STAT 2910 Final Exam Solution
August 1, 2022
1
problem 1
1.1
a
Range R = 11.7 − 2.0 = 9.7.
Positon of median = 0.5(n + 1) = 0.5(50 + 1) = 25.5.
median = 25th+26th
= 7.2+7.2
= 7.2.
2den
2
Position of Q1 = 0.25(n + 1) = .25(50 + 1) = 12.75.
Q1 = 12th + 0.75(13th − 12th) = 5.2 + 0.75(5.3 − 5.2) = 5.275
Postion of Q3 = 0.75(n + 1) = .75(50 + 1) = 38.25.
Q3 = 38th + 0.25(39th − 38th) = 8.2 + 0.25(8.2 − 8.2) = 8.2.
Inerqartile Range = Q3 − Q1 = 2.925.
An approximation for standard deviation is s ≈ R4 = 2.425.
This data refer to a sample.
1.2
b
P
xi
59
n P= 7 = 8.43
2
2
/7
i −x̄)
= 1099−(59)
= 100.
Sample variance s2 = (x
6
√n−1
2
Standard deviation s = s = 10.
Sample mean x̄ =
30−8.43
z-score of 30 z = x−x̄
= 2.157.
s =
10
Since the z-score does not exceed 3 in absolute value, 30 is a suspect as an outlier.
2
problem 2
2.1
a
P (A ∩ B) = P (A) + P (B) − P (A ∪ B) = 0.6 + 0.3 − 0.7 = 0.2.
2.2
b
0.2
2
P (A|B) = P P(A∩B)
(B) = 0.3 = 3
2.3
c
No, since P (A ∩ B) ̸= 0.
2.4
d
No, since P (A ∩ B) = 0.2 ̸= P (A)P (B) = 0.6(0.3) = 0.18
3
problem 3
Let event A = {smoke cigarettes} and event B = {smoke cigars}.
It is given that P (A) = 0.28, P (B) = 0.07 and P (A ∩ B) = 0.05.
1
3.1
a
P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = 0.28 + 0.07 − 0.05 = 0.3
3.2
b
P (B) − P (A ∩ B) = 0.07 − 0.05 = 0.02
4
problem 4
It is a binomial distribution with n = 5, p = 13 and q = 23 .
4.1
a
q = ( 23 )5 = 0.13
5
4.2
b
mean = np = 5( 13 ) = 35 , variance = npq = 5( 13 )( 23 ) = 10
9 .
5
problem 5
It is a possionn distribution with µ = 3.
5.1
a
−µ
x
−3 3
P (x = 3) = e x!µ = e 3!3 = 0.224.
5.2
b
P (x > 1) = 1 − P (x ≤ 1) = 1 − 0.199 = 0.801.
6
problem 6
Since the sample size is large, the sample mean is approximately normally distributed by the Central
Limit Theorem.
6.1
a
mean of x̄ = µ = 40.
6.2
b
4
standard deviation of x̄ = √σn = √100
= 0.4
6.3
c
41−µ
41−40
P (x̄ > 41) = P ( c̄−µ
σ >
σ ) = P (z >
4 ) = P (x > 0.25) = 1−P (z < 0.25) = 1−0.5987 = 0.4013.
7
problem 7
P (x < 1.38) = P (z < 1.38−µ
) = P (z < 1.38−1.2
σ
0.15 ) = P (z < 1.2) = 0.8849.
2
8
problem 8
4−µ
4−µ
P (x > 4) = P (z > 4−µ
σ ) = 1 − P (z < σ ) = 0.9772 P (z < σ ) = 0.0228.
4−µ
So sigma = −2
5−µ
5−µ
Also P (x > 5) = P (z > 5−µ
σ ) = 1 − P (z < σ ) = 0.9332 P (z < σ ) = 0.0668.
5−µ
So sigma = −1.5
Solve the above equations together will get µ = 8, σ = 2.
9
problem 9
9.1
a
H0 : p0 = 0.1 VS Ha : p0 ̸= 0.1
Test statistic is Z = √ p̂−p0
p0 (1−p0 )/n
= √81/900−0.1 = −1
0.1(0.9)/900
p-value=2P (Z < −1) = 2(0.1587) = 0.3174
Since the p-value of 0.3174 exceeds the specified significant level α = 0.02, H0 can not be rejected.
There is not enough evidence at the 2% significant level to infer that the competitor’s claim is wrong.
9.2
b
q
q
p̂)
0.09(0.91)
=
0.09
±
2.575
= 0.09 ± 0.025 or (0.065, 0.115).
The 99% C.I.= 0.09 ± 2.575 p̂(1−
900
900
Since 0.1 is included in the 99% confidence interval, we can not reject H0 . There is no enough evidence
in the data to suggest that the compeotitor’s claim is wrong at 1% level of significance.
