QUESTION: 1081

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

a right-tailed test.
a left-tailed test.
a two-tailed test.
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.

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QUESTION: 1082

In a binomial probability distribution,

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

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QUESTION: 1083

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

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

Answer(s): C

Explanation:

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

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QUESTION: 1084

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:

[866.2; 884.6]
[870.1; 882.2]
[868.6; 883.4]
[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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Free CFA-Level-I Exam Braindumps (page: 271)

QUESTION: 1081

Explanation:

QUESTION: 1082

Explanation:

QUESTION: 1083

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QUESTION: 1084

Explanation: