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Free CFA-Level-I Exam Braindumps (page: 272)

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If the underlying theory does not suggest a particular direction for the value of an estimated variable, it is appropriate to use a:

  1. a right-tailed test.
  2. a left-tailed test.
  3. a two-tailed test.
  4. either a right-tailed or left-tailed test.

Answer(s): C

Explanation:

If there is no suggested directionality to the underlying variable, then you are interested in finding the probability under the null hypothesis that you will find a value beyond a given distance from the null value. For this, you need to use a two-tailed test.



In a binomial probability distribution,

  1. there are 2 distinct modes.
  2. the probability graph is bell-shaped and symmetrical.
  3. each trial can only take values between 0 and 1.
  4. each trial can take only two values.

Answer(s): D

Explanation:

A binomial distribution with "N trials" arises when the same experiment is repeated N times. Each of the trails is known as a "Bernoulli" trial. It is the Bernoulli trial which can only take two distinct values, "success" and "failure." Each outcome of a binomial distribution with N trials is the string of N values generated by the individual Bernoulli trials. Clearly, the binomial distribution with N trials can take 2^N values, even though each individual trial can have only two outcomes. It is important that you understand the difference between the binomial distribution and the underlying Bernoulli trials themselves.



Which of the following are notations for the covariance between X and Y?

  1. Cov(X,Y).
    II. sigma_(X,Y)
    III. [sigma_(X,Y)]^2.
  2. I only.
  3. II only.
  4. I and II.
  5. I and III.

Answer(s): C

Explanation:

I and II are both notations for covariance between X and Y.



A sample of size 225 is drawn from a population. The sample mean equals 876 and the variance of the sample equals 5,924. The 85% confidence interval for the population mean is given by:

  1. [866.2; 884.6]
  2. [870.1; 882.2]
  3. [868.6; 883.4]
  4. [869.3; 882.7]

Answer(s): C

Explanation:

If z is the z-value corresponding to the specified confidence level, the sample mean is M and the standard deviation is D in a sample size N, the confidence interval is specified as [M - z*D/sqrt(N), M + z*D/sqrt(N)]. In the present case, for the 85% confidence interval, the normal probability table gives z = 1.44. The sample standard deviation equals sqrt(5924) = 76.97. Therefore, the confidence interval equals [876 - 1.44*76.97/sqrt (225), 876 + 1.44*76.97/sqrt(225)] = [868.6, 883.4]






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