CFA Quants Questions

*** Got it from CFA Mock Exam: [I most likely drop what i got wrong] A bank account has a stated annual interest rate of 3.5% with quarterly compounding. If the current value of the account is $100,000, the future value of the account two years from now is closest to: $107,123. $107,207. $107,218.

$107,218. [Note: Go and review "continuous compounding' C is correct because FVN = PV(1 + rs/m)mN, where m = number of compounding periods per year = 4, rs = quoted annual interest rate = 3.5%, N = number of years = 2. Thus, FV2 = $100,000(1 + 0.035/4)4(2) = $107,218.17 ≈ $107,218. or PV = 100K, PMT = 0, N = 2*4, I/Y= 3.5/4, CPT FV = 107218... .

*************** Given the following timeline and a discount rate of 4% a year compounded annually, the present value (PV), as of the end of Year 5 (PV5 ), of the cash flow received at the end of Year 20 is closest to: $22,819. $27,763. $28,873.

$27,763. Since it's compounded annually, I didn't really do much on the converting I/Y FV = 50000 I/Y = 4 N=15 PMT = 0 CPT PV = 27763

7. A couple plans to set aside $20,000 per year in a conservative portfolio projected to earn 7 percent a year. If they make their first savings contribution one year from now, how much will they have at the end of 20 years?

(I thought there was a catch. You simply find FC using traditional way) PMT = 20000 N = 20 I/Y = 7 CPT FV = 819,909.85

***14. You are considering investing in two different instruments. The first instrument will pay nothing for three years, but then it will pay $20,000 per year for four years. The second instrument will pay $20,000 for three years and $30,000 in the fourth year. All payments are made at year-end. If your required rate of return on these investments is 8 percent annually, what should you be willing to pay for: A. The first instrument? B. The second instrument (use the formula for a four-year annuity)?

(skip the first 3) Step 1 PMT = 200000, N = 4, I/Y = 8, CPT PV = 66242.54 Step 2 66242.52 / (1.08)^3 = 52, 585.46 Second instrument: (the textbook answer say you separate them) Step 1 PMT = 200000, N = 4, I/Y = 8, CPT PV = 66242.54 Step 2: Since we seperate the 4th year with 10,000 and 20,000. We needt oadd in the last 10,000. => 10,000 / (1.08^4) = 7350.3 66242.54 + 7350 = 73592.84

**** The price of a stock at t = 0 is $208.25 and at t = 1 is $186.75. The continuously compounded rate of return for the stock from t = 0 to t = 1 is closest to: -10.90%. -10.32%. 11.51%.

-10.90%. The continuously compounded return from t = 0 to t = 1 is r0,1 = ln(S1/S0) = ln(186.75/208.25) = -0.10897 = -10.90%.

** A random number between zero and one is generated according to a continuous uniform distribution. What is the probability that the first number generated will have a value of exactly 0.30 0% 30% 70%

0% A is correct. The probability of generating a random number equal to any fixed point under a continuous uniform distribution is zero.

In a discrete uniform distribution with 20 potential outcomes of integers 1-20, the probability that X is greater than or equal to 3 but less than 6, P(3 ≤ X < 6), is: 0.10. 0.15. 0.20.

0.15 (6-3) / (20-1) = 15.7%

*** The cumulative distribution function for a discrete random variable is shown in the following table. X = x. Cumulative Distribution Function. F(x) = P(X ≤ x) 1 .15 2 .25 3 .50 4 .60 5 .95 6 1.00 The probability that X will take on a value of either 2 or 4 is closest to: 0.20. 0.35. 0.85.

0.20. P(2) = 0.25 - 0.15 = 0.1 p(4) = 0.6 - 0.5 = 0.1 0.1 + 0.1 = 0.2

************** A portfolio manager annually outperforms her benchmark 60% of the time. Assuming independent annual trials, what is the probability that she will outperform her benchmark four or more times over the next five years? 0.26 0.34 0.48 (I got it wrong)

0.34 My mistake was: i didn't include the 5th time nCr*P^x*(1-P)^n-x 1) 5C4 * 0.6^4*(0.4)^1 = 0.26 2) 5C5 * 0.6^5*(0.4)^0 = 0.077 3) 0.26 + 0.77 = 0.337

variable is normally distributed with a mean of 5.00 and a variance of 4.00. Using the excerpt above from the cumulative distribution function for the standard normal random variable table, the probability of observing a value of −0.40 or less is closest to: 2.44%. 8.85%. 0.35%.

0.35%. = (-0.4 - 5) /(4)^1/2 = -2.7, or 99.65% CI 1 - 0.9965 = 0.0035

*** Got it from CFA Mock Exam: [I most likely drop what i got wrong] A portfolio manager gathers the following information about her fund: Allocation of corporate bonds. 20% Allocation of bonds that are both corporate and high-yield. 10% If the portfolio manager randomly picks a corporate bond, the probability that the bond is high yield is closest to: 0.02. 0.10. 0.50. [I'm ashamed that I got this type of question wrong]

0.50 [Now I'm looking at it, it's easy now. You're mistake is you don't know the names of these probability methods and you're confusing yourself] C is correct because using the definition of conditional probability P(A | B) = P(AB)/P(B), where P(AB) is the joint probability of A and B and P(B) is the probability of B, then the conditional probability that the portfolio manager randomly picks a high-yield bond, given that it is a corporate bond is equal to P(bond is high yield | corporate bond) = P(bond is high-yield and corporate bond)/P(corporate bond) = 0.1/0.2 = 0.5.

************ A stock is priced at $100.00 and follows a one-period binomial process with an up move that equals 1.05 and a down move that equals 0.97. If 1 million Bernoulli trials are conducted and the average terminal stock price is $102.00, the probability of an up move (p) is closest to: 0.375. 0.500. 0.625.

0.625. I just used hedge ratio: n = C+ - C- / S+ - S- (105 - 100 - 0) / (105 - 97) = 0.625

************** If the probability that a portfolio outperforms its benchmark in any quarter is 0.75, the probability that the portfolio outperforms its benchmark in three or fewer quarters over the course of a year is closest to: 0.26 0.42 0.68

0.68 Same as prior: nCr*P^x*(1-P)^n-x 4C3*0.75^3*(1-0.75)^1 = 0.42. 4C2*0.75^2*(1-0.75)^2 = 0.2 4C1*0.75^1*(1-0.75)^3 = 0.06 4C0*0.75^0*(1-0.75)^4 = 0.004 Sum = 0.68

(This is from Wiley) An investor uses a stock screener to identify stocks on which to perform a more in-depth analysis. The first criterion limited the result to dividend-paying stocks, which eliminated 35% of the stocks in the investment universe. The second criterion looked for stocks that trade at less than three times book value, leaving only 0.50% of the total universe that passed both criteria. The probability that a stock from this investment universe trades at less than three times book value, given that it pays a dividend is closest to: 0.77%. 15.0%. 43.0%.

0.77%. 0.005 / (1 - 0.35) = 0.769

(Check the graph) The probability of a profit greater than or equal to $1 and less than or equal to $4 is closest to: 0.4. 0.6. 0.8. (I made a silly mistake)

0.8. 0.2 * 4 = 0.8

*** Year 1. 19.5 Year 2. −1.9 Year 3. 19.7 Year 4. 35.0 Year 5. 5.7 The geometric mean return for Fund Y is closest to: 14.9%. 15.6%. 19.5%. great reminder

14.9%. Fund Y = [(1 + 0.195) × (1 − 0.019) × (1 + 0.197) × (1 + 0.350) × (1 + 0.057)]^(1/5) − 1 = 14.9%.

A sample of 25 observations has a mean of 8 and a standard deviation of 15. The standard error of the sample mean is closest to: 1.60. 3.00. 3.06.

15/√ 25=3.00

Roy's Safety-First Ratio: Given Portfolio 1: 0.35, Portfolio 2: 0.64. An investor wants to maximize the possibility of earning at least 5% on her investments each year. Given portfolio 3: Expected return: 22% and Standard deviation: 40%. Using Roy's safety-first criterion, the most appropriate choice for the investor is portfolio: 3. 2. 1.

2 P3: (0.22 - 0.05)/ 0.4 = 0.425

Year. Return 1 4.5% 2 6.0% 3 1.5% 4 −2.0% 5 0.0% 6 4.5% 7 3.5% 8 2.5% 9 5.5% 10 4.0% he harmonic mean return over the 10 years is closest to: 2.94%. 2.97%. 3.00%. The geometric mean return over the 10 years is closest to: A.2.94%. B.2.97%. C.3.00%. *** The target semideviation of the returns over the 10 years if the target is 2% is closest to: 1.42%. 1.50%. 2.01%.

