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Solutions to STAT 100 Sample Exams: Probability and Statistics, Exams of Probability and Statistics

University of MarylandProbability and Statistics

Solutions to sample exams for a statistics 100 course, covering topics such as probability distributions, hypothesis testing, and confidence intervals.

Typology: Exams

Pre 2010

Uploaded on 02/13/2009

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Nate Strawn
Fall 08
SAMPLE FINALS
Solutions
Cremins
STAT 100
Solutions to Sample Exam Spring 08
1. Ais Strep, Bis Allergy
(a) P(A∪B) = .45, P(AB) = 0.1, P(AB = 0.15, P(AB) = 0.3
(b) P(A∪B) = P(A) + P(B)−P(AB) = 0.25 + 0.4−0.1 = 0.55
(c) P(A|B) = P(AB)/P (B) = 0.1/0.4 = 0.25
(d) Since P(A) = 0.25, P(A|B) = P(A) and independence holds
2. (a)
cP(X=c)
1 3/6
5 2/6
10 1/6
(b) E(X) = (1)(3/6) + (5)(2/6) + (10)(1/6) = 32/5 = 6.4
3. X=number of girls that expect a female president has binomial distribution with parameters
n= 16, p= 0.4
(a) P(X≥9) = 0.142
(b) P(16 −X≤7) = P(X≥9) = 0.142
(c) E(X) = np = (16)(0.4) = 6.4
(d) P(X < 60) = P(X≤59) ≈P(Z≤(59.5−200(0.4))/p200(0.4)(0.6)) = P(Z≤ −2.96) =
0.0015
4. X∼N(2084,300).
(a) P(X≥2500) = P(Z≥(2500 −2084)/300) = P(Z≥1.39) = 1 −P(Z < 1.39) = 0.0823
(b) P(1800 ≤X≤2400) = P(−0.95 ≤Z≤1.17) = P(Z≤1.17) −P(Z < −0.95) = 0.7079
(c) P(X≤Q0.8) = 0.8, so Q0.8=.85
(d) X∼N(2084,300/√40), so P(X≥2200) = P(Z≥(220 −2084)/(300/√40)) = P(Z≥
2.45) = 0.0071.
5. (a) H0:p= 0.8 vs H1:p6= 0.8. n= 200, X= 156, so ˆp= 156/200 = 0.78.
Z=ˆp−p0
qp0(1−p0)
n
=0.78 −0.80
q0.8(0.2)
200
=−0.71
Rejection region is R:|Z| ≥ zα/2=z0.02/2=z0.01 = 2.33. Znot in rejection region, so
RETAIN H0.
(b) P(|Z|>0.71) = 2P(Z≤ −0.71) = 0.4778.
6. (a) n= 8, so SMALL SAMPLE. Thus, 90% CI is
X±tα/2
S
√n= 10.05 ±2.362.4854
√8= 10.05 ±2.07
pf3

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Fall 08 Solutions STAT 100

Solutions to Sample Exam Spring 08

  1. A is Strep, B is Allergy (a) P (A ∪ B) = .45, P (AB) = 0.1, P (AB = 0.15, P (AB) = 0. 3 (b) P (A ∪ B) = P (A) + P (B) − P (AB) = 0.25 + 0. 4 − 0 .1 = 0. 55 (c) P (A|B) = P (AB)/P (B) = 0. 1 / 0 .4 = 0. 25 (d) Since P (A) = 0.25, P (A|B) = P (A) and independence holds
  2. (a)

c P (X = c) 1 3/ 5 2/ 10 1/ (b) E(X) = (1)(3/6) + (5)(2/6) + (10)(1/6) = 32/5 = 6. 4

  1. X =number of girls that expect a female president has binomial distribution with parameters n = 16, p = 0. 4 (a) P (X ≥ 9) = 0. 142 (b) P (16 − X ≤ 7) = P (X ≥ 9) = 0. 142 (c) E(X) = np = (16)(0.4) = 6. 4 (d) P (X < 60) = P (X ≤ 59) ≈ P (Z ≤ (59. 5 −200(0.4))/

