1. Given a random sample of size 4,900 from a binomial distribution with P=0.50, complete parts (a)
through (e) below.
a) Find the probability that the number of successes is greater than 2,540.
Answer:
n=4,900; P=0.5; μ=nP=(4,900)(0.5)=2450
σ 2 =nP(1–P)=(4,900)(0.5)(1 -0.5)=1225; 𝜎 = √𝜎 2 = √1225 = 35
nP(1–P)=(4,900)(0.5)(1 -0.5)=1225>5, so the follo wing formula can be used;

a  nP 
P( X  a )  P  Z 


nP(
1

P
)



2,535   0.5  4 ,900  
  2.57
P( X  2,535 )  P  Z 

 0.5 4,900 1  0.5  

P(Z>2.57)=1 – P(Z<2.57)
=1 – F(2.57)
=1 – 0.9925
=0.0075.
b)
Hence P(X>2,540)=0.0075
Find the probability that the number of successes is fewer than 2,425.
Answer:

a  nP 
P( X  a )  P  Z 


nP(
1

P
)



P( X  2, 425 )  P  Z 


2, 425   4 ,900  0.50  

 4,900  0.50 1  0.50  

2 , 425   4 ,900  0.50  

 PZ 


4
,
900
0
.
50
1

0
.
50






 P  Z  0.86 
P  Z  0.86   P  Z  0.86 
 1  P  Z  0.86 
 1  0.7611
 0.2389
c)
Find the probability that the number of successes is between 2,440 and 2,485.
Answer:
 a  nP
b  nP 
P( a  X  b )  P 
Z

 nP( 1  P )
nP( 1  P ) 

