Download Exam of Statistics 6C

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Università di Venezia - Corso di Laurea Economics & Management
Exam of Statistics 6C - Prof. M. Romanazzi
May 31st, 2016
Full Name
Matricola
• Total (nominal) score: 30/30 (2/30 for each question).
• Pass score: 18/30. Highest (29, 30) grades must be confirmed by oral discussion.
• Pocket calculator and portable computer are allowed, textbooks or class notes are not.
• Detailed solutions to questions must be given on the draft sheet (foglio di brutta copia);
final answers/results must be copied on the exam sheet, beside the small squares.
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Exercise 1 The stem-and-leaf display in Table 1 shows the distribution of a sample of n = 80 ENI stock price
variations (%) from 1998 to 2013.
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-4
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-1
-0
0
1
2
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n = 80
−1|2
Pn is read −1.2%
xi = −14.632
Pi=1
n
2
i=1 xi = 228.503
60
7
41
5310
8544332211
99998888776666444333211
0011113355555566777788
14556678
023499
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Table 1: Stem-and-leaf of ENI stock price variations (%).
Q1 Compute the percentiles of the order 25% and 75%.
x0.25 = (x(20) + x(21) )/2 = −0.9, x0.75 = (x(60) + x(61) )/2 = 0.7
Q2 Consider a variation equal to −3.0%. What is the position of such a data with respect to the
observed sample (central, left or right tail, outlier)?
Left tail, but not an outlier as Tukey’s lower fence is −3.3.
Q3 Compute the sample mean and the sample standard deviation of ENI stock price variations.
x̄n = −0.183, sX = 1.691
Q4 Derive from the stem-and-leaf display the % of data in the interval x̄n ∓ 2sX . How does it
compare with the corresponding % from the empirical rule of areas under normality?
% of data in x̄n ∓ 2sX : 76 observations, corresponding to 95%
Comparison with empirical rule: result very close to empirical rule.
Exercise 2
1. Is there a relation between ENI stock price variations (X, %) and FTSE MIB index variations 1
(Y , %)? Figure 1 shows the joint distribution of X and Y in the sample of 80 units of Ex. 1. Table
2 gives the summary statistics.
Q1 Compute means, standard deviations, covariance and correlation and report the results in
Table 2.
See Table 2.
Q2 Count from the scatter plot the number of points with both X and Y lower than −3.
2 points.
Q3 Compute the coefficients of the least square line ŷ = α̂ + β̂x and the goodness-of-fit measure.
α̂ = 0.185 , β̂ = 0.689
GOF: R2 = (rX,Y )2 = 0.605
1 FTSE
MIB index is a summary index of Milan stock market.
n
80
x
−0.183
Pn
xi
−14.632
y
0.0594
i=1
Pn
i=1
4.751
sX
1.691
yi
Pn
x2i
228.503
sY
1.498
i=1
Pn
yi2
177.530
sX,Y
1.970
i=1
Table 2: Summary statistics.
Pn
i=1 xi yi
154.792
rX,Y
0.778
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FTSE MIB index variation (%)
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−6
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0
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ENI stock price variation (%))
Figure 1: Scatter plot.
Q4 Compute the prediction of FTSE MIB index variation for a variation of ENI stock price equal
to +1% and evaluate the corresponding prediction error.
√
Prediction: ŷ(x = 1) = 0.874, Prediction error: se = sY 1 − R2 = 0.941
Exercise 3 Suppose to sample at random n values X1 , ..., Xn from a standard normal distribution.
Q1 Put n = 20. What is the probability that exactly 10 observations out of 20 are positive?
Due to properties of standard normal distribution, the problem
amounts to compute the
20
probability of 10 heads when a fair coin is flipped 20 times, 20
(0.5)
= 0.1761971, using the
10
binomial distribution.
Pn
Q2 Put again n = 20 and let Tn = =1 Xi . What are the expectation and the standard deviation
of Tn ?
√
√
Since the standard normal has µ = 0 and σ = 1, E(Tn ) = nµ = 0, SD(Tn ) = nσ = 20 '
4.472.
Pn
Q3 Put n = 100 and let X̄n = ( =1 Xi )/n. Evaluate the probability of the event |X̄n | < 0.05.
√
Using sampling theory, E(X̄n ) = µ = 0 and SD(X̄n ) = σ/ n = 1/10. By the CLT,
P (|X̄n | < 0.05) = P (−0.05 < X̄n < 0.05) ' P (−0.5 < N (0, 1) < 0.5) = 0.383.
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Exercise 4 Consider again the sample of FTSE MIB index variations in Ex. 2 and denote with µ the average
variation in the reference population.
Q1 What is the point estimate of µ and what is the measure of the sampling error?
d Ȳn ) = sY /√n = 0.167.
Point estimate of µ: ȳn = 0.0594, measure of sampling error: SE(
Q2 Derive the confidence interval for µ (confidence level 95%).
d Ȳn ) = (−0.268, 0.387)
ȳn ∓ z0.975 SE(
Q3 Consider the test H0 : µ = 0, H1 : µ 6= 0 (significance level 5%). What is the observed
value of the test statistic and what is the rejection region of the test? Do the results involve
rejection of H0 ? Explain carefully.
d Ȳn )|H0 = 0.356, Rejection region: |T S| ≥ z0.975
Observed value of test statistic: (ȳn −µ)/SE(
The results do not involve rejection of H0 because the observed value of test statistic T S
does not belong to the rejection region of the test. In fact, once the sampling error is taken
into account, the observed value of sample average is fully consistent with the null hypothesis
µ = 0.
Exercise 5 Consider again the sample of ENI stock price variations in the stem-and-leaf display of Table 1 and
denote with pA the proportion of positive variations in the reference population. We want to test
the null hypothesis H0 : pA = 0.5 against the alternative H1 : pA 6= 0.5 (significance level 5%).
Compute the observed value of the test statistic and describe the rejection region of the test. What
is the final decision, rejection or non rejection of H0 ?
From Table 1, the point estimate of pA is p̂A = 38/80 = 0.475
Observed value of test statistic: (p̂A − pA )/SE(p̂A )|H0 ' −0.447, Rejection region: |T S| ≥ z0.975
Final decision: Do not reject H0