AMA1501 Examination 2012/2013 Semester 1 Outline Suggested Solution
Department of Applied Mathematics AMA1501 Introduction to Statistics for Business Examination 2012/2013 Semester 1 Outline Suggested Solution
1 (a)
Class mark (x) 2 4 5.5 6.5 7.5 9 11
f 80 Mean =
fx 561.5 fx
561.5 80
Mode = 6
2
Frequency, f 1 2 14 29 20 9 5
4143.75
7.0188 minutes 29 14
29 14 29 20
Standard deviation =
7 6 6.625 minutes
80 4143.75 561.5 2 80 80 1
1.6019 minutes
1 (b)
Time spent less than (minutes) 1 3 5 6 7 8 10 12 64 4 6 8 7 7.9 minutes D8 7 20
1 (c)
Estimated Pr(time spent between 6.5 minutes and 8.5 minutes) is 6.5 8.5 8 7 6. 7 6 29 20 10 8 9 80 0.4594 Let X be the number of students complete the online test within 6.5 minutes and 8.5 minutes, out of 5 X~B(5,0.4594) 5 5 5 x x Pr X 3 1 0.4594 1 0.4594 0.8592
x 4
1 (d)
x
A 95% confidence interval for is 7.0188 1.96 1.6019
2 (a)
Cumulative frequency 0 1 3 17 46 66 75 80
5
C1 4 C1 3 C 1 12
C 3
80 , i.e. 6.6677 7.3698 (minutes)
5 4 3 220
3 11
1
AMA1501 Examination 2012/2013 Semester 1 Outline Suggested Solution
2 (b) (i)
A: participant rated ‘the course is useful’ B: participant has registered advanced level courses Pr(A) = 0.85 Pr(B) = 0.75 Pr(A|B) = 0.9 Pr A B 1 0.85 0.75 0.75 0.9 0.075
2 (b)(ii)
Pr B A
2 (c)
0.75 0.9 0.85
0.7941
A: company A is chosen B: company B is chosen C: company C is chosen L: Late delivery Pr A = 0.4 Pr B = 0.35 Pr C =0.25
Pr L A 0.015
Pr L B 0.025 =
Pr L C 0.03
0.35 0.025
Pr B L
0.3933 0.4 0.015 0.35 0.025 0.25 0.03 Let X be the number deliveries that was handled by company B, given that these three are late deliveries X~B(3,0.3933) 3 2 Pr X 1 0.3933 1 0.3933 0.4343 1 3 (a)(i)
Let X be the random variable of the monthly repair and maintenance expenses. Pr 80000 X 110000 Pr 1.67 Z 0.83
1 0.0475 0.2033 0.7492 3 (a)(ii) 3 (a)(iii)
P(Z ≥ a) = 0.05 => a = 1.645 100000+(1.645)12000 = 119740 < 128000 so it is on the list Pr X 78000 Pr Z 1.83 0.0336 Let Y be the number of months that the expenses are below $78000 Y ~ b(6, 0.0336) 6 6 5 6 0 Pr Y 5 0.0336 0.9664 0.0336 0.9664 2.4975 10 7 5 6
3 (b)
Let X be the number of citizens who will vote on the Election Day. X ~ b(200, 0.45) Since n>30, 200(0.45) = 90 > 5, 200(0.55) = 110 > 5 and 0.1 < p < 0.9, normal approximation can be used.
