Chapter 17 - Sampling and Estimation: Solutions To Exercise 17A
Chapter 17 - Sampling and Estimation: Solutions To Exercise 17A
Chapter 17 - Sampling and Estimation: Solutions To Exercise 17A
739
Solutions to Exercise 17B
5 1
1 a p= =
10 2
1 2
b 0, , , 1
3 3
55
0 3 1
c Pr(P̂ = 0) = 10 =
12
535
1 2 1 5
Pr(P̂ = ) = 10 =
3 12
3
5 5
2 2 1 5
Pr(P̂ = ) = 10 =
3 12
535
3 0 1
Pr(P̂ = 1) = 10 =
12
3
1 2
p̂ 0 1
3 3
1 5 5 1
Pr(P̂ = p̂)
12 12 12 12
5 1 1
d Pr(P̂ > 0.5) = + =
12 12 2
12 3
2 a p= =
20 5
1 2 3 4
b Values of P̂ : 0, , , , , 1
5 5 5 5
1 2
p̂ 0 5 5
740
Pr(0 < P̂ < 0.7)
e Pr(P̂ < 0.7|P̂ > 0) = = 0.6924
Pr(P̂ > 0)
3 a p = 0.5
1 2
b Values of P̂ : 0, , , 1
2 3
1 2
p̂ 0 1
c 3 3
Pr(P̂ = p̂) 0.1 0.4 0.4 0.1
4 a p = 0.4
1 2
b Values of P̂ : 0, , , 1
3 3
1 2
p̂ 0 1
c 3 3
1 1 3 1
Pr(P̂ = p̂)
6 2 10 30
1
d Pr(P̂ > 0.5) =
3
e Pr(P̂ < 0.5|P̂ > 0)
Pr(0 < P̂ < 0.5) 4
= =
Pr(P̂ > 0) 5
5 a p = 0.5
1 1 3
b Values of P̂ : 0, , , , 1
4 2 4
1 1 3
p̂ 0 1
c 4 2 4
1 1 3 1 1
Pr(P̂ = p̂)
16 4 8 4 16
5
d Pr(P̂ > 0.7) =
16
741
1 2 3 4
6 a Values of P̂ : 0, , , , , 1
5 5 5 5
b 1 2 3 4
p̂ 0 1
5 5 5 5
1 5 5 5 5 1
Pr(P̂ = p̂)
32 32 16 16 32 32
3
c Pr(P̂ < 0.4) =
16
d Pr(P̂ > 0|P̂ < 0.8)
Pr(0 < P̂ < 0.8) 25
= =
Pr(P̂ < 0.8) 26
1 1 3
7 a Values of P̂ : 0, , , , 1
4 2 4
1 1 3
p̂ 0 1
b 4 2 4
256 256 96 16 1
Pr(P̂ = p̂)
625 625 625 625 625
1 1 3
p̂ 0 1
8 4 2 4
1 1 3 1 1
Pr(P̂ = p̂)
16 4 8 4 16
1 1 1 1 3 3 1 3
E(X) = 0 × + × + × + × +1× = 0.5
16 4 4 2 8 4 4 16
1 1 2 1 1 2 3 3 2 1 2 1 5
E(X 2 ) = 02 × + × + × + × + 1 × =
516 41 2 41 2 8 4 4 16 16
∴ Var(X) = − =
16 2 16
1
∴ sd(x) =
4
1 2 3 4
p̂ 0 1
9 5 5 5 5
1 5 5 5 5 1
Pr(P̂ = p̂)
32 32 16 16 32 32
1 1 5 2 5 3 5 4 5 1
E(X) = 0 × + × + × + × + × +1× = 0.5
32 5 32 5 16 5 16 5 32 32
742
1 1 2 5 2 2 5 3 2 5 4 2 5 2 1
E(X ) = 0 × +
2 2
× + × + × + × + 1 × = 0.340176
32 5 32 5 16 5 16 5 32 32
1 2
∴ Var(X) = 0.340176 − = 0.050176
2
∴ sd(X) = 0.224
µ = 0.5, σ = 0.224
1 1 3
p̂ 0 1
10 4 2 4
256 256 96 16 1
Pr(P̂ = p̂)
625 625 625 625 625
256 1 256 3 16 1
E(X) = 0 × + × + × +1× = 0.2
625 4 625 4 625 625
256 1 2 256 3 2 16 2 1
E(X ) = 0 ×
2 2
+ × + × + 1 × = 0.08
625 4 625 4 625 32
2
∴ Var(X) = 0.08 − 0.02 = 0.04
∴ sd(X) = 0.2
µ = 0.2, σ = 0.2
r r
p(1 − p) 0.3 × 0.7
11 n = 30, p = 0.4, µ = p = 0.3, σ = = = 0.084
n 30
