Department of Statistics
& Operations Research
College of Science, King Saud University
STAT 145
Final Exam
First Semester
1434 – 1435 H
:‫اسم الطالب‬
‫رقم التسلسل‬
‫الرقم الجامعي‬
‫اسم الدكتور‬
‫رقم الشعبة‬
INSTRUCTIONS:
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Answer all questions.
Mobile phones are not allowed in the classroom.
Time allowed is 3 Hours.
For each question, put the code of the correct answer in capital letters:
A, B, C, and D in the following table beneath the question number.
Avoid overwritings / corrections in answers
Use boll or ink pen only
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Q1:
Suppose that the proportions of Diabetic (‫ )سكري‬in Saudia and in Emirate are
PS  0.25 , PE  0.30 . If two random samples of nS 100 , nE  150 persons were drawn from
the two populations and let pˆ S , pˆ E be their sample Diabetic proportions; then
1) Find  p̂ S : The mean of the sampling distribution of p̂ S :
(A) 0.0025
(B) 100
(C) 25
(D) 0.25
2) Find  p2ˆ S :The variance of the sampling distribution of p̂ S :
(A) 0.433
(B) 0.001875
(C) 0.0433
(D) 0.1875
3) The approximate distribution of p̂ S :
(A) Normal
(B) t
(C) Poisson
(D) Binomial
4) The probability P( pˆ S  0.22) is
(A) 0.4721
(B) 0.7549
(C) 0.5279
(D) 0.2451
5) If pˆ  pˆ E  pˆ S , then find  p̂ the mean of the sampling distribution of p̂ :
(A) 0.05
(B) 0.55
(C) 0.075
(D) -0.05
6) Find  2p̂ , the variance of the sampling distribution of p̂ :
(A) 0.876
(B) 0.3216
(C) 0.05723
(D) 0.003275
(C) 0.6368
(D) 0.3632
7) The probability P( pˆ E  pˆ S  0.03 ) is
(A) 0.7241
(B) 0.75
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Q2: If the patient stay time (‫ )مدة بقاء المريض‬in a Hospital A and in a Hospital B is normally
distributed with means  A  5.5 ,  B  4.2 days and standard deviations  A 1.3 ,  B 1.1 days. If
X A , X B are the mean stay times of two samples of sizes n A  5 , n B  7 patients from Hospital A
and from Hospital B, respectively, then:
8) The expectation E( X A ) is equal to:
(A) 0.338
(B) 5.5
(C) 1.3
(D) 0.5814
(C) 1.3
(D) 0.5814
9) The variance Var ( X A ) is equal to:
(A) 0.338
(B) 5.5
10) The probability P( 5.1  X A  6.2) is equal to:
(A) 0.5
(B) 0.6398
(C) 0.2341
(D) 0.3602
1
11) If X  X A  X B , then the mean  X is equal to:
(A) 0.338
(B) 5.5
(C) 1.3
(D) 0.5814
12) The variance  2X is equal to:
(A) 0.646
(B) 0.715
(C) 0.417
(D) 0.511
13) The probability P( X  1) is equal to:
(A) 0.2776
(B) 0.7224
(C) 0.6628
(D) 0.3372
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Q3: Let 
be the mean patient stay time (‫ )مدة بقاء المريض‬in a central Hospital, and statistical
inference about  is wanted, so a random sample of size 49 patients stayed in the Hospital was
drawn. The sample gave the mean x  3.2 days and the standard deviation S  2.5 , then:
14) The point estimate of  is:
(A) 49
(B) 0.357
(C) 3.2
(D) 2.5
15) The estimation of standard error of the sampling distribution of x is equal to:
(A) 49
(B) 0.357
(C) 3.2
(D) 2.5
16) The lower bound of 95% confidence interval for 𝜇 is equal to:
) ‫إذا لم تجدي قيمة درجات الحرية بالجدول نأخذ أقرب قيمة له‬: ‫)(مالحظة‬
(df=48 you take df=50)
(A) 2.5
(B) 3.2
(C) 2.843
(D) 2.613
17) The upper bound of 95% confidence interval for  is equal to:
(A) 3.79
 To test H 0 :   3.5
(B) 3.557
(C) 4.0
(D) 3.9
, H 1 :   3.5 with level   0.05 , then
18) The value of the test statistic is equal to:
(A) z = -0.84
(B) t = - 0.84
(C) z = -0.12
(D) t 48 = -0.12
19) The rejection area is equal to:
(A) (,  1.96)
(B) (1.96 , )
(C) (1.6759 , )
(D) (,  1.645)
20) The decision is:
(A) Reject H 0
(B) Accept H 0
(C) No decision
(D) Non is correct
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2
Q4: Two random samples were independently selected from two normal populations with equal
variances. The results are summarized as follows:
Sample size ( 𝑛 )
Sample mean ( x )
Sample variance ( 𝑆 2 )
First Sample
Second Sample
16
9
10.7
3.5
8.5
5
Let 𝜇1 and 𝜇2 be the true means of the first and second populations, respectively
21) A point estimate for 1   2 (or  ) is:
(A) 7
(B) 2.2
22) The pooled variance S p2
(A) 5
(C) 3.5
(D) 5
is:
(B) 3.5
(C) 4.022
(D) 4.24
23) The distribution involved here in the statistical inference about  is:
(A) t 23
(B) t 8
(C) t15
(D) Z
24) The formula of 95% confidence interval for  is equal to:
(A) x  z
1-
(C) x  t