2.94%. You do: = 10 / (1/1.045 + 1/1.06 .. 1/0.98 + 1 + 1/1.045.....) = 10 / 9.7141 = 1.0294 --- B.2.97%. --- we only need the 1.5%, -2%, 0% becasue they are under 2%. (2%-1.5%)^2 = 0.000025 (2%-(-2%))^2 = 0.0016 (0%-2%)^2= 0.0004 Sum = 0.002025 √(0.002025 / 9) = 1.5%.

************** For a sample size of 65 with a mean of 31 taken from a normally distributed population with a variance of 529, a 99% confidence interval for the population mean will have a lower limit closest to: 23.64. 25.41. 30.09.

23.64. estimator +- factor * SE 1) SE = (529)^1/2 / (65)^1/2 = 2.852 2) 31 + 2.57 *2.852 = 31 +- 7.332 3) 23.67

**** An analyst determines that 60% of all US pension funds hold hedge funds. In evaluating this probability, a random sample of 10 US pension funds is taken. Using the binomial probability function, the probability that exactly 6 of the 10 firms in the sample hold hedge funds is closest to: 60.0%. 11.2%. 25.1%.

25.1%. 10C6*0.6^5*(1 - 0.6)^4= 25.08%

The following exhibit shows the annual MSCI World Index total returns for a 10-year period. Year 1 15.25% Year 6 30.79% Year 2 10.02% Year 7 12.34% Year 3. 20.65% Year 8 −5.02% Year 4 9.57% Year 9 16.54% Year 5 −.40.33%. Year 10 27.37% The fourth quintile return for the MSCI World Index is closest to: 20.65%. 26.03%. 27.37% For Year 6-Year 10, the mean absolute deviation of the MSCI World Index total returns is closest to: 10.20%. 12.74%. 16.40%.

26.03%. B is correct. Quintiles divide a distribution into fifths, with the fourth quintile occurring at the point at which 80% of the observations lie below it. The fourth quintile is equivalent to the 80th percentile. To find the yth percentile (Py), we first must determine its location. The formula for the location (Ly) of a yth percentile in an array with n entries sorted in ascending order is Ly = (n + 1) × (y/100). In this case, n = 10 and y = 80%, soL80 = (10 + 1) × (80/100) = 11 × 0.8 = 8.8. With the data arranged in ascending order (−40.33%, −5.02%, 9.57%, 10.02%, 12.34%, 15.25%, 16.54%, 20.65%, 27.37%, and 30.79%), the 8.8th position would be between the 8th and 9th entries, 20.65% and 27.37%, respectively. Using linear interpolation, P80 = X8 + (Ly − 8) × (X9 − X8), P80 = 20.65 + (8.8 − 8) × (27.37 − 20.65) = 20.65 + (0.8 × 6.72) = 20.65 + 5.38 = 26.03%. --- Step 1) Sum annual returns and divide by n to find the arithmetic mean of 16.40%. Step 2) Calculate the absolute value of the difference between each year's return and the mean from Column 1. Sum the results and divide by n to find the MAD. (14.39% + 4.06% + 21.42% + 0.14% + 10.97%) / 5 = 10.2%

(This is from Wiley) A reporting service indicates that the following historical performance occurred: Monthly Data: January 2002 to December 2005 Mean Return (%) Standard Deviation (%) Sharpe Ratio Small-cap value funds 1.25 2.34 0.410 An analyst wants to determine the probability that small-cap value funds will earn a monthly return lower than the monthly risk-free rate of 0.29%. He is basing his estimate of future-period monthly return parameters on the sample mean and standard deviation provided in the table above and has assumed that the fund returns represent a normal distribution. Excerpts from the table of the cumulative standard distribution function for the standard normal probability is included below: (It was given a table) The probability that the small-cap value fund return is below the risk-free rate of return is closest to: 34%. 41%. 66%.

34% The first step is to find the Z value for the 0.29% return: Z=X−μ / σ = 0.29% − 1.25% / 2.34% = −0.41 Using the table provided, the area that corresponds to the Z value of −0.41 is 34.09%. This means that there is a 34% chance that the monthly return will be below the risk-free rate of 0.29%.

The initial value of the stock is $80. The probability of an up move in any given period is 75%, and the probability of a down move in any given period is 25%. Using the binomial model, the probability that the stock's price will be $79.20 at the end of two periods is closest to: A. 37.50%. B. 56.25%. C. 18.75%.

37.50%. 2C1*0.75*0.25= 37.5%

*** A portfolio has an expected return of 7%, with a standard deviation of 13%. For an investor with a minimum annual return target of 4%, the probability that the portfolio return will fail to meet the target is closest to: 33%. 41%. 59%.

41%. B is correct. By using Excel's NORM.S.DIST() function, we get NORM.S.DIST((4% - 7%)/13%) = 40.87%. The probability that the portfolio will underperform the target is about 41%.

************* An analyst determines that approximately 99% of the observations of daily sales for a company are within the interval from $230,000 to $480,000 and that daily sales for the company are normally distributed. If approximately 99% of all the observations fall in the interval μ ± 3σ, then using the approximate z-value rather than the precise table, the standard deviation of daily sales for the company is closest to: $41,667. $62,500. $83,333.

41,667. 1) Mean = ($230,000 + $480,000)/2 = $355,000 2) Z = (X − μ)/σ, ($480,000 − $355,000)/3.0 = $41,667. Z value is 3 because is given by μ ± 3σ [More explanation on CFA question]

(I didn't bring the graph in here) The initial value of the stock is $100. The probability of an up move in any given period is 40%, and the probability of a down move in any given period is 60%. Using the binomial model, the probability that the stock's price will be $101.20 at the end of two periods is closest to: 48%. 24%. 16%. (Kinda of trick question)

48%. 2C1*(0.4)^1*(1 - 0.4)^1 = 0.48 (I thought it has something to do with binomial tree or hedge ratio)

A hypothesis test for a normally distributed population at a 0.05 significance level implies a: 95% probability of rejecting a true null hypothesis. 95% probability of a Type I error for a two-tailed test. 5% critical value rejection region in a tail of the distribution for a one-tailed test. [I got it wrong]

5% critical value rejection region in a tail of the distribution for a one-tailed test.

*** The total number of parameters that fully characterizes a multivariate normal distribution for the returns on two stocks is: 3. 4. 5.

5. C is correct. A bivariate normal distribution (two stocks) will have two means, two variances, and one correlation. A multivariate normal distribution for the returns on n stocks will have n means, n variances, and n(n - 1)/2 distinct correlations.

(This is from Wiley) ***** A trader has two limit orders in place for each of two stocks, Iridium (IRDM) and Verizon (VZ). She estimates that the probability of the IRDM trade executing before the end of the day is 0.50. However, there is only a 15% chance that the VZ trade will execute over the same period. If the probability of both trades executing before the close is 0.09, what is the probability that either one or the other will execute? 65% 56% 33%

56% The union probability is found using the addition rule of probability: P(A or B) = P(A) + P(B) − P(AB) = 0.50 + 0.15 − 0.09 = 0.56 = 56%

**************** The weekly closing prices of Mordice Corporation shares are as follows: DateClosing Price (€) 1 August 112 8 August 160 15 August 120 The continuously compounded return of Mordice Corporation shares for the period August 1 to August 15 is closest to: 6.90%. 7.14%. 8.95%.

6.90%. Using my own note: We're applying Logarithmic steps: 1) Find HPR: (V of End - V beg) / V of End 2) (1 + hpr)^1/n -1 = EAR 3) In (1 + EAR) -1 = continuously compounded return

*** A portfolio has an expected mean return of 8% and standard deviation of 14%. The probability that its return falls between 8% and 11% is closest to: 8.5%. 14.8%. 58.3%.

8.5%. For the first term, NORM.S.DIST((11% - 8%)/14%) = 58.48%. To get the second term immediately, note that 8% is the mean, and for the normal distribution, 50% of the probability lies on either side of the mean. Therefore, N(Z corresponding to 8%) must equal 50%. So, P(8% ≤ Portfolio return ≤ 11%) = 0.5848 - 0.50 = 0.0848, or approximately 8.5%.

