Quantitative Methods for Accounting & Finance

Quantitative Methods for Accounting & Finance

Answer 1
Part a)
Sample size n = 16, population mean μ=1000 and population standard deviation σ=200
i.
P ( X
−
>1050 ) =P
( X
−
−μ
σ
√ n
>1050−μ
σ
√ n )=P
( Z >1050−1000
200
√ 16 ) =P ( Z >1 ) =1−P ( Z≤1 )
Now, P ( Z≤1 ) =0 . 8413 implies that P ( X
−
>1050 ) =0 .1587
ii.
P ( 960< X
−
<1050 )=P
( 960−μ
σ
√ n
< X
−
−μ
σ
√ n
<1050−μ
σ
√ n ) =P
( 960−1000
200
√ 16
<Z <1050−1000
200
√ 16 )
P ( 960< X
−
< 1050 )=P ( −0 . 8< Z< 1 ) =P ( Z <1 ) −P ( Z<−0 .8 ) =P ( Z <1 ) + P ( Z < 0. 8 ) −1
Again, P ( Z≤0. 8 ) =0 .7881 implied that P ( 960< X
−
< 1050 )=0 . 8413+0 . 7881−1=0. 6294
Part b)
Let X denotes the length of the subcomponent, where X ~ N ( 117 ,(5. 2 )2 )
i. Required probability
P ( X >120 ) =P ( X−μ
σ >120−μ
σ )=P ( Z >120−117
5 . 2 )=P ( Z> 0. 58 )
=> P ( X >120 ) =P ( Z >0 . 58 ) =1−P ( Z≤0 . 58 ) =1−0 .719=0 .281
Hence, required probability for one selected subcomponent to be longer than 120 cm is
0.281.
ii. Let X denotes the average length of four subcomponents.
Then, the required probability is,
P ( X
−
>120 ) =P
( X
−
−μ
σ
√ n
>120−μ
σ
√ n )=P
( Z >120−117
5 . 2
√ 4 )=P ( Z>1. 15 ) =1−P ( Z≤1 .15 )

=> P ( X
−
>120 ) =1−P ( Z≤1 .15 ) =1−0 . 8749=0 . 1251
Hence, the probability of mean length of four subcomponents exceeding 120cm is 0.1251
Answer 2
Part a)
i. The standard deviation of the population was provided in the problem, the population was
assumed to follow a normal distribution with mean 10, and standard deviation 10.
At 5% level of significance, the test statistics is calculated as
Z= X
−
−μ
σ
√ n
=10−10
10
√ 100
=0
The two tail rejection region for α =0 . 05 has been indicated in the following figure 1 as |
Z|>1.96 (Ott, and Longnecker, 2015).
Figure 1: Rejection Region at Alpha = 0.05 for Z test
Hence, P (|z|> 0 ) =0. 5 implied that p-value = 1 (p-value > 0.05), which implied that the
null hypothesis failed to get rejected at 5% level of significance.