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Chapter 5 Confidence Interval & Hypothesis Test Review: Example Solutions

Confidence Intervals

Examples

Example 3. Suppose X ∼ χ2 (5). Find P(X > 15.09).

Solution:

In this case, there are 5 degrees of freedom, so v = 5. We have that χ2α (5) = 15.09. We need to

identify α.

Option 1: Go to the Table in the textbook. Look for the v value. The body of the table contains

the chi-square values. The areas are along the top. If the exact value is not in the table,

the most you can do is to give an interval for α. In our scenario, α = 0.01 .

Option 2: In your calculator, there is a function that allows you to calculate chi-square probabilities.

It works similar to the normalcdf function, except that it has one additional input, the

degrees of freedom value.

P X ≥ χ2α (v) = χ2 cdf χ2α (v), 999, v = α,

and

P X ≤ χ2α (v) = χ2 cdf 0, χ2α (v), v = 1 − α.

For our example,

P (X ≥ 15.09) = χ2 cdf(15.09, 999, 5) = 0.01 .

Example 4. Find the χ2 value with 7 degrees of freedom leaving an area of 0.05 to the right.

Solution:

This question is asking us to find χ20.05 (7), where we have

P X > χ20.05 (5) = 0.05.

There is no inverse function for the chi-square distribution on your calculator, so you need to

know how to use the table. Our solution is

χ20.05 (5) = 14.067 .

1

Example 5. Suppose F ∼ F (5, 8).

1. Find k such that P(F > k) = 0.05.

Solution:

In this case, α = 0.05. r1 = 5 and r2 = 8. We find that k = 3.69 = F0.05 (5, 8) .

2. Find k such that P(F > k) = 0.01.

Solution:

In this case, α = 0.01. r1 = 5 and r2 = 8. We find that k = 6.63 = F0.01 (5, 8) .

3. Find F0.95 (8, 5).

Solution:

In this case, α = 0.95, which is NOT on one of our tables. However, we can still find this

value.

1

1

=

= 0.27 .

F0.95 (8, 5) =

F0.05 (5, 8)

3.69

4. Find F0.05 (8, 5).

Solution:

In this case, α = 0.05. r1 = 8 and r2 = 5. We find that F0.05 (8, 5) = 4.82 .

Example 6. Suppose we are interested in the accelerated life testing of paints. Suppose we have

independent normal distributions where

n1 = 10 s21 = 400

n2 = 12 s22 = 340

Find a 90% confidence interval for σ12 /σ22 .

Solution:

In this case, α = 0.10, so α/2 = 0.05. We have

Fα/2 (n1 − 1, n2 − 1) = F0.05 (9, 11) = 2.90

Fα/2 (n2 − 1, n1 − 1) = F0.05 (11, 9) = 3.10

The 90% confidence interval for σ12 /σ22 is given by

1

400

400

, 3.10

= (0.41, 3.65) .

2.90 340

340

Note that the interval we found contains 1. This means that it is reasonable to believe that σ12

may be equal to σ22 .

2

Hypothesis Testing

Examples

Example 7. We want to test whether the variability in incoming material from a new supplier

(σ22 ) is smaller than that coming from the current supplier (σ12 ). We obtain the following data:

n1 = 13 x̄ = 26.2 s2x = 29.1

n2 = 11 ȳ = 25.9 s2y = 14.6

Perform a hypothesis test with α = 0.05.

Solution:

• State your hypotheses:

H0 : σ12 = σ22

• Test Statistic:

F =

versus

H1 : σ12 > σ22

29.1

s2x

=

= 1.99

s2y

14.6

• Critical Region:

Reject H0 if F > F0.05 (12, 10) = 2.91

• Critical Region Decision:

Since 1.99 6> 2.91, Do Not Reject H0 .

• p-value Calculation:

p-value = P (F > 1.99) = Fcdf(1.99, 999, 12, 10) = 0.1421

• p-value Decision:

Since p-value 6< α = 0.05, Do Not Reject H0 .
3
Additional Examples
Example 8. A random sample of 100 recorded deaths in the US during the last year showed an
average lifespan of 71.8 years. Assuming a population standard deviation of 8.9 years, does this
seem to indicate that the mean life span today is greater than 70 years?
(a) Perform a hypothesis test. Use the critical region method with α = 0.04 to make your
decision.
Solution:
• State your hypotheses.
H0 : µ = 70
versus
H1 : µ > 70.

• Calculate your test statistic.

We have x̄ = 71.8, σ = 8.9, and n = 100. Since σ is known, use z.

z=

x̄ − µ0

71.8 − 70

√

√ =

= 2.02.

