION 1.86% 1. Formulate null and alternative hypotheses consistent with a two-sided test that the mean difference between the worldwide and only US strategies equals 0. 2. Identify the test statistic for conducting a test of the hypotheses in Part 1. 3. Identify the rejection point or points for the hypothesis tested in Part 1 at the 0.01 level of significance. 4. Determine whether the null hypothesis is rejected or not rejected at the 0.01 level of significance. (Use the tables in the back of this volume.) 5. Discuss the choice of a paired comparisons test.

1. H0: μd = 0 versus Ha: μd ≠ 0. 2. t-test with 60 − 1 = 59 degrees of freedom 3. In the table for the t-distribution, the closest entry to df = 59 is df = 60. We look across the row for 60 degrees of freedom to the 0.005 column, to find 2.66. We will reject the null if we find that t > 2.66 or t < −2.66. 4. sample error = 1.86 / sqroot 60 = .24012497 test = .22/ .24012491 = .92 5. Several US stocks that are part of the S&P 500 index are also included in the Vanguard Total World Stock Index ETF. The profile of the World ETF indicates that nine of the top ten holdings in the ETF are US stocks. As a result, they are not independent samples; in general, the correlation of returns on the Vanguard Total World Stock Index ETF and SPDR S&P 500 ETF should be positive. Because the samples are dependent, a paired comparisons test was appropriate.

You continue with your analysis of Sendar Equity Fund, a midcap growth fund that has been in existence for only 24 months. Recall that during this period, Sendar Equity achieved a sample standard deviation of monthly returns of 3.60 percent. You now want to test a claim that the particular investment disciplines followed by Sendar result in a standard deviation of monthly returns of less than 4 percent. 1. Formulate null and alternative hypotheses consistent with the verbal description of the research goal. 2. Identify the test statistic for conducting a test of the hypotheses in Part 1. 3. Identify the rejection point or points for the hypothesis tested in Part 1 at the 0.05 level of significance. 4. Determine whether the null hypothesis is rejected or not rejected at the 0.05 level of significance. (Use the tables in the back of this volume.)

1. H0: σ2 ≥ 16.0 versus Ha: σ2 < 16.0. 2.The test statistic is χ2 with 24 − 1 = 23 degrees of freedom. 3. The lower 0.05 rejection point is found on the line for df = 23, under the 0.95 column. The rejection point is 13.091. We will reject the null if we find that χ2 is less than 13.091. 4. test statistic = = n-1(sample SD)^2 / new SD^2 = (24-1) (3.60)^2 / 4^2 = 298.08 / 16 = 18.64 Because 18.63 (the calculated value of the test statistic) is not less than 13.091, we do not reject the null hypothesis. We cannot conclude that Sendar's investment disciplines result in a standard deviation of monthly returns of less than 4 percent.

You gather data for 156 weeks of returns during 2004 to 2006 and 156 weeks of returns during 2010 to 2012. You have specified a 0.01 level of significance. Before crisis n 156 Mean weekly return 0.358% Variance of returns 7.240 After crisis n 156 mean weekly return 0.110 Variance of returns 6.269 1. Formulate null and alternative hypotheses consistent with the verbal description of the research goal. 2. Identify the test statistic for conducting a test of the hypotheses in Part 1. 3. Determine whether or not to reject the null hypothesis at the 0.01 level of significance.

1. H0:Before=After Ha:Before≠After 2. 156 -1 =155 Numerator dof 3. The "before" sample variance is larger, so following a convention for calculating F-statistics, the "before" sample variance goes in the numerator: F = 7.240/6.269 = 1.155. Because this is a two-tailed test, we use F-tables for the 0.005 level (= 0.01/2) to give a 0.01 significance level. In the tables in the back of the volume, the closest value to 155 degrees of freedom is 120 degrees of freedom. At the 0.01 level, the rejection point is 1.61. Because 1.155 is less than the critical value 1.61, we cannot reject the null hypothesis.