2
(B) x 1 - x 2   z
n

(D) x1 - x 2   t
S
1- ,n 1
2
n
1-
 12

n1
2


1- ,n1  n2  2 
2
 22
n2
Sp
1
1

n1 n2
25) The lower bound of 95% confidence interval for  is equal to:
(A) 0.543
(B) 0.441
(C) 0.4752
(D) 0.4711
26) The upper bound of 95% confidence interval for  is equal to:
(A) 3.929
(B) 3.925
(C) 4.0
(D) 3.90
 To Test H 0 : 1   2 H A : 1   2 with level   0.05 , give :
27) The value of the test statistic
(A) Z  2.2
(B) t 23  2.2
(C) t15  2.2
(D) t 23  2.633
28) The acceptance region of H 0 is:
(A) (-1.645, 1.645)
(B) (-1.96, 1.96)
(C) (-2.069, 2.069)
(D) (-1.714, 1.714)
29) The decision is:
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(A) Accept H 0
Q5: Let P
R
(B) Reject H 0
(C) No decision
(D) Non is correct
, PJ be the proportions of vaccinated children in Riyadh and in Jeddah, and statistical
inference about PR , PJ is wanted, so two random samples were drawn from the two populations
of sizes n R  400 , n J  500 children which showed 350 children from Riyadh and 415 children
from Jeddah were vaccinated. Denote by pˆ R , pˆ J to the sample proportions of vaccinated
children in Riyadh and in Jeddah, then:
30) A point estimate of the proportion PR is:
(A) 400
(B) 350
(C) 0.875
(D) 0.125
31) Find ( S pˆ R ), the estimation of the standard error of the of p̂R :
(A) 0.0002734
(B) 0.01654
(C) 0.875
(D) 0.125
32) The lower bound of 95% confidence interval for PR is equal to:
(A) 0.8426
(B) 0.8478
(C) 0.875
(D) 0.125
33) The upper bound of 95% confidence interval for PR is equal to:
(A) 0.9022
(B) 0.875
(C) 0.125
(D) 0.9074
 To Test H 0 : PR  0.90 H A : PR  0.90 with level   0.05 , give :
34) The value of the test statistic is equal to:
(A) Z  1.67 (B) t  1.67
(C) t  1.511
(D) Z  1.511
35) The rejection region of H 0 is equal to:
(A) (1.645 , )
(B) (,  1.28)
(C) (,  1.96)
(D) (,  1.645)
36) The decision is:
(A) Accept H 0
(B) Reject H 0
(C) No decision
(D) Non is correct
H A : PR  PJ with level   0.05 , give :
37) The value of the pooled proportion p is equal to:
 To Test H 0 : PR  PJ
(A) 0.85
(B) 0.8525
(C) 0.875
(D) 0.83
38) The value of the test statistic is equal to:0.02395
(A) Z  1.91 (B) Z  1.91
(C) Z  1.88
(D) Z  1.88
39) The acceptance region of H 0 is equal to:
(A) (1.645 ,1.645)
(B) (,  1.645)
(C) (1.96 ,1.96)
(D) (1.96 , )
40) The decision is:
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(A) Accept H 0
(B) Reject H 0
(C) No decision
(D) Non is correct
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