µ = 0.3, σ = 0.084
r r
p(1 − p) 0.4 × 0.6
12 n = 100, p = 0.4, µ = p = 0.4, σ = = = 0.049
n 100
µ = 0.4, σ = 0.049
r r
p(1 − p) 0.2 × 0.8
13 n = 100, p = 0.2, µ = p = 0.2, σ = = = 0.04 µ = 0.2,
n 100
σ = 0.04
14 a p = 0.65, n = 20
Pr(P̂ = 0.65) = Pr(X = 13) = 0.1844
r r
p(1 − p) 0.65 × 0.35
b µ = 0.65, σ = = = 0.1066
n 20
µ − σ = 0.543
µ + σ = 0.757
743
Pr(0.543 < P̂ < 0.757) = Pr(10.86 < X < 15.14)
= Pr(11 ≤ X ≤ 15)
= 0.7600
c µ − 2σ = 0.4368
µ + 2σ = 0.8632
Pr(0.4368 < P̂ < 0.8632) = Pr(8.74 < X < 17.26)
= Pr(9 ≤ X ≤ 17)
= 0.9683
744
Solutions to Exercise 17C
1 p = 0.5, n = 50
r 5 p = 0.3, n = 50
r
p(1 − p) p(1 − p)
µ = 0.5, σ = = µ = 0.3, σ =
r n r n
0.5 × 0.5 0.3 × 0.7
= 0.0707 = = 0.0648
50 50
0.46 − 0.5 The calculation can also be done directly
Pr(P̂ < 0.46) ≈ Pr(Z ≤ )=
0.7070 with calculator:
Pr(Z ≤ −0.5658)
Pr(P̂ < 0.2) ≈ 0.0614
The calculation can also be done directly
with calculator:
Pr(P̂ < 0.46) ≈ 0.2858 6 p = 0.6, n = 100
r
p(1 − p)
µ = 0.6, σ =
n
2 p = 0.12, n = 300
r
0.6 × 0.4
= = 0.0490
r
p(1 − p)
µ = 0.12, σ = 100
n
a Pr(P̂ < 0.8) ≈ 1
r
0.12 × 0.88
= = 0.018762
300
The calculation can also be done directly b Pr(0.6 < P̂ < 0.8) ≈ 0.5
with calculator:
Pr(P̂ > 0.1) ≈ 0.8568 c Pr(0.7 < P̂ < 0.8|P̂ > 0.6)
Pr(0.7 < P̂ < 0.8)
= ≈ 0.0412
Pr(P̂ > 0.6)
3 p = 0.5, n = 25
r
p(1 − p)
µ = 0.5, σ = 7 p = 0.5, n = 100
n r
r p(1 − p)
0.5 × 0.5 µ = 0.5, σ =
= = 0.1 n
25 r
The calculation can also be done directly 0.5 × 0.5
= = 0.05
with calculator: 100
The calculation can also be done directly
Pr(P̂ > 0.6) ≈ 0.1587
with calculator:
Pr(0.4 < P̂ < 0.6) ≈ 0.9545
4 p = 0.1, n = 200
r
p(1 − p)
µ = 0.1, σ = 8 p = 0.1, n = 1000
n r
r p(1 − p)
0.1 × 0.9 µ = 0.1, σ =
= = 0.0212 n
200 r
The calculation can also be done directly 0.1 × 0.9
= = 0.0095
with calculator: 1000
Pr(P̂ > 0.15) ≈ 0.0092 a Pr(0.08 < P̂ < 0.12) ≈ 0.9650
745
b Pr(0.08 < P̂ < 0.12|P̂ > 0.10) = 10 p = 0.9, n = 250
Pr(0.08 < P̂ < 0.12)
≈ 0.9650 212
Pr(P̂ < 0.12) a p̂ = = 0.848
250
r
9 p = 0.52, n =400 p(1 − p)
b µ = 0.9, σ = = 0.0190
n
230 Pr(P̂ ≤ 0.848) ≈ 0.0031
a p̂ = = 0.575
400
r c Yes, because the chance of the battery
p(1 − p)
b µ = 0.52, σ = = 0.0350 lasting only this short period of time
n is very small if the manufacturers
Pr(P̂ ≥ 0.575) ≈ 0.0139 claim is correct.