We want to test if the difference in recovery rates between those two types of firms is statistically significant. With μ1 denoting the population mean recovery rate for the bonds of electricity firms and μ2 denoting the population mean recovery rate for the bonds of retail firms, the hypotheses are H0: μ1 − μ2 = 0 versus Ha: μ1 − μ2 ≠ 0. ELECTRICITY Number of observations 39 Average price $48.03 Standard deviation $22.67 RETAIL Number of Observations 33 Average Price $33.40 Standard deviation $34.19 We assume that the populations (recovery rates) are normally distributed and that the samples are independent. Based on the data in the table, address the following: 1. Discuss whether we should choose a test based on equations. 2. Calculate the test statistic to test the null hypothesis given above. 3. What is the value of the test's modified degrees of freedom? 4. Determine whether to reject the null hypothesis at the 0.10 level.

1. The SD are not the same therefore the variances are not the same so we should use t test 2. plug into test statistic to get 2.099 3. plug into modified equation to get 55.61 and round to 56 4. Look at DOF of 56 and closest is 60 at .05 because two tailed value is 1.671 So reject the null hypothesis

Steps in hypothesis testing

1. state the hypothesis 2. identify the appropriate test and its probability distribution 3. specifying the significance level 4. stating decision 5. collection data and calculating the test statistic 6. making statistical decision 7. making economic or investment decision

Significance level .10 = 1.645 > 1.28 < -1.28 significance level .05 = 1.96 > 1.645 < -1.645 significance level .01 = 2.575 > 2.33 < -2.33

:)

1. Identify the appropriate test statistic or statistics for conducting the following hypothesis tests. (Clearly identify the test statistic and, if applicable, the number of degrees of freedom. For example, "We conduct the test using an x-statistic with y degrees of freedom.") A. H0: μ = 0 versus Ha: μ ≠ 0, where μ is the mean of a normally distributed population with unknown variance. The test is based on a sample of 15 observations. B. H0: μ = 0 versus Ha: μ ≠ 0, where μ is the mean of a normally distributed population with unknown variance. The test is based on a sample of 40 observations. C. H0: μ ≤ 0 versus Ha: μ > 0, where μ is the mean of a normally distributed population with known variance σ2. The sample size is 45. D. H0: σ2 = 200 versus Ha: σ2 ≠ 200, where σ2 is the variance of a normally distributed population. The sample size is 50. E. Ho: Sd1 = Sd2 and Ha: Sd1 ≠ Sd2 where they are independent random samples F. H0: (Population mean 1) − (Population mean 2) = 0 versus Ha: (Population mean 1) − (Population mean 2) ≠ 0, where the samples are drawn from normally distributed populations with unknown variances. The observations in the two samples are correlated. G. H0: (Population mean 1) − (Population mean 2) = 0 versus Ha: (Population mean 1) − (Population mean 2) ≠ 0, where the samples are drawn from normally distributed populations with unknown but assumed equal variances. The observations in the two samples (of size 25 and 30, respectively) are independent.

A. 15-1 = 14 degrees of freedom. Since it is so small, and do not know variance use a T test B. 40-1 = 39 since it does not have variance use a t test can also use a z test because has more than 30 n C. Known variance use the z test with 41 dof D. chi squared test E. F statistic because ratio of sample variances F. Samples are correlated so use a t test for paired observations G. T test

5. ANALYST A N = 101 Mean = 0.05 SD = 0.10 ANALYST B N = 121 Mean = 0.02 SD = 0.09 Investment analysts often use earnings per share (EPS) forecasts. One test of forecasting quality is the zero-mean test, which states that optimal forecasts should have a mean forecasting error of 0. (Forecasting error = Predicted value of variable − Actual value of variable.) You have collected data (shown in the table above) for two analysts who cover two different industries: Analyst A covers the telecom industry; Analyst B covers automotive parts and suppliers. a. With μ as the population mean forecasting error, formulate null and alternative hypotheses for a zero-mean test of forecasting quality. b. For Analyst A, using both a t-test and a z-test, determine whether to reject the null at the 0.05 and 0.01 levels of significance. c. For Analyst B, using both a t-test and a z-test, determine whether to reject the null at the 0.05 and 0.01 levels of significance.

