Download Exam 1 and solution

The University of Texas at San Antonio
Department of Management Science and Statistics
STA 3523, Mathematical Statistics
Instructor: Victor De Oliveira
February 11, 2010
Name:
EXAM 1: SOLUTION
1. A study on the pedaling technique of endurance cyclists reported the following data on single-led
power at a high workload: 244 191 160 187 180 176 174 205 211 183 211 180 194 200
Compute
(a) The sample mean 192.571
(c) The sample standard deviation
(b) The sample median 189
(d) The sample interquartile range
20.861
25
(4 points)
2. Identify the following as a parameter, statistic or neither
(a) X̄ + 2.011S statistic
(c) X̄ − µ neither
(b) σ/µ parameter
(d) P (X > 1) parameter
(2 points)
3. Suppose that the scores for a particularly difficult exam are normally distributed with mean 60 and
standard deviation 10.
(a) What is the probability that the average score of 49 students is larger than 62 ?
(2 points)
(b) The instructor decides to curve the grades so that the top 10% of the class receives a grade of A.
What is the ‘cut off’ score that separates the A’s from the B’s ?
(2 points)
Answer.
(a) P (X̄ > 62) = 1 − P (X̄ ≤ 62) = 1 − P (Z ≤ 62−60
10/7 ) = 1 − Φ(1.4) = 0.0808.
a−60
a−60
(b) 0.9 = P (X ≤ a) = Φ( 10 ), so 10 = 1.285 and a = 72.85.
4. A certain machine makes electrical resistors having a mean resistance of 40 ohms and a standard
deviation of 2 ohms. What is (approximately) the probability that a random sample of 36 of these
resistors will have a combined (total/sum) resistance of more than 1458 ohms ?
(2 points)
Answer.
Let X1 , . . . , X36 be the individual resistances, with E(X i ) = 40 and var(Xi ) = 4, and T = 36
i=1 Xi be
approx
the combined resistance. By the CLT we have
that T ∼ N (1440, 144). Hence
P (T > 1458) = 1 − P (T ≤ 1458) = 1 − P Z ≤ 1458−1440
≈ 1 − Φ(1.5) = 1 − 0.9332 = 0.0668.
12
P
5. Use the appropriate table to compute the following:
(a) t0.025,12 2.179
(b) χ20.8,15 10.307
(1 point)
2
STA 3523, Mathematical Statistics
EXAM 1: SOLUTION
6. Decide whether the following statements are true or false:
(a) Let CI95 and CI99 be confidence intervals for θ computed from the same data set, with confidence
levels 95% and 99%, respectively. Then CI 95 is narrower than CI99 . True
(b) Based on 1000 samples of size 30 (from the same population) we compute 1000 90% confidence
intervals for the population mean µ. Then exactly 900 of those intervals will contain the true value of
µ. False
(c) Let T1 and T2 be two unbiased estimators of a parameter θ. Then 14 T1 + 43 T2 is also an unbiased
estimator of θ. True
(3 points)
7. A random sample of 110 lightning flashes in a certain region resulted in a sample mean radar echo
duration of 0.81 sec and a sample standard deviation of 0.34 sec.
(a) Calculate a 99% confidence interval for the true mean echo duration µ.
(2 points)
(b) Calculate an upper confidence bound for µ using a confidence level of 90%.
(2 points)
(c) Suppose that you had believed a priori that the population standard deviation was about 0.4 sec.
Based on this supposition, how many lightning flashes would have been recorded in that region to
estimate µ within 0.05 sec with 95% confidence ?
(2 points)
Answer.
z0.005
√ S = 0.81 ± (2.575)(0.34)
√
n
110
(1.28)(0.34)
z√
0.1 S
√
=
=
0.81
+
n
110
2
(a) x̄ ±
(b) x̄ +
(c) 2(1.96)(0.4)/0.1
= (0.727, 0.893).
0.851.
= 245.86, so n = 246.
8. Let X1 , . . . , Xn be a random sample from the geometric distribution with pmf
fθ (x) =
(
θ(1 − θ)x if x = 0, 1, 2, . . .
,
0
otherwise
θ ∈ (0, 1).
Find the maximum likelihood estimate of θ.
(3 points)
Answer.
L(θ) =
Qn
Pn
xi
n
θ)
i=1 θ(1 − θ) = θ (1 − P
solving for θ̂ in 0 =
l0 (θ̂)
=
n
θ̂
−
i=1
n
i=1
1−θ̂
xi
xi
so l(θ) = log(L(θ)) = n log(θ) + (
, we get θ̂ =
1
1+x̄ .
Pn
i=1
xi ) log(1 − θ). Hence