10
problem 10
10.1
a
The point estimate for the average weight is
µ̂ = x̄ = 0.53
10.2
b
The point estimate of σ 2 is σˆ2 = s2 = 0.05592 = 0.0031
10.3
c
H0 : µ = 0.5
Ha : µ > 0.5
x̄ − µ
0.53 − 0.5 .
√ = 1.3146
Test statistic : t = √ =
s/ n
0.0559/ 6
Reject Region: reject H0 if t > tα , where tα = t0.05 = 2.015 with df = 5
Conclusion: Since the calculated value of the test statistic 1.3146 does not fall in the reject region, we
can not reject H0 . The data do not present sufficient evidence to indicate that the mean diamond weight
exceeds 0.5 karat at α = 0.05
10.4
d
H0 : σ 2 = 0.015
Ha : σ 2 ̸= 0.015
(6 − 1) · 0.05592
(n − 1)S 2
=
= 1.0416
Test statistic: X 2 =
2
σ0
0.015
2
2
2
Reject Region: reject H0 if X 2 > Xα/2
= 11.0705 or X 2 < X(1−α/2)
= 1.1455. Because X 2 < X1−α/2
,
we reject H0 .
Conclusion: There is sufficient evidence to say σ 2 ̸= 0.015 at α = 0.10.
3
11
problem 11
11.1
a
n1 = 8
n2 = 9
s2 =
11.2
s1 = 3
s2 = 5
(n1 − 1) · s21 + (n2 − 1)s22
(8 − 1) · 32 + (9 − 1) · 52 .
=
= 17.5333
(n1 + n2 − 2)
(8 + 9 − 2)
b
H0 : σ12 = σ22
Ha : σ12 ̸= σ22
S2
25
Test Statistic: F = 12 =
= 2.78
df1 = n1 − 1 = 8
df2 = n2 − 1 = 7
S1
9
Reject Region: reject H0 if F > F(8,7,0.025) = 4.9. We don’t reject H0 because F < 4.9.
Conclusion: It is reasonable to assume equality of variances in this problem at α = 0.05.
11.3
c
H0 : µ1 = µ2
Ha : µ1 > µ2
(x̄1 − x̄2 )
(90 − 84)
.
Test Statistic: t = q
=q
= 2.9489 Reject Region: reject H0 if t > tα .
1
1
1
1
s2 ( n1 + n2 )
17.5333 ∗ ( 8 + 9 )
tα = 1.753 with df = 8 + 9 − 2 = 15. We can reject H0 because t > 1.753.
Conclusion : There is sufficient evidence at α = 0.05 to say that the top speed of model A is greater than
model B.
12
problem 12
Let µ1 denote the number of cases per hour before the implementation of the program.
Let µ2 denote the number of cases per hour after the implementation of the program.
10 + 5 + 0 + 5 + 3
Σdi
=
= 4.6
d¯ =
n
5
Sd = 3.6469
ud = µ2 − µ1
H0 : ud = 0
df = 5 − 1 = 4
Ha : ud > 0.
d¯ − 0
4.6
.
√ = 2.818
Test Statistic : t =
=
sd
3.6496/ 5
Since 0.01 < P (t > 2.818) < 0.025 which means P (t > 2.818) < 0.1. we reject H0 and conclude that
the time management is effective at 10% level of significance.
13
problem 13
x̄ ∼ N (3, 22 /50)
P (x̄ > d) = 1 − P (x̄ < d) = 1 − P (Z <
P (Z <
d−3
√ ) = 0.6368
2/ 50
d−3
√ ) = 0.3632
2/ 50
d−3
√ = −0.35
2/ 50
.
d = 2.9
14
problem 14
14.1
a
H0 : p1 = p2
Ha : p1 < p2
4
Test Statistic : z = q
pˆ1 − pˆ2
p̂q̂( n11 + n12 )
n1 · p1 + n2 · p2
50 · 0.4 + 50 · 0.64
where p̂ =
=
= 0.52 q̂ = 0.48
n1 + n2
50 + 50
0.4 − 0.64
Therefore, z = q
= −2.4019
1
1
0.52 ∗ 0.48 ∗ ( 50
+ 50
)
Reject H0 when z < −zα = −1.645.
Since the computed value of Z falls in the reject region, we reject H0 and conclude that there is
sufficient evidence to say that the drug is effective.
14.2
b
The 95% confidence interval is
r
(pˆ2 − pˆ1 ) ± z α2 ·
r
= (0.64 − 0.4) ± 1.96 ·
= 0.24 ± 0.19
= (0.05, 0.43)
0.4 · 0.6 0.64 · 0.36
+
50
50
5
pˆ1 qˆ1
pˆ2 qˆ2
+
n1
n2