A. H0: μ = 0 versus Ha: μ ≠ 0. B. T TEST At the 0.05 significance level, we reject the null if t > 1.984 or if t < −1.984. At the 0.01 significance level, we reject the null if t > 2.626 or if t < −2.626. For Analyst A, we have = 0.05/0.00995 = 5.024938 or 5.025. We clearly reject the null hypothesis at both the 0.05 and 0.01 levels. Z TEST The rejection point conditions for a two-tailed test are as follows: z > 1.96 and z < −1.96 at the 0.05 level; and z > 2.575 and z < −2.575 at the 0.01 level. 5.025 is greater than 2.575, we reject the null at the 0.01 level; we also reject the null at the 0.05 level. C. T TEST At the 0.05 significance level, we reject the null if t > 1.980 or if t < −1.980. At the 0.01 significance level, we reject the null if t > 2.617 or if t < −2.617. = 0.02/0.008182 = 2.444444 or 2.44. Because 2.44 > 1.98, we reject the null at the 0.05 level. However, 2.44 is not larger than 2.617, so we do not reject the null at the 0.01 level. Z TEST The rejection point conditions are the same as given in Part B, and we come to the same conclusions as with the t-test. Because 2.44 > 1.96, we reject the null at the 0.05 significance level; however, because 2.44 is not greater than 2.575, we do not reject the null at the 0.01 level.

6. ANALYST A N = 101 Mean = 0.05 SD - 0.10 ANALYST B N = 121 Mean = 0.02 SD - 0.09 Reviewing the EPS forecasting performance data for Analysts A and B, you want to investigate whether the larger average forecast errors of Analyst A are due to chance or to a higher underlying mean value for Analyst A. Assume that the forecast errors of both analysts are normally distributed and that the samples are independent. A. Formulate null and alternative hypotheses consistent with determining whether the population mean value of Analyst A's forecast errors (μ1) are larger than Analyst B's (μ2). B. Identify the test statistic for conducting a test of the null hypothesis formulated in Part A. C. Identify the rejection point or points for the hypothesis tested in Part A, at the 0.05 level of significance. D. Determine whether or not to reject the null hypothesis at the 0.05 level of significance.

A. H0: μ1 - μ2 ≤ 0 versus Ha: μ1 - μ2 > 0 B. n1 + n2 - 2 = 101 + 121 - 2 = 222 - 2 = 220. C. For df = 200 (the closest value to 220), the rejection point for a one-sided test at the 0.05 significance level is 1.653. D. picture

29. Which of the following should be used to test the difference between the variances of two normally distributed populations? a. t-test b. F-test c. Paired comparisons test

B

The second step in hypothesis testing is identifying the appropriate test statistic and its probability distribution.

Test statistic = [sample statistic - value of population parameter under Ho] / standard error

The third step in hypothesis testing is specifying the significance level. When the test statistic has been calculated, two actions are possible: 1) We reject the null hypothesis or 2) we do not reject the null hypothesis. The action we take is based on comparing the calculated test statistic to a specified possible value or values. The comparison values we choose are based on the level of significance selected.

The level of significance reflects how much sample evidence we require to reject the null.

The seventh and final step in hypothesis testing is making the economic or investment decision. The economic or investment decision takes into consideration not only the statistical decision but also all pertinent economic issues. In the sixth step, we found strong statistical evidence that the Canadian risk premium is positive.

The magnitude of the estimated risk premium, 5.3 percent a year, is economically very meaningful as well. Based on these considerations, an investor might decide to commit funds to Canadian equities.

For the value of the test statistic of 3.07 in the risk premium hypothesis test, using a spreadsheet function for the standard normal distribution, we calculate a p-value of 0.00107.

We can reject the null hypothesis at that level of significance. The smaller the p-value, the stronger the evidence against the null hypothesis and in favor of the alternative hypothesis. The p-value for a two-sided test that a parameter equals zero is frequently generated automatically by statistical and econometric software programs.

2. For each of the following hypothesis tests concerning the population mean, μ, state the rejection point condition or conditions for the test statistic (e.g., t > 1.25); n denotes sample size. a. H0: μ = 10 versus Ha: μ ≠ 10, using a t-test with n = 26 and α = 0.05 b. H0: μ = 10 versus Ha: μ ≠ 10, using a t-test with n = 40 and α = 0.01 c. H0: μ ≤ 10 versus Ha: μ > 10, using a t-test with n = 40 and α = 0.01 d. H0: μ ≤ 10 versus Ha: μ > 10, using a t-test with n = 21 and α = 0.05 e. H0: μ ≥ 10 versus Ha: μ < 10, using a t-test with n = 19 and α = 0.10 f. H0: μ ≥ 10 versus Ha: μ < 10, using a t-test with n = 50 and α = 0.05