(This is from Wiley) ***** Assume the corporate bond spread over Treasuries is normally distributed with a population mean of 2.0% and a population standard deviation of 8.0%. Over the past six years, the corporate bond spread over Treasuries has averaged 5.0%. An analyst considers this to be highly unusual and wants to determine whether the recent result is likely 90% of the time, given the population's statistics. The confidence interval around the population mean that the analyst will calculate is closest to: −4.4% to 8.4%. −3.4% to 7.4%. −0.2% to 4.2%. (A very great question)

90% confidence level has a critical Z-value of 1.65, with the Z-value being used because the population standard deviation is known. Given the population data and the six-year period, the confidence interval is calculated as follows: => μ +- zα/2* σ/√n = 2.0 − +1.65 *8.0/√6= −3.39%to7.39% μ +- Reliability factor (Z or T) * SE = CI My mistake was I was confused with the distractor information: 5%.

A variable is normally distributed with a mean of 2.00 and a variance of 16.00. Using the excerpt, the probability of observing a value of 7.40 or less is closest to: 63.3%. 91.2%. 96.8%

91.2%. finding Z score and look for CI. You don't even need to do anything else. We go from right to left and it's only asking for 7.4 or less. Thus, we also include left tail

(This is from Wiley) A sample size of 40 is drawn from a normally distributed population. Which of the following is the appropriate test and degrees of freedom to determine whether the sample exhibits a specific variance? A t-statistic with 39 degrees of freedom A chi-square (χ2) with 38 degrees of freedom A chi-square (χ2) with 39 degrees of freedom

A chi-square (χ2) with 39 degrees of freedom When testing the variance of a single, normally distributed population, the appropriate test statistic is chi-square (χ2) and the test has n − 1 = 40 − 1 = 39 degrees of freedom.

*** Got it from CFA Mock Exam: [I most likely drop what i got wrong] A discrete random variable X has the following probability distribution: Probability. Outcome 0.20 35 0.30 50 0.50 80 The standard deviation of X is closest to: 18.73. 20.00. 22.91. [I'm ashamed that I got this type of question wrong] [A good question for future similar questions]

A is correct because the Step 1) expected value E(X) = Σi=1nP(Xi)Xi = (0.20 × 35) + (0.30 × 50) + (0.50 × 80) = 62. Step 2) The variance σ2(X) = E{[X - E(X)]2} = Σi=1nP(Xi)[X - E(X)]2 = 0.20 × (35 - 62)^2 + 0.30 × (50 - 62)^2 + 0.50 × (80 - 62)^2 = 351. "Standard deviation is the positive square root of variance:" σ = 351^ 1/2 ≈ 18.73.

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

A one-tailed test more strongly reflects the beliefs of the researcher than a two-tailed test. B is correct. One-tailed tests in which the alternative is "greater than" or "less than" represent the beliefs of the researcher more firmly than a "not equal to" alternative hypothesis. The above is the right anwer. From Google: "... It is about being sure about saying something is statistically different from the mean. While the two-tailed test just tries to prove that the hypothesized mean is statistically different (just different, doesn't take a stance) the one-tailed test takes a stance and tries to prove that the hypothesized mean is greater/lesser than the mean. Taking a specific direction is more representative of a firmer research belief..." --- It's not A because if the level of significance goes down, the failed-to-reject region increases.

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

A p-value less than the specified level of significance leads to rejection of the null hypothesis. C is correct. The p-value is the smallest level of significance at which the null hypothesis can be rejected for a given value of the test statistic. The null hypothesis is rejected when the p-value is less than the specified significance level.

Which of the following assets most likely requires the use of a multivariate distribution for modeling returns? A call option on a bond A portfolio of technology stocks A stock in a market index

A portfolio of technology stocks B is correct. Multivariate distributions specify the probabilities for a group of related random variables. A portfolio of technology stocks represents a group of related assets. Accordingly, statistical interrelationships must be considered, resulting in the need to use a multivariate normal distribution.

Sector. A. AA. AAA Communication Services. 25. 32. 27 Consumer Staples. 30. 25. 25 Energy. 100. 85. 30 Health Care. 200. 100. 63 Utilities. 22. 28 14 The relative frequency of AA rated energy bonds, based on the total count, is closest to: A.10.5%. B.31.5%. C.39.5%. [I actually got wrong, it's easy!] --- The marginal frequency of energy sector bonds is closest to: A.27. B.85. C.215.

A.10.5%. Total = 806 85/806 = 0.105 My mistake was I did 85 / 270. (270 is the total of AA only) --- C.215.

Published ratings on stocks ranging from 1 (strong sell) to 5 (strong buy) are examples of which measurement scale? A.Ordinal B.Continuous C.Nominal --- Data values that are categorical and not amenable to being organized in a logical order are most likely to be characterized as: A.ordinal data. B.discrete data. C.nominal data. --- A fixed-income analyst uses a proprietary model to estimate bankruptcy probabilities for a group of firms. The model generates probabilities that can take any value between 0 and 1. The resulting set of estimated probabilities would most likely be characterized as: ordinal data. discrete data. continuous data. --- * An analyst uses a software program to analyze unstructured data—specifically, management's earnings call transcript for one of the companies in her research coverage. The program scans the words in each sentence of the transcript and then classifies the sentences as having negative, neutral, or positive sentiment. The resulting set of sentiment data would most likely be characterized as: ordinal data. discrete data. nominal data.

A.Ordinal --- C.nominal data. --- continuous data. --- ordinal data. A is correct. Ordinal data are categorical values that can be logically ordered or ranked. In this case, the classification of sentences in the earnings call transcript into three categories (negative, neutral, or positive) describes ordinal data, as the data can be logically ordered from positive to negative. B is incorrect because discrete data are numerical values that result from a counting process. In this case, the analyst is categorizing sentences (i.e., unstructured data) from the earnings call transcript as having negative, neutral, or positive sentiment. Thus, these categorical data do not represent discrete data. C is incorrect because nominal data are categorical values that are not amenable to being organized in a logical order. In this case, the classification of unstructured data (i.e., sentences from the earnings call transcript) into three categories (negative, neutral, or positive) describes ordinal (not nominal) data, as the data can be logically ordered from positive to negative.

*** Which of the following statements is correct with respect to the null hypothesis? It can be stated as "not equal to" provided the alternative hypothesis is stated as "equal to." Along with the alternative hypothesis, it considers all possible values of the population parameter. In a two-tailed test, it is rejected when evidence supports equality between the hypothesized value and the population parameter. [CFA answer is wrong. For the past me who was extremely confused, you don't mind this question because I know you will.]

Along with the alternative hypothesis, it considers all possible values of the population parameter. This should be the correct answer! The null hypothesis and the alternative hypothesis are complements of one another and together are exhaustive; that is, the null and alternative hypotheses combined consider all the possible values of the population parameter.

4. The value in six years of $75,000 invested today at a stated annual interest rate of 7% compounded quarterly is closest to: A. $112,555. B. $113,330. C. $113,733.

Ans: . $113,733. -> PV =75k N = 6*4 I/Y= 7/4 CPT FV = 113733.2

Assume that monthly returns are normally distributed with a mean of 1% and a sample standard deviation of 4%. The population standard deviation is unknown. Construct a 95% confidence interval for the sample mean of monthly returns if the sample size is 24.

Answer: - 0.689%, - 2.168% 1) SE = 4%/(24^1/2) = 0.00816 2) 1%+- (reliability factor*0.00816) 3) reliability factor: t -> (24-1)= 23 => 2.069 1% +- (2.68%

Mean plus or minus one standard deviation. ? Mean plus or minus two standard deviations. ? Mean plus or minus three standard deviations. ?

Approximately 68% Approximately 95% Approximately 99%

For a sample size of 17, with a mean of 116.23 and a variance of 245.55, the width of a 90% confidence interval using the appropriate t-distribution is closest to: 13.23. 13.27. 13.68.

B is correct. The confidence interval is calculated using the following equation: X ± t * SE Sample standard deviation (s) = √245.55 = 15.670. SE = 15.670 / √17 = 3.80 For a sample size of 17, degrees of freedom equal 16, so t0.05 = 1.746. The confidence interval is calculated as 116.23±6.6357. or 116.23± (3.8 * 1.746) Therefore, the interval spans 109.5943 to 122.8656, meaning its width is equal to approximately 13.271. (This interval can be alternatively calculated as 6.6357 × 2.)