σ/ n

8.9/ 100

• Define the critical region.

α = 0.04. We have

zα = z0.04 = invnorm(1 − 0.04) = 1.75.

Our critical region is:

Reject H0 if z > 1.75.

• Make your decision and state your conclusion.

Since 2.02 > 1.75, we Reject H0 . There is enough evidence that the mean life span

today is greater than 70 years.

4

(b) Perform a hypothesis test. Use the p-value method to make your decision. Use a cutoff

value of 0.04.

Solution:

• State your hypotheses.

H0 : µ = 70

versus

H1 : µ > 70.

• Calculate your test statistic.

We have x̄ = 71.8, σ = 8.9, and n = 100. Since σ is known, use z.

z=

71.8 − 70

x̄ − µ0

√ =

√

= 2.02.

σ/ n

8.9/ 100

• Calculate the p-value.

p-value = P (Z > 2.02) = normalcdf(2.02, 999) = 0.0217.

• Make your decision and state your conclusion.

Since our p-value is small (even smaller than our cutoff value of 0.04), we would

Reject H0 . There is enough evidence that the mean life span today is greater than

70 years.

5

Example 9. The number of kilowatt hours used annually by various home appliances is published

annually. It is claimed that a vacuum cleaner uses an average of 46 kilowatt hours per year. If a

random sample of 12 homes included in a planned study indicates that vacuums use an average

of 42 kilowatt hours per year with a standard deviation of 11.9 kilowatt hours, does this suggest

at the 0.10 level of significance that vacuum cleaners use, on average, less than 46 kilowatt hours

annually?

(a) Perform a hypothesis test. Use the critical region method to make your decision.

Solution:

• State your hypotheses.

H0 : µ = 46

versus

H1 : µ < 46.
• Calculate your test statistic.
We have x̄ = 42, s = 11.9, and n = 12. Since σ is unknown, use t.
t=
42 − 46
x̄ − µ0
√ =
√ = −1.164.
s/ n
11.9/ 12
This value has v = n − 1 = 11 degrees of freedom.
• Define the critical region.
α = 0.10. Use 11 degrees of freedom. We have
−tα (n − 1) = −t0.10 (11) = −1.363.
Our critical region is:
Reject H0 if t < −1.363.
• Make your decision and state your conclusion.
Since −1.164 6< −1.363, Do Not Reject H0 . There is not enough evidence to conclude
that the mean number of kilowatt hours a vacuum uses per year is significantly less
than 46.
6
(b) Perform a hypothesis test. Use the p-value method to make your decision, with the cutoff
value being the level of significance.
Solution:
• State your hypotheses.
H0 : µ = 46
versus
H1 : µ < 46.
• Calculate your test statistic.
We have x̄ = 42, s = 11.9, and n = 12. Since σ is unknown, use t.
t=
42 − 46
x̄ − µ0
√ =
√ = −1.164.
s/ n
11.9/ 12
This value has v = n − 1 = 11 degrees of freedom.
• Calculate the p-value.
p-value = P (T < −1.164) = tcdf(−999, −1.164, 11) = 0.1345 ∈ (0.10, 0.15) .
• Make your decision and state your conclusion.
Since our p-value is large (it is larger than our cutoff value of 0.10), we Do Not Reject
H0 . There is not enough evidence to conclude that the mean number of kilowatt
hours a vacuum uses per year is significantly less than 46.
7
Example 10. An experiment was performed to compare the abrasive wear of two different laminated materials. We obtain the following data:
Material 1
n1 = 12
x̄1 = 85 units of wear
s1 = 4
Material 2
n2 = 10
x̄2 = 81 units of wear
s2 = 5
Assume the populations to be approximately normal with equal variances. Can we conclude at
the 0.05 level of significance that the abrasive wear of material 1 exceeds that of material 2 by
more than 2 units?
(a) Perform a hypothesis test. Use the critical region method with the stated level of significance to make your decision.
Solution:
• State your hypotheses:
H0 : µ1 − µ2 = 2
versus
H1 : µ1 − µ2 > 2.

• Is the data paired?

No.

• Are the population variances known or unknown? If they are unknown, are they

equal?

They are unknown and assumed to be equal. Use Case 3 information.

• Calculate your test statistic.

Use t since we have an example of Case 3.

401

(12 − 1)(4)2 + (10 − 1)(5)2

=

= 20.05

12

+

10

−

2

20

√

sp = 20.05 = 4.478

(85 − 80) − 2

q

t=

= 1.043.

1

1

4.478 · 12

+ 10

s2p =

• Define the critical region.