a. 2.060, the rejection point for the right tail. By symmetry, −2.060 is the rejection point for the left tail. B. 2.708, the rejection point for the right tail. By symmetry, −2.708 is the rejection point for the left tail. C. 2.426, the rejection point for the right tail. D. 1.725, the rejection point for the right tail. E. With df = 18, the rejection point condition for this one-sided test is t < −1.330. Read across df = 18 to the α = 0.10 column to find 1.330, the rejection point for the right tail. By symmetry, the rejection point for the left tail is −1.330. F. With df = 49, the rejection point condition for this one-sided test is t < −1.677. Read across df = 49 to the α = 0.05 column to find 1.677, the rejection point for the right tail. By symmetry, the rejection point for the left tail is −1.677.

3. For each of the following hypothesis tests concerning the population mean, μ, state the rejection point condition or conditions for the test statistic (e.g., z > 1.25); n denotes sample size. a. H0: μ = 10 versus Ha: μ ≠ 10, using a z-test with n = 50 and α = 0.01 b. H0: μ = 10 versus Ha: μ ≠ 10, using a z-test with n = 50 and α = 0.05 c. H0: μ = 10 versus Ha: μ ≠ 10, using a z-test with n = 50 and α = 0.10 d. H0: μ ≤ 10 versus Ha: μ > 10, using a z-test with n = 50 and α = 0.05

a. 2.575 and -2.575 b. 1.96 and -1.96 C. 1.645 and -1.645 d. 1.645

4. Willco is a manufacturer in a mature cyclical industry. During the most recent industry cycle, its net income averaged $30 million per year with a standard deviation of $10 million (n = 6 observations). Management claims that Willco's performance during the most recent cycle results from new approaches and that we can dismiss profitability expectations based on its average or normalized earnings of $24 million per year in prior cycles. a. With μ as the population value of mean annual net income, formulate null and alternative hypotheses consistent with testing Willco management's claim. b. Assuming that Willco's net income is at least approximately normally distributed, identify the appropriate test statistic. c. Identify the rejection point or points at the 0.05 level of significance for the hypothesis tested in Part A. d. Determine whether or not to reject the null hypothesis at the 0.05 significance level.

a. H0: μ ≤ 24 versus Ha: μ > 24. b. the appropriate test statistic is t with n − 1 = 6 − 1 = 5 degrees of freedom. c. We will reject the null if t > 2.015. d. picture = 1.469694 so do not reject

17. You are interested in whether excess risk-adjusted return (alpha) is correlated with mutual fund expense ratios for US large-cap growth funds. The following table presents the sample. a. Formulate null and alternative hypotheses consistent with the verbal description of the research goal. b. Identify the test statistic for conducting a test of the hypotheses in Part A. c. Justify your selection in Part B. d. Determine whether or not to reject the null hypothesis at the 0.05 level of significance.

a. H0: ρ = 0 versus Ha: ρ ≠ 0 b. spearman rank

8. During a 10-year period, the standard deviation of annual returns on a portfolio you are analyzing was 15 percent a year. You want to see whether this record is sufficient evidence to support the conclusion that the portfolio's underlying variance of return was less than 400, the return variance of the portfolio's benchmark. A. Formulate null and alternative hypotheses consistent with the verbal description of your objective. B. Identify the test statistic for conducting a test of the hypotheses in Part A. C. Identify the rejection point or points at the 0.05 significance level for the hypothesis tested in Part A. D. Determine whether the null hypothesis is rejected or not rejected at the 0.05 level of significance.

a. H0: σ2 ≥ 400 versus Ha: σ2 < 400 b. 10-1 =9 and chi c. -3.325 d.

9. You are investigating whether the population variance of returns on the S&P 500/BARRA Growth Index changed subsequent to the October 1987 market crash. You gather the following data for 120 months of returns before October 1987 and for 120 months of returns after October 1987. You have specified a 0.05 level of significance. Before n 120 mean 1.416 variance 22.367 after n 120 mean 1.436 variance 15.795 A. Formulate null and alternative hypotheses consistent with the verbal description of the research goal. B. Identify the test statistic for conducting a test of the hypotheses in Part A. C. Determine whether or not to reject the null hypothesis at the 0.05 level of significance. (Use the F-tables in the back of this volume.)

a. Ho: before = after Ha: before≠ after b. chi square with 9 DOF c. The "before" sample variance is larger, so following a convention for calculating F-statistics, the "before" sample variance goes in the numerator. F = 22.367/15.795 = 1.416, with 120 − 1 = 119 numerator and denominator degrees of freedom. Because this is a two-tailed test, we use F-tables for the 0.025 level (df = 0.05/2). Using the tables in the back of the volume, the closest value to 119 is 120 degrees of freedom. At the 0.05 level, the rejection point is 1.43. Because 1.416 is not greater than 1.43, we do not reject the null hypothesis that the "before" and "after" variances are equal.