200(0.4)(0.6)) = P (Z ≤ − 2 .96) =

4. X ∼ N (2084, 300).

(a) P (X ≥ 2500) = P (Z ≥ (2500 − 2084)/300) = P (Z ≥ 1 .39) = 1 − P (Z < 1 .39) = 0. 0823 (b) P (1800 ≤ X ≤ 2400) = P (− 0. 95 ≤ Z ≤ 1 .17) = P (Z ≤ 1 .17) − P (Z < − 0 .95) = 0. 7079 (c) P (X ≤ Q 0. 8 ) = 0.8, so Q 0. 8 =. 85 (d) X ∼ N (2084, 300 /

40), so P (X ≥ 2200) = P (Z ≥ (220 − 2084)/(300/

40)) = P (Z ≥

  1. (a) H 0 : p = 0.8 vs H 1 : p 6 = 0.8. n = 200, X = 156, so ˆp = 156/200 = 0.78.

Z = √pˆp^ − 0 (1^ p−^0 p 0 ) n

=^0.^ √^780 −.8(0^0 .2).^80

200

Rejection region is R : |Z| ≥ zα/ 2 = z 0. 02 / 2 = z 0. 01 = 2.33. Z not in rejection region, so RETAIN H 0. (b) P (|Z| > 0 .71) = 2P (Z ≤ − 0 .71) = 0.4778.

  1. (a) n = 8, so SMALL SAMPLE. Thus, 90% CI is

X ± tα/ 2 √^ Sn = 10. 05 ± 2. 362.^ √^48548 = 10. 05 ± 2. 07

Fall 08 Solutions STAT 100

(b) Since 11 is in previous CI, RETAIN H 0.

  1. LARGE SAMPLES, so H 0 : μ 1 − μ 2 = 0 vs H 1 : μ 1 − μ 2 < 0. Then

Z = √ X^ −^ Y S 12 /n 1 + S 22 /n 2

= √^3.^9 −^5.^2

Rejection region is R : Z ≤ −zα = − 1 .645, so REJECT H 0.

  1. D 1 = − 62 , D 2 = − 63 , D 3 = − 55 , D 4 = − 58 , D 5 = −57. Thus, D = −59 and SD = 3.3912. n = 5 is SMALL SAMPLE, so 99% CI is

D ± tα/ 2 √^ SDn = − 59 ± 5. 8413. √^3912 5

  1. H 0 : pmon = 1/ 5 , ptue = 1/ 5 , pwed = 1/ 5 , pthu = 1/ 5 , pf ri = 1/5, χ^2 = 23. 1833 , R : χ^2 ≥ χ^2 α = 13.277, so REJECT H 0.
  2. (a) False. This is Type II error. (b) False. Null is retained in this situation. (c) True. Slope is Sxy/Sxx and r = Sxy/

SxxSyy, so both have the same sign. (d) Can’t be determined. P (A ∪ B) = P (A) + P (B) in this case, so will be true if P (B) = P (A)/(1 − P (A)).

Solutions to Sample Exam Fall 07

  1. (a) x = 1.6615 and S = 0. 6564

(b)

stem leaf

  1. 5
  2. 8444220
  3. 76400 (c) n = 13 is SMALL SAMPLE, so 95% CI is

x ± tα/ 2 √^ Sn = 1. 6615 ± 2. 1790.^ √^656413 = 1. 6615 ± 0. 3967

(d) 2.1 is not in the previous CI, so REJECT H 0.

  1. (a) #B=8, #A=13, #BA=2, so #AB=11 and #AB=6. (b) P (AB) = 6/24 = 1/ 4 (c) P (AB) = 11/ 24 (d) P (A|B) = P (AB)/P (B) = 2/8 = 1/4 and P (A) = 13/24. Thus, P (A|B) 6 = P (A), so NOT independent.