α = 0.05. Use df = v = n1 + n2 − 2 = 12 + 10 − 2 = 20. We have

tα (n1 + n2 − 2) = t0.05 (20) = 1.725.

Our critical region is:

Reject H0 if t > 1.725.

• Make your decision and state your conclusion.

Since 1.043 6> 1.725, Do Not Reject H0 . There is not enough evidence that the actual

abrasive wear of material 1 exceeds that of material 2 by more than 2 units.

8

(b) Perform a hypothesis test. Use the p-value method to make your decision, with the cutoff

value being the level of significance.

Solution:

• State your hypotheses:

H0 : µ1 − µ2 = 2

versus

H1 : µ1 − µ2 > 2.

• Is the data paired?

No.

• Are the population variances known or unknown? If they are unknown, are they

equal?

They are unknown and assumed to be equal. Use Case 3 information.

• Calculate your test statistic.

Use t since we have an example of Case 3.

401

(12 − 1)(4)2 + (10 − 1)(5)2

=

= 20.05

12

+

10

−

2

20

√

sp = 20.05 = 4.478

(85 − 80) − 2

q

= 1.043

t=

1

1

+ 10

4.478 · 12

s2p =

df = n1 + n2 − 2 = 12 + 10 − 2 = 20.

• Calculate the p-value.

p-value = P (T > 1.043) = tcdf(1.043, 999, 20) = 0.1547 ∈ (0.15, 0.20).

• Make your decision and state your conclusion.

Since our p-value is large (larger than our cutoff value), we Do Not Reject H0 . There

is not enough evidence that the actual brasive wear of material 1 exceeds that of

material 2 by more than 2 units.

9

Example 11. The following data represent the running times of films produced by two motionpicture companies:

Company

1

2

102

81

Time (minutes)

86 98 109 92

165 97 134 92 87

114

Test the hypothesis that the average running time of films produced by company 2 is less than

or equal to the average running time of films produced by company 1 by 10 minutes against the

one-sided alternative that the difference is more than 10 minutes. Use a 0.1 level of significance

and assume the distributions of times to be approximately normal with unequal variances.

(a) Perform a hypothesis test. Use the critical region method to make your decision.

Solution:

Option 1:

• State your hypotheses.

H0 : µ2 − µ1 = 10

H1 : µ2 − µ1 > 10

• Is the data paired?

No.

• Are the population variances known or unknown? If they are unknown, are they

equal?

They are unknown and assumed to be unequal. Use Case 2 information.

• Sample Information

n1 = 5 x̄1 = 97.4 σ1 = 8.877

n2 = 7 x̄2 = 110 σ2 = 30.221

• Calculate your test statistic.

Degrees of Freedom

v=

(s21 /n1 + s22 /n2 )2

(s21 /n1 )2

(s2 /n )2

+ n2 2 −12

n1 −1

=

8.8772

5

(8.8772 /5)2

5−1

2

+

30.2212

7

+

(30.2212 /7)2

7−1

=

21384.06612

= 7.376

2899.283387

Use v = 7 degrees of freedom.

Test Statistic

t=

(x̄2 − x̄1 ) − d0

(110 − 97.4) − (10)

q 2

= q

= 0.215

2

s1

s2

8.8772

30.2212

+

+

5

7

n1

n2

10

• Define the critical region.

α = 0.10. We have

−tα (v) = t0.10 (7) = 1.415.

Our critical region is:

Reject H0 if t > 1.415.

• Make your decision and state your conclusion.

Since 0.215 6> 1.415, we Do Not Reject H0 . There is not enough evidence that

the average running time of films produced by company 2 is more than the average

running time of films produced by company 1 by more than 10 minutes.

Option 2:

• State your hypotheses.

H0 : µ1 − µ2 = −10

H1 : µ1 − µ2 < −10
• Is the data paired?
No.
• Are the population variances known or unknown? If they are unknown, are they
equal?
They are unknown and assumed to be unequal. Use Case 2 information.
• Sample Information
n1 = 5 x̄1 = 97.4 σ1 = 8.877
n2 = 7 x̄2 = 110 σ2 = 30.221
• Calculate your test statistic.
Degrees of Freedom
v=
(s21 /n1 + s22 /n2 )2
(s21 /n1 )2
(s2 /n )2
+ n2 2 −12
n1 −1
=
8.8772
5
(8.8772 /5)2
5−1
+
+
30.2212
7
2
(30.2212 /7)2
=
7−1
21384.06612
= 7.376
2899.283387
Use v = 7 degrees of freedom.
Test Statistic
t=
(x̄1 − x̄2 ) − d0
(97.4 − 110) − (−10)
q 2
= q
= −0.215
2
s1
s2
8.8772
30.2212
+
+
5
7
n1
n2
• Define the critical region.
α = 0.10. We have
−tα (v) = −t0.10 (7) = −1.415.
Our critical region is:
Reject H0 if t < −1.415.
11
• Make your decision and state your conclusion.
Since −0.215 6< −1.415, we Do Not Reject H0 . There is not enough evidence that
the average running time of films produced by company 2 is more than the average
running time of films produced by company 1 by more than 10 minutes.
(b) Perform a hypothesis test. Use the p-value method to make your decision, with the cutoff
value being the level of significance.
Solution:
Option 1:
• State your hypotheses.
H0 : µ2 − µ1 = 10
H1 : µ2 − µ1 > 10

