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2015 3 Joh PT A

The document contains statistical calculations and analyses, including interquartile range, probability of events, normal approximation, confidence intervals, hypothesis testing, and chi-square tests. It discusses various scenarios involving weights, grades, and sales of petrol, providing results and conclusions based on statistical methods. The findings indicate relationships between variables and support or reject specific hypotheses based on calculated test statistics.

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Xue Yi Lam
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26 views4 pages

2015 3 Joh PT A

The document contains statistical calculations and analyses, including interquartile range, probability of events, normal approximation, confidence intervals, hypothesis testing, and chi-square tests. It discusses various scenarios involving weights, grades, and sales of petrol, providing results and conclusions based on statistical methods. The findings indicate relationships between variables and support or reject specific hypotheses based on calculated test statistics.

Uploaded by

Xue Yi Lam
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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2015-3-JOH-PT-A

1 a N  15 b) Interquartile range=Q3-Q1=14-8=6
) N 15 3 Upper boundary=14+1.5(6)=23
 3 Lower boundary=8-1.5(6)= -1
4 4 4
Q1  r4  8 
N 15 1 outlier
 7
2 2 2
Q1 Q2 Q3
Q2  r8  10
3 N 3(15) 1
  11
4 4 4
Q3  r12  14
2 1 5 1
P ( A )= , P ( B )= , P ( C ) =
2 9 6
P ( A ∩ C )=0
Events A and C are mutually exclusive.
10 5 1 5
P ( A ∩ B )= = = × =P ( A ) × P ( B )
36 18 2 9
Events A and B are independent.
2
36 1
P ( C|B )= =
5 10
9

3 X B ( n , p)
1
X B 500 ,( 2 )
1
μ= E ( X )=np=500 × =250
2
1 1
σ 2=npq=500× × =125
2 2
1
Since, n>50, p= and np> 5, then the normal approximation is used.
2
X N ( 250 ,125 )
P (|X −E ( X )|≤ 25 )
¿ P (|X −250|≤25 )
¿ P (−25 ≤ X −250 ≤−25 )
¿ P ( 225≤ X ≤ 275 )
¿ P ( 225−0.5 ≤ X ≤275+ 0.5 ) [Taking continuity correction]
¿ P ( 224.5< X <275.5 )
224.5−250 X−250 275.5−250
¿P ( √125
<
√ 125
<
√ 125)
¿ P (−2.281< Z <2.281 )
= 0.977

4 ¿
7280
μ=
130
a) ¿ 56
5005
σ^2= =38.8
129
(Use the symbol for unbiased estimate for variance σ^2.
If students don’t write ^μ orσ^2, -1 m.)

6 . 229
(
b) 95% confidence interval = 56 ± 1. 96
√ 130 )= (54.93, 57.07)
c) 55 kg is in the 95% confidence interval,
therefore do not reject H0, that means the principal does not understate the mean weight.
5 2.04+1.97+ 1.99+ 2.03+2.04+2.10+ 2.01+1.98+ 2.07
x́=
9

x́ 2.026

H 0 :μ=2.00
H 1 : μ ≠ 2.00
Significance level, α =0.05
Critical regions: z ←1.96and z >1.96
Test statistic:
2.206−2.00
Z= =3.83
0.02
√9
Since z > 1.96, ∴reject H 0,
That is the machine’s setting has been altered.

6 H 0: There is no association between the grades


H 1:There is an association between the grades

Oi Ei ( Oi−E i )
2
( Oi−Ei )
2

Ei
17 21.11 16.892 0.800
38 33.17 23.329 0.703
12 12.73 0.5329 0.042
28 26.77 1.5129 0.056
45 42.08 8.526 0.203
12 16.15 17.223 1.066
18 15.12 8.294 0.549
16 23.76 60.218 2.534
14 9.12 23.814 2.611
Total 8.565

Test Statistics = 8.565


Degree of freedom = 2x2 =4
Chi-square (5 % ) = 9.488
Since the test statistic < 9.488
Accept H o ,There is no association between the grades
7 a) 78−80

(
P ( x́<78 )=P Z<
3.5
√9 ) =0.0433

b) 95%, Z α =1.96
2
(
Confidence interval = 78 ± 1.96 ( 3.5
√ 9 ))
=( 75.713 , 80.287 )

81 is not in the interval.


Reject the claim μ= 81 at 95% confidence level.

c)
P (|x́−μ|< 0.2 )=0.90
90%, Z α =1.645
2
3.5
1.645 ( )
√n
=0.2

n=829

8 8.(a) A 90% C.I. for the sales of leaded petrol


=8.72±1.6449(3.25)
=8.72±5.345925
=(3.37,14.1)/(3.374 ,14 .07)
(b) A 95% C.I. for the sales of unleaded petrol

=9 . 71±1. 96
(3.√10025 )
=9 . 71±0. 637
=(9.073 , 10.347 )
(c)
H0: μ=9.10
H1: μ>9.10

Significance level, α =0.05


Critical region: Z > 1.6449

3.252
(
X́ N 9.10 ,
100 )
Test statistic:
9.71−9.10
Z=
3.25
√100
=1.8769
Since Z = 1.8769 > 1.6449,
∴there is insufficient evidence to accept H0,
that is, there is evidence that the mean sales of unleaded petrol in 1990 is greater than in 1989.

(d)
σ
Zα <0 .50
2 √n
2
Zα
n> ( )
0 .50
σ2

2
1. 960
n> (
0 .50
×3.25 )
n>162 .3
∴ n=163

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