7. Let μd stand for the population mean value of difference between S&P 500 returns and small-cap stock returns. Use a significance level of 0.05 and suppose that mean differences are approximately normally distributed. a. Formulate null and alternative hypotheses consistent with testing whether any difference exists between the mean returns on the S&P 500 and small-cap stocks. b. Determine whether or not to reject the null hypothesis at the 0.05 significance level for the January 1960 to December 1999 period. c. Determine whether or not to reject the null hypothesis at the 0.05 significance level for the January 1960 to December 1979 subperiod. d. Determine whether or not to reject the null hypothesis at the 0.05 significance level for the January 1980 to December 1999 subperiod.

a. We test H0: μd = 0 versus Ha: μd ≠ 0. B. This is a paired comparisons t-test with n − 1 = 480 − 1 = 479 degrees of freedom. At the 0.05 significance level, we reject the null hypothesis if either t > 1.96 or t < −1.96. C. At the 0.05 significance level, because neither rejection point condition is met, we do not reject the null hypothesis

Probability of a Type 1 error

alpha, is the level of significance of the test

Paired observations

are observations that are dependent because they have something in common

A hypothesis

asks "is the value of the parameter (say the mean) 45?" It is a statement about one or more populations

this test is

asymmetrical

15. A Type II error is best described as: a. rejecting a true null hypothesis. b. failing to reject a false null hypothesis. c. failing to reject a false alternative hypothesis.

b

16. The level of significance of a hypothesis test is best used to: a. calculate the test statistic. b. define the test's rejection points. c. specify the probability of a Type II error.

b

18. All else equal, is specifying a smaller significance level in a hypothesis test likely to increase the probability of a: Type I error? Type II error? A No No B No Yes C Yes No

b

19. The probability of correctly rejecting the null hypothesis is the: a. p-value. b. power of a test. c. level of significance

b

23. Which of the following tests of a hypothesis concerning the population mean is most appropriate? a. A z-test if the population variance is unknown and the sample is small b. A z-test if the population is normally distributed with a known variance c. A t-test if the population is non-normally distributed with unknown variance and a small sample

b

25. For a small sample from a normally distributed population with unknown variance, the most appropriate test statistic for the mean is the: a. z-statistic. b. t-statistic. c. χ2 statistic.

b

26. A pooled estimator is used when testing a hypothesis concerning the: a. equality of the variances of two normally distributed populations. b. difference between the means of two at least approximately normally distributed populations with unknown but assumed equal variances. c. difference between the means of two at least approximately normally distributed populations with unknown and assumed unequal variances.

b

27. When evaluating mean differences between two dependent samples, the most appropriate test is a: a. chi-square test. b. paired comparisons test. c. z-test.

b

12. Which of the following statements regarding a one-tailed hypothesis test is correct? a. The rejection region increases in size as the level of significance becomes smaller. b. A one-tailed test more strongly reflects the beliefs of the researcher than a two-tailed test. c. The absolute value of the rejection point is larger than that of a two-tailed test at the same level of significance

b.

20. When making a decision in investments involving a statistically significant result, the: a. economic result should be presumed meaningful. b. statistical result should take priority over economic considerations. c. economic logic for the future relevance of the result should be further explored

c

21. Which of the following statements is correct with respect to the p-value? a. It is a less precise measure of test evidence than rejection points. b. It is the largest level of significance at which the null hypothesis is rejected. c. It can be compared directly with the level of significance in reaching test conclusions.

c

22. Which of the following represents a correct statement about the p-value? a. The p-value offers less precise information than does the rejection points approach. b. A larger p-value provides stronger evidence in support of the alternative hypothesis. c. A p-value less than the specified level of significance leads to rejection of the null hypothesis.

c