Free CFA-Level-I Exam Braindumps (page: 245)

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Walter Jennings, a quantitative analyst with Smith, Kleen & Beetchnutty Brokerage, has just been informed of an important error in one of his recent statistical endeavors. Specifically, in one hypothesis test, Mr. Jennings failed to reject a null hypothesis that later was proven to be false. Which of the following best describes this type of error in hypothesis testing? Further, if the confidence level of the test were increased, would the probability of this error increase, decrease, or is this probability difficult to determine?

  1. Type II error; difficult to determine
  2. Type I error; difficult to determine
  3. Type I error; decrease
  4. Type II error; decrease
  5. Type I error; increase
  6. Type II error; increase

Answer(s): A

Explanation:

In this example, Mr. Jennings has incorrectly failed to reject a null hypothesis. This type of error in hypothesis testing is called a Type II error. In hypothesis testing, the Type I error is given more attention than the Type II error. A Type I error is defined as the act of incorrectly rejecting the null hypothesis. In most hypothesis tests, the probability of a null hypothesis is equal to the significance level of the test. A significance level of 0.01, for example, indicates that a 1% chance exists that the null hypothesis will be rejected when it is indeed true.
Another way to think of the probability of a Type I error is to observe the following relationship:
{Probability of a Type I error = (1 - confidence level)}.
For example, a confidence level of 95% leaves a 5% probability of a Type I error occurring. If this confidence level were to increase to say, 98%, then the probability of a Type I error would reduce to 2%. While a relationship exists between the confidence level for a hypothesis test and the probability of a Type I error, the relationship between confidence levels and Type II errors is not as explicit. The probability of a Type II error is inherently difficult to quantify, and as such, few hypothesis tests call for a determination of the probability of a Type II error. If the confidence level of a hypothesis test to be increased, it would be difficult to determine what effect, if any, this would have on the probability of a Type II error.



If you owe a debt of $1,000 today and also owe $500 in 24 months, what single payment could you make 12 months from today that would pay off both of these debts, if interest is assessed at 8% per year, compounded monthly?

  1. $751.62
  2. $1,426.30
  3. $1,500.00
  4. $1,544.68
  5. $2,041.93

Answer(s): D

Explanation:

To solve this question, set the problem up as the sum of two compound interest calculations. Move the $1,000 from today over to month 12 and add it to the $500 brought back from month 24 to month 12. On the BAII Plus, press 12 N, 8 divide 12 = I/Y, 1000 PV, 0 PMT, CPT FV, which yields $1,083.00. Then press STO 1. Then press 500 FV, CPT PV which yields $461.68. Finally press + RCL 1 = to see the answer. On the HP12C, press 12 n, 8 ENTER 12 divide i, 1000 PV, 0 PMT, FV. Then press STO 1. Then press 500 FV, PV. Finally press RCL 1 + to see the answer.



On a very hot summer day, 5 percent of the production employees at Midland States Steel are absent from work. The production employees are to be selected at random for a special in depth study on absenteeism. What is the probability of selecting 10 production employees at random on a hot summer day and finding that none of them are absent?

  1. 0.599
  2. 0.100
  3. 0.002
  4. 0.344
  5. None of these answers

Answer(s): A

Explanation:

This is a binomial probability. The probability of getting r successes out of n trials where the probability of success each trial is p and probability of failure each trial is q (where q = 1-p) is given by: n!(p^r)[q^(n-r)]/r!(n-r)!.
Here n = 10, r = 0,p = 0.05 and q = 0.95. Therefore we have 10!(0.05^0)(0.95^10)/0!10! = 0.599.



If an analyst is trying to estimate the value of a stock, using the formula E(Y | E) = [y_1 * P(y_1 |E) + y_2 * P (y_2|E) + ... y_n * P(y_n|E), the analyst is making use of:

  1. subjective probability.
  2. the multiplication rule for independent events.
  3. unconditional expectation.
  4. conditional expectation.

Answer(s): D

Explanation:

Since the stock value is conditioned on an uncertain event and the likelihood of certain stock price outcomes given the occurrence of that uncertain event, the analyst is using conditional expectation. The expected value is conditioned upon the occurrence of the event.






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