*** 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. No Yes B is correct. Specifying a smaller significance level decreases the probability of a Type I error (rejecting a true null hypothesis) but increases the probability of a Type II error (not rejecting a false null hypothesis). As the level of significance decreases, the null hypothesis is less frequently rejected.

(This is from Wiley) ***** A company is known for engaging in numerous restructurings, which involve taking "one-time" charges. An analyst determines that in the past 10 years, the company has restructured four times and not restructured six times. Assuming this pattern will continue in the future, the analyst wants to determine the probability that the company will restructure exactly twice in the next 10 years. Based on the data provided, the analyst should conclude that this probability is closest to: 1%. 12%. 20%.

Based on the past, the probability of a restructuring in any given year is 40% (4 out of 10). The following calculation shows the determination of the probability of exactly two restructurings in the next 10 years: P(X=x)=n!(n−x)!x! * p^x* (1−p)^n−x = 10!/(10−2)!2!* 0.40^2 * (1−0.40)^ 10−2 = 12%

************ An analyst develops the following capital market projections. Stocks. Bonds Mean Return 10% 2% Standard Deviation 15% 5% Assuming the returns of the asset classes are described by normal distributions, which of the following statements is correct? Bonds have a higher probability of a negative return than stocks. On average, 99% of stock returns will fall within two standard deviations of the mean. The probability of a bond return less than or equal to 3% is determined using a Z-score of 0.25.

Bonds have a higher probability of a negative return than stocks. A is correct. The chance of a negative return falls in the area to the left of 0% under a standard normal curve. By standardizing the returns and standard deviations of the two assets, the likelihood of either asset experiencing a negative return may be determined: Z-score (standardized value) = (X - μ)/σ. Z-score for a bond return of 0% = (0 - 2)/5 = -0.40. Z-score for a stock return of 0% = (0 - 10)/15 = -0.67. For bonds, a 0% return falls 0.40 standard deviations below the mean return of 2%. In contrast, for stocks, a 0% return falls 0.67 standard deviations below the mean return of 10%. A standard deviation of 0.40 is less than a standard deviation of 0.67. Negative returns thus occupy more of the left tail of the bond distribution than the stock distribution. Thus, bonds are more likely than stocks to experience a negative return. [Don't quite understand the explanation]

*** Homoskedasticity is best described as the situation in which the variance of the residuals of a regression is: A.zero. B.normally distributed. C.constant across observations.

C.constant across observations.

*** When evaluating mean differences between two dependent samples, the most appropriate test is a: A.z-test. B.chi-square test. C.paired comparisons test.

C.paired comparisons test. . A paired comparisons test is appropriate to test the mean differences of two samples believed to be dependent.

(This is from Wiley) Assume that quarterly returns are normally distributed and that the population mean and standard deviation are unknown. A random sample of 17 quarterly returns had a mean of 1.5% and a standard deviation of 6.5%. Table 2 below is an excerpt from Student's t-distribution. Table 2: Student's t-Distribution (One-Tailed Probabilities). [It was given a table] The 95% confidence interval for the population mean of quarterly returns is closest to −1.25% to 4.25%. −1.84% to 4.84%. −1.94% to 4.94%.

CI=¯X ± tα/2 * s√n = 1.5%±2.120*(6.5%/√17) = −1.842; 4.842

10. To cover the first year's total college tuition payments for his two children, a father will make a $75,000 payment five years from now. How much will he need to invest today to meet his first tuition goal if the investment earns 6 percent annually?

FV = 75000, N = 5, I/Y=6, CPT PV= 56044.3

*** A sample mean is computed from a population with a variance of 2.45. The sample size is 40. The standard error of the sample mean is closest to: 0.039. 0.247. 0.387.

Formula for the standard error of the sample mean (σX), based on a known sample size (n), is =σ/√n √2.45 = 1.565 1.565 /√40 =0.247

*** Got it from CFA Mock Exam: [I most likely drop what i got wrong] An individual can invest $19,000 today and receive $20,000 in one year's time. If her required rate of return is 5%, the rate of return on the investment is: less than the required rate of return. equal to the required rate of return. Greater than the required rate of return. [Drop it for practice, and for the rest which didn't mark as I got it wrong]

Greater than the required rate of return. I did: PV = 19000, FV= 20000, N = 1, PMT = 0, CPT I/Y = 5.26 C is correct because the return on the investment can be calculated as (20,000 - 19,000)/19,000 = 1,000/19,000 = 5.26%. Therefore, it exceeds the required return of 5%.

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

It can be compared directly with the level of significance in reaching test conclusions. C is correct. When directly comparing the p-value with the level of significance, it can be used as an alternative to using rejection points to reach conclusions on hypothesis tests. If the p-value is smaller than the specified level of significance, the null hypothesis is rejected. Otherwise, the null hypothesis is not rejected.

Which one of the following statements about Student's t-distribution is false? It is symmetrically distributed around its mean value, like the normal distribution. It has shorter (i.e., thinner) tails than the normal distribution. As its degrees of freedom increase, Student's t-distribution approaches the normal distribution.

It has shorter (i.e., thinner) tails than the normal distribution. (CFA answer is wrong) B is correct. Student's t-distribution has longer (fatter) tails than the normal distribution and, therefore, it may provide a more reliable, more conservative downside risk estimate.

**************** A population has a non-normal distribution with mean µ and variance σ2. The sampling distribution of the sample mean computed from samples of large size from that population will have: the same distribution as the population distribution. its mean approximately equal to the population mean. its variance approximately equal to the population variance.

Its mean approximately equal to the population mean. B is correct. Given a population described by any probability distribution (normal or non-normal) with finite variance, the central limit theorem states that the sampling distribution of the sample mean will be approximately normal, with the mean approximately equal to the population mean, when the sample size is large.

Which of the following is characteristic of the normal distribution? Asymmetry Kurtosis of 3 Definitive limits or boundaries

Kurtosis of 3

*************** A couple plans to set aside $20,000 per year in a conservative portfolio projected to earn 7 percent a year. If they make their first savings contribution one year from now, how much will they have at the end of 20 years? (I got confused with the timeline)

N = 20 1/Y = 7 PMT = 20K PV = 0 CPT FV = 819,909.85

*** An analyst is examining a large sample with an unknown population variance. Which of the following is the most appropriate test to test the hypothesis that the historical average return on an index is less than or equal to 6%? One-sided t-test Two-sided t-test One-sided chi-square test [ A great explanation here to solve your Z and T-test on unknown variacne]

One-sided t-test A is correct. If the population sampled has unknown variance and the sample is large, a z-test may be used. Hypotheses involving "greater than" or "less than" postulations are one sided (one tailed). In this situation, the null and alternative hypotheses are stated as H0: μ ≤ 6% and Ha: μ > 6%, respectively. A one-tailed t-test is also acceptable in this case, and the rejection region is on the right side of the distribution.

(This is from Wiley) ***** A merger arbitrage hedge fund manager is considering using a scoring model to test the likelihood that a merger deal will close within six months. Back-testing the model produced the following results: P(Closed on time) = 0.70 P(Passed test) = 0.76 P(Passed test | Closed on time) = 0.82 The manager uses the model to score a new deal. If the deal fails the scoring test, what is the updated probability that it will not close within six months? P(no close | failed) = 0.245 P(no close | failed) = 0.475 P(no close | failed) = 0.705 (Reminding future self to add more this type of question here!!)

P(no close | failed) = 0.475 My own calculation:

. A client can choose between receiving 10 annual $100,000 retirement payments, starting one year from today, or receiving a lump sum today. Knowing that he can invest at a rate of 5 percent annually, he has decided to take the lump sum. What lump sum today will be equivalent to the future annual payments?