Pr X 100 Pr Z 3 (c)
Pr( Z 1.35) 0.0885 200(0.45)(0.55) 99.5 90
Let X be the number of customers coming to the company per day Then X ~ Po(4)
40 41 4 2 4 3 4 4 Pr X 5 e 0.6288 0! 1! 2! 3! 4! 4
Let Y be the number of days that branch A has less than 5 customers per day Y ~ b(7, 0.6288) 7
6
Pr Y 6 7 C7 0.6288 7 C 6 0.6288 1 0.6288 0.1995 2
AMA1501 Examination 2012/2013 Semester 1 Outline Suggested Solution
4 (a)
Let X i be the score of mathematics competence test of Faculty i student
82
10
X A X B ~ N 86 80,
92
10
1 Pr 5 X A X B 5 1 Pr 2.89 Z 0.26
1 0.3974 0.00193 0.60453 4 (b)
Let X be the amount of expenditure in an invoice ($), X ~ N , 2 n H 0 : 240 H 1 : 240
0.05 Critical region: t t 0.05;11 1.796
Under H 0 , test statistic t
280 240 18
12
7.698
Decision: reject H 0 4 (c)
Let p be the population proportion of students who would like to study complementary course A. H 0 : p 0.3 H1 : p 0.3
0.01 Critical region: z z0.01 2.33 pˆ
180 500
Under H 0 , test statistic z
0.36 0.3
2.93
0.3 0.7 500
Decision: Reject H 0 4 (d)
Sample mean number of cars passing through a junction = 565 200 2.825 H 0 : number of cars passing through a junction follows a Poisson distribution H 1 : number of cars passing through a junction does not follow a Poisson
distribution 0.05 Number of cars Oi Ei
0 1 2 3 4 18 28 43 40 36 11.86 33.51 47.33 44.57 31.48
Critical region: 2 12.592, 8 1 1 6 Under H 0 , test statistic 2
Oi E i E i
Decision: Reject H 0
3
2
13.785
5 23 17.79
6 12 8.37
7 0 5.09
AMA1501 Examination 2012/2013 Semester 1 Outline Suggested Solution
5 (a)
Let pmale and p female be the population proportion of male and female, respectively, participating in company activities. pˆ male 72 100 pˆ female 45 80 pˆ 72 45 100 80 117 180 H 0 : pmale p female H1 : pmale p female
0.05 Critical region: z 1.645
72 45 100 80 2.2014 117 63 1 1 180 180 100 80
Under H 0 , test statistic z
Decision: Reject H 0 5 (b)
Let D be the difference in computational time (d = new – existing) 2 D ~ N D , D n d: -9 -13 -6 -8 -13 -17 -14 -8 -16 104 11.5556 d 9 sd
9 1324 104 9 9 1
2
3.9087
H 0 : D 0 H 1 : D 0
0.05 Critical region: t t 0.05;8 1.86
Under H 0 , test statistic t
11.5556 0 3.9087
9
8.869
Decision: Reject H 0 5 (c)
H 0 : grade of employees and their opinion are independent H 1 : grade of employees and their opinion are not independent
0.05 Critical region: 2 9.488, 3 13 1 4
Expected frequencies: Disagree 14.80 23.74 35.46 74
Senior management Middle management Frontline staff Total
Under H 0 , test statistic 2
Oi E i E i
Decision: Reject H 0
4
Neutral 14.20 22.78 34.02 71 2
23.2399
Agree 19.00 30.48 45.52 95
Total 48 77 115 240
AMA1501 Examination 2012/2013 Semester 1 Outline Suggested Solution
6 (a)(i)
b
13 206725 450 5830 13 17250 450
2
2.9391
5830 450 2.9391 346.7241 13 13 yˆ 346.7241 2.9391x a
6 (a)(ii)
2
13 206725 450 5830 R 0.31 13 17250 450 2 13 2661150 5830 2 2
69% variation in weekly sales cannot be explained by the fitted equation on weekly advertising expenditures. 6 (b)(i) 6 (b)(ii)
Sales 5.5957 15.5804 Newspaper+4.1866 Coupon a 0.9438 2144.5875 2024.0617 b 2144.5875 2024.0617 120.5258 c 2 d 12 2 1 9 e 12 1 11
f g h
6 (b)(iii)
2024.0617 2 120.5258 9 1012.03
1012.03
13.3918
75.5712 13.3918 H 0 : 1 2 0 H 1 : at least one i 0
0.05 Critical region: f F 0.05;2,9 4.26
Under H 0 , test statistic f = 75.5712 Decision: Reject H 0 6 (b)(iv)
H 0 : 2 0 H 1 : 2 0
0.01 Critical region: t t0.005;9 3.25 and t 3.25
Under H 0 , test statistic t
4.1866 0.9450
Decision: Reject H 0
5
4.43
