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Multiple Choice Questions from Different Exercises.
Typology: Exercises
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BUSINESS STATISTICS - Second Year Academic Year 2010- June 28, 2011
Example: (^12545) ︸ ︷︷ ︸ PEREZ, Ernesto Exam type (^) ︸︷︷︸ 0 Resit
MULTIPLE CHOICE QUESTIONS (Time: 1 hour and 30 minutes)
(A) Paris (B) Sebastopol (C) Madrid (D) London (E) Pekin
Questions 2 and 3 refer to the following exercise: In a given sports activity, the annual rate of accidents is of about 4 per thousand people. An insurance company has 3 000 insured clients that practice such sports activity.
(A) Poisson (λ = 12) (B) Binomial (p = 0. 4 , n = 3 000) (C) Poisson (λ = 48) (D) N (12, σ^2 = 12) (E) Binomial (p = 0. 5 , n = 3 000)
Questions 4 to 6 refer to the following exercise: The number of clients that enter a given store each hour follows a Poisson distribution with mean 3.25. We assume independence between the different hours.
(A) 0.8118 (B) 0.1172 (C) 0.9168 (D) 0.0252 (E) 0.
(A) 3 (B) 4 (C) 2 (D) 2.25 (E) 3.
(A) 0.6255 (B) 0.6844 (C) 0.3745 (D) 0.3446 (E) 0.
Questions 7 to 9 refer to the following exercise: The waiting time for clients in a given restaurant follows an exponential distribution with mean equal to 20 minutes. We assume independence between the different days.
is:
(A) t 1 (B) F 2 , 1 (C) χ^22 (D) F 1 , 2 (E) All false
Questions 18 and 19 refer to the following exercise: Let X be a r.v. with probability mass function given by:
θ, P (X = 0) = 1 −
θ, P (X = 2) =
θ
In order to estimate the parameter θ, a r.s. of size n = 10 has been taken, providing the following results: three times the value -2, five times the value 0 and two times the value 2.
(A) 1.4 (B) 0 (C) 0.75 (D) 0.25 (E) 0.
(A) 0.37 (B) 0 (C) 1.4 (D) 0.25 (E) 0.
Questions 20 and 21 refer to the following exercise: Let X be a r.v. with probability density function given by:
f (x; θ) =
e−^
(^12) (x−θ) , x > θ,
and mean equal to 2 + θ. In order to estimate the parameter θ, a r.s. of size n, X 1 ,... , Xn has been taken.
(A) max {xi} (B) X + 2 (C) X (D) X − 2 (E) min {xi}
(A) (^) X^1 (B) X (C) X 2 (D) X + 2 (E) X − 2
Questions 22 and 23 refer to the following exercise: We have a r.s. of size n from a Poisson distribution of parameter λ. In order to estimate λ, the estimator ˆλ = X + 1 n^2 is proposed.
(A) unbiased and asymptotically biased (B) biased and asymptotically unbiased (C) unbiased and asymptotically unbiased (D) biased and asymptotically biased (E) All false
(A) No (B) − (C) Yes (D) − (E) −
Questions 24 and 25 refer to the following exercise: Let X be a r.v. having a N (m, σ^2 = 1) distribution. In order to be able to test the null hypothesis H 0 : m = 5, against the alternative hypothesis H 1 : m > 5, using X as the test statistic for it, we take the critical region as: CR = (C, ∞).
(A) 0.9406 (B) 0.6103 (C) 0.5714 (D) 0.3897 (E) 0.
Questions 26 to 28 refer to the following exercise: In a marketing study we are interested in testing the null hypothesis H 0 indicating that al least 40% of the population is willing to buy a new product. In order to do so, a random sample of size n = 20 is taken, and the test statistic Z: number of people in the sample who are willing to buy the new product.
(A) z ≤ 4 (B) z ≤ 3 (C) All false (D) z > 4 (E) z > 3
(A) 0.8929 (B) 0.7625 (C) 0.1071 (D) 0.2375 (E) All false
Questions 29 and 30 refer to the following exercise: We wish to test the null hypothesis that the benefits (in millions) of all firms in a given industry follow a normal distribution with mean 10 and variance 4. In order to do so, we have observed the benefits for 100 firms, from which 30 had benefits smaller than 8, 50 had benefits between 8 and 13, and the remaining 20 firms had benefits larger than 13.
(A) - (B) Do not reject H 0 (C) - (D) - (E) Reject H 0
Exercise A
It corresponds to a goodness of fit test to a completely specified distribution. Based on the information provided in the sample, we build the corresponding table as follows:
ni pi npi (ni − npi) (ni−npi)
2 npi
Monday 49 0.2 40 +9 2.
Tuesday 35 0.2 40 -5 0.
Wednesday 32 0.2 40 -8 1.
Thursday 39 0.2 40 -1 0.
Friday 45 0.2 40 +5 0.
200 1 200 0 z =4.
Under the null hypothesis of fit to the totally specified distribution stated by the firm’s manager, we have that the test statistics
i
(ni−npi)^2 npi ∼^ χ
2 (K−1), where^ K^ is the number of categories in which the sample has been divided (i.e., K = 5).
At the approximate 5% significance level, the decision rule will be to reject the null hypothesis if:
z > χ^2 (5−1) , 0. 05 = χ^24 , 0. 05
In this case, we have that:
z = 4. 9 < 9 .49 = χ^24 , 0. 05
so that, at the 5% significance level, we do not reject the null hypothesis of fit to the distribution specified by the firm’s manager. That is, we can state that, based on the information provided in the sample, the probability that employees visit the firm’s medical consulting room is the same for every working day in the week.
Exercise C
We have a r.v. X with probability density function given by:
f (x; a) =
a^3 2
x^2 e−ax, x > 0 , a > 0
We wish to test: H 0 : a = 1/2 against H 1 : a = 2. In order to do so, a r.s. of size n = 1 is taken. i) In order to be able to determine the most powerful critical region (CR) for this test, we use the Neyman- Pearson Theorem, so that CR would be given by:
f (x; a = 1/2) f (x; a = 2)
1 2
2
x^2 e−^ (^12) x 1 2 ( (^3) ) x (^2) e− 2 x ≤^ K
e−(^
(^12) x− 2 x) ≤ K 1 3 2
x ≤ K 2
x ≤ C
CR = (0 , C]
The most powerful critical region for this test will reject the null hypothesis if X ≤ C, C > 0.
ii) Under the null hypothesis, the distribution of the test statistic X is:
X ∈ γ
≡ γ
≡ χ^26
iii) For an α = 5% significance level:
α = 0. 05 ≥ P (Reject H 0 | H 0 ) = = P (X ≤ C | X ∈ χ^26 ) ⇒ C = χ^26 , 0. 95 = 1. 64 ⇒ CR = (0 , 1 .64]