 2, 445  0. 5 4900
  Z
P( 2 ,440 X  2, 485) P
  0. 5 4900 1 0. 5

 P  0 .14  Z  1
,2 485
  . 05 4900
 
 .0 5 4900
 1 . 0 5
P  0.14  Z  1  F 1  1  F  0.29  
 0.8413  1  0.6141
 0.4554
d) With pro bability 0.40, the nu mber of successes is fewer than how many ?
Answer:
Since the probability is less than 0.5, the values of X are less than the mean, so the
corresponding value o f Z is negative. Use the symmetry o f the normal distributio n to find
an equivalent value for F(z).
P(Z<– z)=P(Z>z)
P(Z<– z)= 1 – P(Z<z)
0.40=1 – F(z)
F(z) = 1– 0.40=0.60
z=0.25
No w sub stitute this value for Z in the formula shown belo w and solve for X
X  nP
Z 
nP 1  P 
Substitute 0.25 for Z, 2,450 for nP, 35 for √𝑛𝑃(1 − 𝑃
0.25 
X  2, 450
35
X =2441.25
e) With pro bability 0.09, the nu mber of successes is greater than how ma ny?
Answer:
Use the symmetr y of the nor mal distrib utio n to find an eq uivalent value for F(z).
P(Z> z)=1 – P(Z<z)
0.09=1 – F(z)
F(z) = 1– 0.09=0.91
z=1.34
No w sub stitute this value for Z in the formula shown belo w and solve for X
Z 
X  nP
nP 1  P 
Substitute 1.34 for Z, 2,450 for nP, 35 for √𝑛𝑃(1 − 𝑃
1.34 
X  2 , 450
35
X =2496.9
2. A professor claims that exam scores for non-accounting majors are more variable than for accounting majors.
Random samples of 16 non-accounting majors and 29 accounting majors are taken from a final exam. The sample
variances are 1516.541 and 429.631 respectively for non-accounting and accounting majors. Test at the 1% level
the null hypothesis that two population variances are equal against the alternative that the true variance is higher
for the non-accounting majors. Assume the populations are normally distributed.
Answer:
n1=16; n2 = 29; 𝑠𝑥2 = 1516.541 ; ; 𝑠𝑦2 = 429.631
H0: 𝜎𝑥2 = 𝜎𝑦2 ;
𝐹=
𝑠𝑥2
2
𝑠𝑦
=
H1 𝜎𝑥2 > 𝜎𝑦2 :
1516.541
429.631
= 3.5299 where 𝑠𝑥2 denotes the larger sample variance.
nx – 1 = 16 – 1=15; ny– 1 = 29– 1=28.
F15, 28, 0.01 = 2.75
F=3.5299 > F15, 28, 0.01 = 2.75
Therefore, H0 will be rejected.
3. Suppose that we have a population proportion P=0.40 and a random sample of size n=300drawn from
the population. Complete (a) through (c) below.
a) What is the probability that the sa mple proportion is greater than 0.42?
Answer:
The normal distribution can be used to approximate this distribution because nP = 300(0.40) and
n(1 – P) = 300(1 – 0.40) are both least 5.
Z
p̂  P
P 1  P 
n
0. 42 0. 40
Z
0. 40 1 0. 60
300
 0. 71
P  ˆp  0.42   P  Z  0.71
 1  P  Z  0.71
 1  F  0.71
 1  0.8888
 0.1112
b)
What is the probability that the sa mple proportion is less tha n 0.37?
Answer:
Z
p̂  P
P 1  P 
n
Z
0. 35 0. 40
0. 40 1 0. 60
300
 1. 77
P  ˆp  0.35   P  Z  1.77 
 P  Z  1.77 
 1  F 1.22 
 1  0.0385
 0.9615
c)
What is the probability that the sa mple proportion is between 0.36 and 0.41?
Answer:
Z
p̂  P
Z 
P 1  P 
n
0. 41 0. 40
0. 40 1 0. 60
300
P  ˆp  0.41  P  Z  0.35
 0. 35
Z 
0. 36 0. 40
0. 40 1 0. 60
300
 1. 41
P  0.36  ˆp   P  Z  1.41
 0.6382
 P  Z  1.41
 1  F( 1.41 )
 0.0793
P  0.36  ˆp  0.41  0.6382  0.0793
 0.5589
4. Late payment of medical claims can add to the cost of healthcare. For one insurance company, 74.4 % of the
claims were paid in full when first submitted. The insurance company developed a new payment system in an
effort to on this percentage. A sample of 290 claims processed under this system revealed that 223 of the claims
were paid in full when first submitted.
(a) At the 0.05 level of significance, is there evidence that the population proportion of claims processed under
the new system is higher than the article reported for the previous system?
(b) Compute p-value and interpret its meaning.
Answer:
(a) H0: P ≤ 0.744
H1: P > 0.744
α=0.05; n = 290.
nP(1 –P) = 290. 0.744(1 - 0.744) = 55.23456>5. Thus, Z test statistic and standard normal distribution can be used
𝑧=
𝑝−𝑃
√𝑃(1 − 𝑃)/𝑛
This is one-tail test. Therefore, the critical value;
z0.05=1.64
223
(
) − 0.744
290
𝑧=
=
= 0.97
√𝑃(1 − 𝑃)/𝑛 √0.744(1 − 0.744)/(290)
𝑝−𝑃
Since z = 0.97< z0.05=1.28, do not reject H0.
(b) For this one-tail test, the p-value is the probability that z>0.97
p-value = P(z>0.97) = 1 – P(z < 0.97)= 1 – F(0.97) = 1 – 0.8340 = 0.1660.
The p-value is the probability of getting a test statistic equal to or more extreme than the sample result, given that the
null hypothesis is true.
5. Fit a simple linear regression model to the oxygen purity data given in the following table.
Observation
Number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Hydrocarbon
Level
x (%)
0.99
1.02
1.15
1.29
1.46
1.36
0.87
1.23
1.55
1.4
1.19
1.15
0.98
1.01
1.11
1.2
1.26
1.32
1.43
0.95
Purity y
(%)
90.01
89.05
91.43
93.74
96.73
94.45
87.59
91.77
99.42
93.65
93.54
92.52
90.56
89.54
89.85
90.39
93.25
93.41
94.98
87.33
Answer:
n
x
 xi
i 1
n
20
 xi

i 1
20
n
 1.1960
y
n
b1 
 (xi  x)(yi  y)
i 1
n
 (x  x)
2
i 1
20
 yi

i 1
n
y
i
i 1
20
 92.1605
20
sy
Cov(x,y)

 rxy 
2
sx
sx
 (x  1.1960)(y  92.1605)
i 1
i
i
b0  y  b1 x=92.1605 - (14.947)(1.1960)=74.283
i
20
 (x  1.1960)
2
i 1
i
 14.947