PMT = 10000, N = 10, I/Y = 5 CPT PV = 772173.4

***16. An investment pays €300 annually for five years, with the first payment occurring today. The present value (PV) of the investment discounted at a 4% annual rate is closest to: A. €1,336. B. €1,389. C. €1,625. 17. At a 5% interest rate per year compounded annually, the present value (PV) of a 10-year ordinary annuity with annual payments of $2,000 is $15,443.47. The PV of a 10-year annuity due with the same interest rate and payments is closest to: A. $14,708. B. $16,216. C. $17,443

PMT = 300, N = 5, I/Y = 4, CPT PV = 1335.5 1335.5 * (1.04) = 1388.97 => annuity due 17. $15,443.47 * 1.05 = 16215.64

* A bar chart that orders categories by frequency in descending order and includes a line displaying cumulative relative frequency is referred to as a: Pareto Chart. grouped bar chart. frequency polygon. --- * Which visualization tool works best to represent unstructured, textual data? Tree-Map Scatter plot Word cloud --- * A tree-map is best suited to illustrate: underlying trends over time. joint variations in two variables. value differences of categorical groups. --- * A line chart with two variables—for example, revenues and earnings per share—is best suited for visualizing: the joint variation in the variables. underlying trends in the variables over time. the degree of correlation between the variables. --- * A heat map is best suited for visualizing the: frequency of textual data. degree of correlation between different variables. shape, center, and spread of the distribution of numerical data. --- * Which valuation tool is recommended to be used if the goal is to make comparisons of three or more variables over time? Heat map Bubble line chart Scatter plot matrix

Pareto Chart. A is correct. A bar chart that orders categories by frequency in descending order and includes a line displaying cumulative relative frequency is called a Pareto Chart. A Pareto Chart is used to highlight dominant categories or the most important groups. B is incorrect because a grouped bar chart or clustered bar chart is used to present the frequency distribution of two categorical variables. C is incorrect because a frequency polygon is used to display frequency distributions. --- Word cloud C is correct. A word cloud, or tag cloud, is a visual device for representing unstructured, textual data. It consists of words extracted from text with the size of each word being proportional to the frequency with which it appears in the given text. --- value differences of categorical groups. C is correct. A tree-map is a graphical tool used to display and compare categorical data. --- underlying trends in the variables over time. --- degree of correlation between different variables. B is correct. A heat map is commonly used for visualizing the degree of correlation between different variables. --- B is correct. A bubble line chart is a version of a line chart where data points are replaced with varying-sized bubbles to represent a third dimension of the data. A line chart is very effective at visualizing trends in three or more variables over time. A is incorrect because a heat map differentiates high values from low values and reflects the correlation between variables but does not help in making comparisons of variables over time. C is incorrect because a scatterplot matrix is a useful tool for organizing scatterplots between pairs of variables, making it easy to inspect all pairwise relationships in one combined visual. However, it does not help in making comparisons of these variables over time.

*** Got it from CFA Mock Exam: [I most likely drop what i got wrong] An analyst gathers the following information about three portfolios: Portfolio X. Portfolio Y. Portfolio Z Expected annual return 10% 12% 14% Standard deviation of returns 12% 16% 20% If the shortfall level is 5%, which portfolio has the largest safety-first ratio? Portfolio X Portfolio Y Portfolio Z

Portfolio Z Portfolio X: (0.10 - 0.05)/0.12 = 0.417 Portfolio Y: (0.12 - 0.05)/0.16 = 0.438 Portfolio Z: (0.14 - 0.05)/0.20 = 0.450 Therefore, Portfolio Z is the best alternative according to the safety-first criterion.

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

Power of a test B is correct. The critical value in a decision rule is the rejection point for the test. It is the point with which the test statistic is compared to determine whether to reject the null hypothesis, which is part of the fourth step in hypothesis testing.

*** Which of the following is a Type I error? Rejecting a true null hypothesis Rejecting a false null hypothesis Failing to reject a false null hypothesis [You should be able to pick this up very quick by now]

Rejecting a true null hypothesis

Suppose we take a random sample of 30 companies in an industry with 200 companies. We calculate the sample mean of the ratio of cash flow to total debt for the prior year. We find that this ratio is 23%. Subsequently, we learn that the population cash flow to total debt ratio (taking account of all 200 companies) is 26%. What is the explanation for the discrepancy between the sample mean of 23% and the population mean of 26%? Sampling error. Bias. A lack of consistency.

Sampling error. A is correct. The discrepancy arises from sampling error. Sampling error exists whenever one fails to observe every element of the population, because a sample statistic can vary from sample to sample. As stated in the reading, the sample mean is an unbiased estimator, a consistent estimator, and an efficient estimator of the population mean. Although the sample mean is an unbiased estimator of the population mean—the expected value of the sample mean equals the population mean—because of sampling error, we do not expect the sample mean to exactly equal the population mean in any one sample we may take.

Which parameter equals zero in a normal distribution? Kurtosis Skewness Standard deviation

Skewness A normal distribution has a skewness of zero (it is symmetrical around the mean). A non-zero skewness implies asymmetry in a distribution.

*************** Two years from now, a client will receive the first of three annual payments of $20,000 from a small business project. If she can earn 9 percent annually on her investments and plans to retire in six years, how much will the three business project payments be worth at the time of her retirement?

Step 1 N = 3 PMT =20000 PV= 0 I/Y = 9 CPT FV = 65562 Step 2: PV = 65562 N = 2 i/y = 9 PMT = 0 CPT PV = 71462

*************** A client requires £100,000 one year from now. If the stated annual rate is 2.50% compounded weekly, the deposit needed today is closest to: £97,500. £97,532. £97,561.

Step 1: Find EAR (1 + 0.025/52)^52 = 1.025308, -1 = 0.025308 N = 1 I/Y = 2.5308 FV = 100000 PMT = 0 CPT PV = 97531.8

***** 8. Two years from now, a client will receive the first of three annual payments of $20,000 from a small business project. If she can earn 9 percent annually on her investments and plans to retire in six years, how much will the three business project payments be worth at the time of her retirement?

Step 1: PMT = 20000, N = 3, I/Y = 9 CPT FV = $65,562.00 Step 2: PV = $65,562.00, N = 2, I/Y = 9, CPT FV= $77,894.21

*** 6. Given a €1,000,000 investment for four years with a stated annual rate of 3% compounded continuously, the difference in its interest earnings compared with the same investment compounded daily is closest to: A. €1. B. €6. C. €455.

Step 1: compounded daily 1 mil * (1 +0.03/365)^365*4 - 1 mil = 127,491.29 Step 2: continuous compounding 1,000,000*e^0.03(4) - €1,000,000 = 127,496.85 or => 0.03*4, => [2nd] e => 1127486.85 - 1000000 => 127,496. Step 3 = 127,491.29 - 127496 = 5.56 or 6

*** The following table shows the significance level (α) and the p-value for two hypothesis tests. α. p-Value Test 1 0.02 0.05 Test 2 0.05 0.02 In which test should we reject the null hypothesis? Test 1 only Test 2 only Both Test 1 and Test 2 [I got it wrong, silly mistake]

Test 2 only B is correct. The p-value is the smallest level of significance (α) at which the null hypothesis can be rejected. If the p-value is less than α, the null is rejected. In Test 1, the p-value exceeds the level of significance, whereas in Test 2, the p-value is less than the level of significance.

*** Which of the following statements about hypothesis testing is correct? The null hypothesis is the condition a researcher hopes to support. The alternative hypothesis is the proposition considered true without conclusive evidence to the contrary. The alternative hypothesis exhausts all potential parameter values not accounted for by the null hypothesis. [i got it wrong the second time] (I'm up to module 6, so start with Module 6-7 then you go bacl to M1,2,3)

The alternative hypothesis exhausts all potential parameter values not accounted for by the null hypothesis. C is correct. Together, the null and alternative hypotheses account for all possible values of the parameter. Any possible values of the parameter not covered by the null must be covered by the alternative hypothesis (e.g., H0: μ ≤ 5 versus Ha: μ > 5).

Why is the central limit theorem important?

The central limit theorem states that if the sample size is large, regardless of the shape of the underlying population, the distribution of the sample mean is approximately normal. Therefore, even in these instances, we can still construct confidence intervals (and conduct tests of inference) as long as the sample size is large (generally n ≥ 30).

Julie Moon is an energy analyst examining electricity, oil, and natural gas consumption in different regions over different seasons. She ran a simple regression explaining the variation in energy consumption as a function of temperature. The total variation of the dependent variable was 140.58, and the explained variation was 60.16. She had 60 monthly observations. Calculate the coefficient of determination. Calculate the F-statistic to test the fit of the model. Calculate the standard error of the estimate of the regression estimation. Calculate the sample standard deviation of monthly energy consumption.

The coefficient of determination is 0.4279: Explained variation / Total variation = 60.16/ 140.58 = 0.4279. F= {(60.16/1) / [(140.58 − 60.16)/(60 − 2)]}= 60.16/ 1.3866 = 43.3882. Begin with the sum of squares error of 140.58 − 60.16 = 80.42. Then calculate the mean square error of 80.42 ÷ (60 − 2) = 1.38655. The standard error of the estimate is the square root of the mean square error: se = √1.38655 = 1.1775. Total variation / (n − 1) = 140.5860 − 1 = 2.3827.