746
Solutions to Exercise 17D
1 a 0.08 Since n must be an integer larger than
the calculated value to ensure the margin
b (0.0268, 0.1332) of error is no more than 0.05, n = 246
2 a 0.192 9 p∗ = 0.30
b (0.1432, 0.2408) M = 0.03
1.96 2
n= × 0.3 × 0.7 = 896.37
0.03
Since n must be an integer larger than
3 a 0.2
the calculated value to ensure the
b (0.1216, 0.2784) margin of error is no more than 0.03,
n = 897
1.96 2
4 (0.28395, 0.3761) ba M = 0.02, × 0.3 × 0.7 = 2017
0.02
c Reducing margin of error by 1%
5 a (0.4761, 0.5739)
requires the sample size to be more
b (0.5095, 0.5405) than doubled
747
11 90%: (0.5194, 0.6801), 12 90%: (0.5111, 0.5629),
95%: (0.5034, 0.6940), 95%: (0.5061, 0.5679),
99%: (0.4738, 0.7262); Interval 99%: (0.4964, 0.5776); Interval
width increases as confidence level width increases as confidence level
increases increases
748
Solutions to Technology-free questions
1 a All employees of the company 0.588
∴M= √
n
b p = 0.35
c Margin of√error would decrease by a
c p̂ = 0.40 factor of 2
k 6 a 50 × 0.95 = 45
3 a
100
r b
b p̂ ± 1.96
p̂(1 − p̂ Pr(Y ≥ 49) = Pr(Y = 49) + Pr(Y = 50)
n ! !
50 50
= (0.1) (0.9) +
1 49
(0.1)0 (0.9)50
s
k k
k 100 (1 − 49 50
= ± 1.96 100
100 100 = 5(0.9)49 + (0.9)50
√
k 1.96 k(100 − k) = 5.9(0.9)49
= ±
100 1000
7 a p̂ = 0.60
4 a p̂ = 0.9
r b M = 0.10
p̂(1 − p̂
b M = 1.96
100 c Increase sample size
r
0.9 × 0.1
= 1.96 ×
n
0.3
= 1.96 × √
n
749
Solutions to multiple-choice questions
1 B This class is a sample of the whole 8 E I the centre of a confidence
school population, so any statistics interval is a sample parameter not a
determined from this sample is population parameter
called a sample statistic. II the bigger the margin of error the
bigger the confidence interval
2 C When the statistics is calculated III a point estimate is a single value
from the whole population it is estimate like p̂
known as a population parameter. IV the sample proportion a point
estimate
3 D All we can say about a 95%
confidence interval is that 95% of 9 C Since the width of the confidence
such intervals will capture the true interval is inversely proportional to
mean. Statement B is a common in- the square root of the sample size,
correct interpretation of a confidence increasing the sample size by a factor
interval. of 4 decreases the width by a factor
r
p̂(1 − p̂ of 2.
4 E M = 1.96 1.96 2
100
r 10 E M = 0.03 n = × 0.3 × 0.7 =
0.3 × 0.7 0.03
= 1.96 × 896.37 ≈ 897
50
= 0.1270 11 A See definitions
750
Solutions to extended-response questions
1.96 2
1 a n= p∗ (1 − p∗ ) 0 ≤ p∗ ≤ 1
M
1.96 2
= p∗ (1 − p∗ )
0.02
= 96042 p∗ (1 − p∗ )
n
2401
0 p*
0.5 1
b From the graph, the maximum occurs when p∗ = 0.5
c If they use the maximum samples size (2401) then they will ensure the margin of
error stays within the desired range of ±2%
2 p = 0.6, n = 100 r
p̂(1 − p̂
µ = p̂ = 0.6, σ = = 0.0490
1n
3 a p̂ = 0.57, n = 100
95% CI = (0.4730, 0.6670)
b i Pr(Y = 5) = 55 (0.95)5 (0.05)0 = 0.7738
5
ii Pr(Y = 0) = 0 (0.95)0 (0.05)5 = 0.0000003
iv 0.95 × 5 = 4.75
c n = 500
X = 57 + 67 + 72 + 55 + 60 = 311
751
311
p̂ = = 0.622
500
CI = (0.5795, 0.6645)
500
4 a p=
N
60
b p̂ = = 0.15
400
500 500
c = 0.15 N ≈ = 3333.33 ≈ 3333
N 0.15
d 95% CI for r
p̂ r
0.15 × 0.85 0.15 × 0.85
0.15 − 1.96 < p < 0.15 + 1.96
r 400 r 400
0.1275 0.1275
0.15 − 1.96 < p < 0.15 + 1.96
400 400
500
e 0.1150 < < 0.1850
N
N
5.4056 < < 8.6951
500
2703 < N < 4348
752