Which one of the following statements concerning chi-square and F-distributions is false? They are both asymmetric distributions. As their degrees of freedom increase, the shapes of their pdfs become more bell curve-like. The domains of their pdfs are positive and negative numbers.

The domains of their pdfs are positive and negative numbers. Both bounded by 0. They can't have negative number.

Which of the following is best described as a discrete random variable? The expected percentage change in a country's gross national product for the next year The number of days on which the DJIA experienced an increase since 2013 The expected annual return on the Nikkei 225 Index over the next year

The expected percentage change in a country's gross national product for the next year B is correct. A discrete random variable is a random variable that can take on at most a countable number of possible values. The number of days on which the DJIA experienced an increase since 2013 is the only choice with a discrete number of possible values. A is incorrect. A percentage change is an example of a continuous random variable. C is incorrect. A return is an example of a continuous random variable.

When analyzing investment returns, which of the following statements is correct? The geometric mean will exceed the arithmetic mean for a series with non-zero variance. The geometric mean measures an investment's compound rate of growth over multiple periods. The arithmetic mean measures an investment's terminal value over multiple periods.

The geometric mean measures an investment's compound rate of growth over multiple periods.

*** Which of the following statements on p-value is correct? The p-value indicates the probability of making a Type II error. The lower the p-value, the weaker the evidence for rejecting the H0. The p-value is the smallest level of significance at which H0 can be rejected.

The p-value is the smallest level of significance at which H0 can be rejected. C is correct. The p-value is the smallest level of significance (α) at which the null hypothesis can be rejected.

Which of the following is a continuous random variable? The value of a futures contract quoted in increments of $0.05 The total number of heads recorded in 1 million tosses of a coin The rate of return on a diversified portfolio of stocks over a three-month period

The rate of return on a diversified portfolio of stocks over a three-month period C is correct. The rate of return is a random variable because the future outcomes are uncertain, and it is continuous because it can take on an unlimited number of outcomes.

*** Got it from CFA Mock Exam: [I most likely drop what i got wrong] Which of the following is a continuous random variable? The number of trades at a stock exchange The price of a stock that is quoted in increments of $0.01 The rate of return on an investment over a six-month period [A good question for future similar questions] [You get confused with these variables]

The rate of return on an investment over a six-month period A is incorrect because this is a discrete random variable. B is incorrect because this is a discrete random variable.

*** In which of the following situations would a nonparametric test of a hypothesis most likely be used? The sample data are ranked according to magnitude. The sample data come from a normally distributed population. The test validity depends on many assumptions about the nature of the population. [area you're not familair]

The sample data are ranked according to magnitude. A is correct. A nonparametric test is used when the data are given in ranks.

*** Which of the following statements is correct regarding the chi-square test of independence? The test has a one-sided rejection region. The null hypothesis is that the two groups are dependent. If there are two categories, each with three levels or groups, there are six degrees of freedom. [area you're not familair]

The test has a one-sided rejection region.

***** A saver deposits the following amounts in an account paying a stated annual rate of 4%, compounded semiannually: Year End of Year Deposits ($) 1 4,000 2 8,000 3 7,000 4 10,000 At the end of Year 4, the value of the account is closest to: A. $30,432 B. $30,447 C. $31,677

To solve for the future value of unequal cash flows, compute the future value of each payment as of Year 4 at the semiannual rate of 2% 1) 4000 * (1.02)^6 = 4,504.65 2) 8,000 * (1.02)^ 4= 8,659.46 3) 7,000* (1.02)^2 = 7,282.80 4) 10,000* (1.02)^0= 10,000.00 30,446.91

** 5. A bank quotes a stated annual interest rate of 4.00%. If that rate is equal to an effective annual rate of 4.08%, then the bank is compounding interest: A. daily B. quarterly. C. semiannually

[I tried to find a quicker way, I assume you just have to try it out one by one] EAR = (1 + Periodic interest rate)^n- 1 EAR = (1 + 0.04/365)^365 - 1 EAR = (1.0408) - 1 = 0.04081 ≈ 4.08%

*** Got it from CFA Mock Exam: [I most likely drop what i got wrong] An analyst runs a simple linear regression to test whether the variation in the demand for corn explains the variation in the supply of wheat. In this model, the demand for corn is a(n): indicator variable. Explained variable. independent variable. [Must review Regression thoroughly with its conceptual analysis]

[Must review Regression thoroughly with its conceptual analysis] C is correct because the variation in the demand for corn is being used to explain the variation in the supply of wheat. "We refer to the variable whose variation is being used to explain the variation of the dependent variable as the independent variable, or the explanatory variable; it is typically denoted by X."

*** An analyst is examining the monthly returns for two funds over one year. Both funds' returns are non-normally distributed. To test whether the mean return of one fund is greater than the mean return of the other fund, the analyst can use: a parametric test only. a nonparametric test only. both parametric and nonparametric tests. [area you're not familair]

a nonparametric test only. B is correct. There are only 12 (monthly) observations over the one year of the sample and thus the samples are small. Additionally, the funds' returns are non-normally distributed. Therefore, the samples do not meet the distributional assumptions for a parametric test. The Mann-Whitney U test (a nonparametric test) could be used to test the differences between population means.

A chi-square test is most appropriate for tests concerning: a single variance. differences between two population means with variances assumed to be equal. differences between two population means with variances assumed to not be equal. [I got it wrong]

a single variance.

If an estimator is consistent, an increase in sample size will increase the: accuracy of estimates. efficiency of the estimator. unbiasedness of the estimator.

accuracy of estimates. A is correct. A consistent estimator is one for which the probability of estimates close to the value of the population parameter increases as sample size increases. More specifically, a consistent estimator's sampling distribution becomes concentrated on the value of the parameter it is intended to estimate as the sample size approaches infinity.

*** Got it from CFA Mock Exam: [I most likely drop what i got wrong] With regard to probability concepts, an event represents: a single outcome only. a set of outcomes only. both a single outcome and a set of outcomes. [I'm ashamed that I got this type of question wrong]

both a single outcome and a set of outcomes. C is correct because "[a]lthough we may be concerned about a single outcome, frequently our interest may be in a set of outcomes. The concept of 'event' covers both."

In generating an estimate of a population parameter, a larger sample size is most likely to improve the estimator's: consistency. unbiasedness. efficiency.

consistency. A is correct. A consistent estimator is one for which the probability of estimates close to the value of the population parameter increases as the sample size increases. Unbiasedness and efficiency are properties of an estimator's sampling distribution that hold for any size sample. B is incorrect. Unbiasedness and efficiency are properties of an estimator's sampling distribution that hold for any size sample. C is incorrect. Unbiasedness and efficiency are properties of an estimator's sampling distribution that hold for any size sample.

************************ A bank quotes a stated annual interest rate of 4.00%. If that rate is equal to an effective annual rate of 4.08%, then the bank is compounding interest: daily. quarterly. semiannually.

daily. The equation is: EAR = (1 + rnom/n)n − 1, and you want to solve it for n. The good news is: you can't. Algebraically, there's no way to solve that equation as a closed formula for n. (Why is that good news? Because you won't waste any more time trying to do the impossible.) The other good news is that you don't have to solve it for n. Because this is a multiple-choice question, you know for certain that the correct value for n is either 2, 4, or 365. In this case, you should try n = 4 (the middle number) to see if you get 4.08%. If you do, then 4 is the correct answer. If you get a number less than 4.08%, then 365 is the correct answer. If you get a number greater than 4.08%, then 2 is the correct answer. By the way, the real answer is n ≈ 77. They rounded off to too few digits. --- You plug it in: 4% / 365 = 0.0001095 (1 + 0.0001095)^365 = 1.0408... Go for the others

*** An increase in sample size is most likely to result in a: wider confidence interval. decrease in the standard error of the sample mean. lower likelihood of sampling from more than one population.

decrease in the standard error of the sample mean. B is correct. All else being equal, as the sample size increases, the standard error of the sample mean decreases and the width of the confidence interval also decreases.

*** The level of significance of a hypothesis test is best used to: calculate the test statistic. define the test's rejection points. specify the probability of a Type II error. [You should be able to pick this up very quick by now]

define the test's rejection points.

*** A pooled estimator is used when testing a hypothesis concerning the: equality of the variances of two normally distributed populations. difference between the means of two at least approximately normally distributed populations with unknown but assumed equal variances. difference between the means of two at least approximately normally distributed populations with unknown and assumed unequal variances. [Not seen pooled estimator on Wiley]

difference between the means of two at least approximately normally distributed populations with unknown but assumed equal variances.

The best approach for creating a stratified random sample of a population involves: drawing an equal number of simple random samples from each subpopulation. selecting every kth member of the population until the desired sample size is reached. drawing simple random samples from each subpopulation in sizes proportional to the relative size of each subpopulation.

drawing simple random samples from each subpopulation in sizes proportional to the relative size of each subpopulation. C is correct. Stratified random sampling involves dividing a population into subpopulations based on one or more classification criteria. Then, simple random samples are drawn from each subpopulation in sizes proportional to the relative size of each subpopulation. These samples are then pooled to form a stratified random sample.

When making a decision about investments involving a statistically significant result, the: economic result should be presumed to be meaningful. statistical result should take priority over economic considerations. economic logic for the future relevance of the result should be further explored. [I got it wrong]

economic logic for the future relevance of the result should be further explored. C is correct. When a statistically significant result is also economically meaningful, one should further explore the logic of why the result might work in the future.

The nominal risk-free rate is best described as the sum of the real risk-free rate and a premium for: maturity. liquidity. expected inflation. --- Which of the following risk premiums is most relevant in explaining the difference in yields between 30-year bonds issued by the US Treasury and 30-year bonds issued by a small private issuer? Inflation Maturity Liquidity --

expected inflation. C is correct. The sum of the real risk-free interest rate and the inflation premium is the nominal risk-free rate. --- Liquidity C is correct. US Treasury bonds are highly liquid, whereas the bonds of small issuers trade infrequently and the interest rate includes a liquidity premium. This liquidity premium reflects the relatively high costs (including the impact on price) of selling a position.

*** A Type II error is best described as: rejecting a true null hypothesis. failing to reject a false null hypothesis. failing to reject a false alternative hypothesis. [You should be able to pick this up very quick by now]

failing to reject a false null hypothesis.

*** Got it from CFA Mock Exam: [I most likely drop what i got wrong] The central limit theorem is best described as stating that the sampling distribution of the sample mean will be approximately normal for large-size samples: if the population distribution is normal. for populations described by any probability distribution. if the population distribution is symmetrical.

for populations described by any probability distribution. B is correct. The central limit theorem holds without regard for the distribution of the underlying population.

(Check graph) The conditions for a probability function are satisfied by: f(x). g(x). h(x).

g(x). B is correct. The function g(x) satisfies the conditions of a probability function. All of the values of g(x) are between 0 and 1, and the values of g(x) all sum to 1.

*** Year 1 62.00 Year 2 76.00 Year 3 84.00 Year 4 90.00 The average price is best represented as the: harmonic mean of €76.48. geometric mean of €77.26. arithmetic average of €78.00. great reminder

harmonic mean of €76.48. harmonic mean: 4 / [1/62 + 1/76 + 1/84 + 1/90] = 76.48

*** Got it from CFA Mock Exam: [I most likely drop what i got wrong] Any normal distribution: has a kurtosis of zero. has a skewness of zero. is symmetric around zero. [I'm ashamed that I got this type of question wrong]

has a skewness of zero. B is correct because "[t]he normal distribution has a skewness of 0 (it is symmetric)." A is incorrect because "[t]he normal distribution has a kurtosis of 3." C is incorrect because, while the normal distribution is symmetric, it is symmetric around its mean, which may not be equal to zero.

In contrast to normal distributions, lognormal distributions: are skewed to the left. have outcomes that cannot be negative. are more suitable for describing asset returns than asset prices. ------------------------------------------------- The lognormal distribution is a more accurate model for the distribution of stock prices than the normal distribution because stock prices are: symmetrical. unbounded. non-negative.

have outcomes that cannot be negative. non-negative.

*** Got it from CFA Mock Exam: [I most likely drop what i got wrong] The 'box' in a box and whisker plot represents the: interquartile range. range between the median and the arithmetic average. range between the highest and lowest values of the distribution.

interquartile range. A is correct because "[a] box and whisker plot consists of a 'box' with 'whiskers' connected to the box.... The 'box' represents the lower bound of the second quartile and the upper bound of the third quartile, with the median or arithmetic average noted as a measure of central tendency of the entire distribution. ... In Exhibit 44, visually, the interquartile range is the height of the box." "The interquartile range (IQR) is the difference between the third quartile and the first quartile, or IQR = Q3 − Q1."

Which of the following is the least likely characteristic of the normal probability distribution? The normal probability distribution: is more suitable as a model for asset prices than for returns. has the same value for mean, median, and mode. has kurtosis of 3.0

is more suitable as a model for asset prices than for returns. Logarithmic is suitable for asset price.

****** Perkiomen Kinzua, a seasoned auditor, is auditing last year's transactions for Conemaugh Corporation. Unfortunately, Conemaugh had a very large number of transactions last year, and Kinzua is under a time constraint to finish the audit. He decides to audit only the small subset of the transaction population that is of interest and to use sampling to create that subset. The most appropriate sampling method for Kinzua to use is: judgmental sampling. systematic sampling. convenience sampling. -------------------------- Which one of the following statements is true about non-probability sampling? There is significant risk that the sample is not representative of the population. Every member of the population has an equal chance of being selected for the sample. Using judgment guarantees that population subdivisions of interest are represented in the sample.

judgmental sampling. There is significant risk that the sample is not representative of the population.

The value of the cumulative distribution function F(x), where x is a particular outcome, for a discrete uniform distribution: sums to 1. lies between 0 and 1. decreases as x increases.

lies between 0 and 1. The value of the cumulative distribution function lies between 0 and 1 for any x: 0 ≤ F(x) ≤ 1.

If a stock's continuously compounded return is normally distributed, then the distribution of the future stock price is best described as being: A. normal. B. a Student's t. C. lognormal.

lognormal.

************ A call option on a stock index is valued using a three-step binomial tree with an up move that equals 1.05 and a down move that equals 0.95. The current level of the index is $190, and the option exercise price is $200. If the option value is positive when the stock price exceeds the exercise price at expiration and $0 otherwise, the number of terminal nodes with a positive payoff is: one. two. three.

one. Only the top node value of $219.9488 exceeds $200.

*** Got it from CFA Mock Exam: [I most likely drop what i got wrong] A dataset consisting of 10 years of annual returns for each stock within a single industry is best described as: panel data. time-series data. cross-sectional data.

panel data. A is correct because "[p]anel data are a mix of time-series and cross-sectional data that are frequently used in financial analysis and modeling. Panel data consist of observations through time on one or more variables for multiple observational units." Here, the multiple observational units are the different stocks within the industry, and 10 yearly observations for each stock are present in the dataset.

A two-dimensional rectangular array would be most suitable for organizing a collection of raw: panel data. time-series data. cross-sectional data. --- In a frequency distribution, the absolute frequency measure: represents the percentages of each unique value of the variable. represents the actual number of observations counted for each unique value of the variable. allows for comparisons between datasets with different numbers of total observations.

panel data. --- represents the actual number of observations counted for each unique value of the variable.

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

power of a test.

*** Got it from CFA Mock Exam: [I most likely drop what i got wrong] Sampling error is the difference between the observed value of a statistic and the: mean of the sample. quantity it is intended to estimate. observed value of the random variable. [I'm ashamed that I got this type of question wrong]

quantity it is intended to estimate. B is correct because "[s]ampling error is the difference between the observed value of a statistic and the quantity it is intended to estimate as a result of using subsets of the population.

(This is from Wiley) Balance sheet and income statement data denominated in Indian rupees would best be described as: ratio-scaled. ordinal-scaled. nominal-scaled.

ratio-scaled. Figures used to denominate currency are ratio-scaled data because they can be meaningfully ranked and used to perform calculations. Currency can be meaningfully added, subtracted, and used to compute ratios.

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

reject the null hypothesis.

************************* An analyst tests the profitability of a trading strategy with the null hypothesis that the average abnormal return before trading costs equals zero. The calculated t-statistic is 2.802, with critical values of ±2.756 at significance level α = 0.01. After considering trading costs, the strategy's return is near zero. The results are most likely: statistically but not economically significant. economically but not statistically significant. neither statistically nor economically significant. [I believe this might be the type of question that will appear on the test]

statistically but not economically significant. A is correct. The hypothesis is a two-tailed formulation. The t-statistic of 2.802 falls outside the critical rejection points of less than −2.756 and greater than 2.756. Therefore, the null hypothesis is rejected; the result is statistically significant. However, despite the statistical results, trying to profit on the strategy is not likely to be economically meaningful because the return is near zero after transaction costs.

A report on long-term stock returns focused exclusively on all currently publicly traded firms in an industry is most likely susceptible to: look-ahead bias. survivorship bias. intergenerational data mining.

survivorship bias.

For a binomial random variable with five trials and a probability of success on each trial of 0.50, the distribution will be: skewed. uniform. symmetric.

symmetric. C is correct. The binomial distribution is symmetric when the probability of success on a trial is 0.50, but it is asymmetric or skewed otherwise. Here, it is given that p = 0.50.

(This is from Wiley) An analyst is using a hypothesis test concerning a population mean (μ). The null and alternative hypotheses for the test are: H0:μ=0.0058 Ha:μ≠0.0058 The sample size (n) is 39 and the alpha (α) is 0.10. For this hypothesis test, the acceptance point or range is closest to: t = 1.304. t < −1.304 and t > 1.304. t < −1.686 and t > 1.686. (It was given a t distribution table)

t < −1.686 and t > 1.686. The degrees of freedom are n − 1 = 39 − 1 = 38. The null hypothesis is that the mean is equal to 0.0058, so in the test, we are concerned whether the mean is significantly larger or significantly smaller than 0.0058. This means that the 10% alpha is evenly split between the two tails (5% in each tail). So in Table 1 we use the column labels p = 0.05 and row 38. The resulting t-critical of 1.686 is the number of standard deviations to the right of the mean (+1.686) or left of mean (−1.686). Observations outside of those critical t-values would invalidate the null hypothesis, so the acceptance range is −1.686 to 1.686.

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

t-statistic.

*** Got it from CFA Mock Exam: [I most likely drop what i got wrong] Two populations are normally distributed with unknown variances that are assumed to be equal. If a large independent random sample is drawn from each population, the most appropriate test statistic for testing the difference between the population means is the: t-statistic. z-statistic. F-statistic. [I'm ashamed that I got this type of question wrong]

t-statistic. A is correct because, for tests concerning differences between population means, "[w]hen we can assume that the two populations are normally distributed and that the unknown population variances are equal, a t-test based on independent random samples is given by" t = [(X1 - X2) - (μ1 - µ2)]/√(sp2/n1 + sp2/n2), where (X1 - X2) is the difference between the sample means, (μ1 - µ2) is the hypothesized difference between the population means, sp2 is a pooled estimator of the common variance, and n1 and n2 are the corresponding sample sizes.

A Monte Carlo simulation can be used to: directly provide precise valuations of call options. simulate a process from historical records of returns. test the sensitivity of a model to changes in assumptions—for example, on distributions of key variables. ----------------------------------- A limitation of Monte Carlo simulation is: its failure to do "what if" analysis. that it requires historical records of returns. its inability to independently specify cause-and-effect relationships.

test the sensitivity of a model to changes in assumptions—for example, on distributions of key variables. C is correct. A characteristic feature of Monte Carlo simulation is the generation of a large number of random samples from a specified probability distribution or distribution to represent the role of risk in the system. Therefore, it is very useful for investigating the sensitivity of a model to changes in assumptions—for example, on distributions of key variables. ----------------------- its inability to independently specify cause-and-effect relationships.

*** Got it from CFA Mock Exam: [I most likely drop what i got wrong] When two random events X and Y are independent: X and Y are mutually exclusive. the conditional probability of X given Y is less than the unconditional probability of X. the joint probability of X and Y equals the product of the individual probabilities of X and Y. [I'm ashamed that I got this type of question wrong]

the joint probability of X and Y equals the product of the individual probabilities of X and Y. C is correct because "[w]hen two events are independent, the joint probability of A and B equals the product of the individual probabilities of A and B." A is incorrect because "[w]hen two events are not independent, they are dependent. ... For a different example, if two events are mutually exclusive, then knowledge that one event has occurred gives us information that the other (mutually exclusive) event cannot occur," thus the events are dependent (as their joint probability is equal to zero irrespective of their individual probabilities). "When two events are independent, the joint probability of A and B equals the product of the individual probabilities of A and B," and not zero.

The power of a hypothesis test is: equivalent to the level of significance. the probability of not making a Type II error. unchanged by increasing a small sample size.

the probability of not making a Type II error. B is correct. The power of a hypothesis test is the probability of correctly rejecting the null when it is false. Failing to reject the null when it is false is a Type II error. Thus, the power of a hypothesis test is the probability of not committing a Type II error.

(This is from Wiley) Standard error of the estimate, se, would least likely be described as: the square root of sum of the squared errors (SSE). a goodness-of-fit measure for the estimated regression line. an absolute measure of the distance of the observed dependent variable from the regression line. (I got this wrong) (Wiley Question ends here for now)

the square root of sum of the squared errors (SSE). Standard error of the estimate is the square root of the mean squared error rather than the sum of squared errors: Standard error of the estimate (se)= ∑√[(Yi −ˆYi)^2 / n−2]

*** Got it from CFA Mock Exam: [I most likely drop what i got wrong] A bubble line chart without any color-coding can represent a maximum of: two dimensions of data. three dimensions of data. four dimensions of data. [I'm ashamed that I got this type of question wrong]

three dimensions of data. B is correct because "[w]hen an observational unit ... has more than two features (or variables) of interest, it would be useful to show the multi-dimensional data all in one chart to gain insight from a more holistic view. How can we add an additional dimension to a two-dimensional line chart? We can replace the data points with varying-sized bubbles to represent a third dimension of the data. Moreover, these bubbles may even be color-coded to present additional information. This version of a line chart is called a bubble line chart." Hence, if color-coding is not used, the maximum number of dimensions that can be represented is three (x-axis, y-axis, and size of bubbles).

Time Period Index A Index B Index C Year t-3 15.56%. 11.84% −4.34% Year t-2 −4.12% −6.96% 9.32% Year t-1 11.19%. 10.29% −12.72% Year t 8.98% 6.32% 21.44% Each individual column of data in the table can be best characterized as: panel data. time-series data. cross-sectional data. [i actually got this wrong] --- Each individual row of data in the table can be best characterized as: panel data. time-series data. cross-sectional data.

time-series data. [My mistake is I didn't know it was talking about each column of data!] B is correct. Time-series data are a sequence of observations of a specific variable collected over time and at discrete and typically equally spaced intervals of time, such as daily, weekly, monthly, annually, and quarterly. In this case, each column is a time series of data that represents annual total return (the specific variable) for a given country index, and it is measured annually (the discrete interval of time). A is incorrect because panel data consist of observations through time on one or more variables for multiple observational units. The entire table of data is an example of panel data showing annual total returns (the variable) for three country indexes (the observational units) by year. --- cross-sectional data.

An estimator with an expected value equal to the parameter that it is intended to estimate is described as: efficient. unbiased. consistent.

unbiased. An unbiased estimator is one for which the expected value equals the parameter it is intended to estimate.

Compared with bootstrap resampling, jackknife resampling: is done with replacement. usually requires that the number of repetitions is equal to the sample size. produces dissimilar results for every run because resamples are randomly drawn.

usually requires that the number of repetitions is equal to the sample size. B is correct. For a sample of size n, jackknife resampling usually requires n repetitions. In contrast, with bootstrap resampling, we are left to determine how many repetitions are appropriate.

Peter Biggs wants to know how growth managers performed last year. Biggs assumes that the population cross-sectional standard deviation of growth manager returns is 6% and that the returns are independent across managers. How large a random sample does Biggs need if he wants the standard deviation of the sample means to be 1%? How large a random sample does Biggs need if he wants the standard deviation of the sample means to be 0.25%?

we have 1% = 6%/√n Squaring this value, we get a random sample of n = 36. ... (same)

**************** Given a €1,000,000 investment for four years with a stated annual rate of 3% compounded continuously, the difference in its interest earnings compared with the same investment compounded daily is closest to: €1. €6. €455. great question and I'm proud I got it right, I wasn't guessing!

€6. Step 1: = €1,000,000* e0.03(4) = 1,127,496.68 Step 2: (0.03/365) = 0.000082192 we find EAR: (1 + 0.000082192)^(365*4) = 1.127491 1.127491* 1,000,000 = 1127491 127496.68 - 127491 = 5.... ≈ 6