• Is the data paired?

No.

• Are the population variances known or unknown? If they are unknown, are they

equal?

They are unknown and assumed to be unequal. Use Case 2 information.

• Sample Information

n1 = 5 x̄1 = 97.4 σ1 = 8.877

n2 = 7 x̄2 = 110 σ2 = 30.221

• Calculate your test statistic.

Degrees of Freedom

v=

(s21 /n1 + s22 /n2 )2

(s21 /n1 )2

n1 −1

+

(s22 /n2 )2

n2 −1

=

8.8772

5

(8.8772 /5)

5−1

2

2

+

30.2212

7

+

(30.2212 /7)2

7−1

=

21384.06612

= 7.376

2899.283387

Use v = 7 degrees of freedom.

Test Statistic

t=

(x̄2 − x̄1 ) − d0

(110 − 97.4) − 10

q 2

= q

= 0.215

2

s1

s2

8.8772

30.2212

+

+

5

7

n1

n2

• Calculate the p-value.

(

= tcdf(0.215, 999, 7) = 0.4179

p-value = P (T > 0.215)

> 0.40

12

via calculator

via table

• Make your decision and state your conclusion.

Since our p-value is large (even larger than our cutoff value of 0.10), we Do Not Reject

H0 . There is not enough evidence that the average running time of films produced

by company 2 is more than the average running time of films produced by company

1 by more than 10 minutes.

Option 2:

• State your hypotheses.

H0 : µ1 − µ2 = −10

H1 : µ1 − µ2 < −10
• Is the data paired?
No.
• Are the population variances known or unknown? If they are unknown, are they
equal?
They are unknown and assumed to be unequal. Use Case 2 information.
• Sample Information
n1 = 5 x̄1 = 97.4 σ1 = 8.877
n2 = 7 x̄2 = 110 σ2 = 30.221
• Calculate your test statistic.
Degrees of Freedom
v=
(s21 /n1 + s22 /n2 )2
(s2 /n )2
(s21 /n1 )2
+ n2 2 −12
n1 −1
=
8.8772
5
(8.8772 /5)2
5−1
+
+
30.2212
7
2
(30.2212 /7)2
7−1
=
21384.06612
= 7.376
2899.283387
Use v = 7 degrees of freedom.
Test Statistic
t=
(x̄1 − x̄2 ) − d0
(97.4 − 110) − (−10)
q 2
= q
= −0.215
2
s1
s2
8.8772
30.2212
+
+
5
7
n1
n2
• Calculate the p-value.
(
= tcdf(−999, −0.215, 7) = 0.4179
p-value = P (T < −0.215)
> 0.40

via calculator

via table

• Make your decision and state your conclusion.

Since our p-value is large (even larger than our cutoff value of 0.10), we Do Not Reject

H0 . There is not enough evidence that the average running time of films produced

by company 2 is more than the average running time of films produced by company

1 by more than 10 minutes.

13

Example 12 (Math Finals). Having done poorly on their math final exam in June, 6 students

repeat the course in summer school and take another exam in August. Assume these students

are representative of all students who might attend his summer school. Assume a Normal

Distribution. We obtain the following data:

Student

1

2

3

4

5

6

June Percentage

54

49

68

66

62

62

August Percentage

50

65

74

64

68

72

We want to test the hypothesis that the average difference in student test scores in June are

different from the test scores in August. Use a significance level of 0.05.

(a) Perform a hypothesis test. Use the critical region method to make your decision.

Solution:

• State your hypotheses.

H0 : µD = 0

versus

µD 6= 0.

• Is the data paired?

Yes.

• Sample Information

Student

1

2

3

4

5

6

June Percentage

54

49

68

66

62

62

August Percentage August – June

50

−4

65

16

74

6

64

−2

68

6

72

10

¯

d

5.333

sd

7.448

• Calculate your test statistic.

Degrees of Freedom

v =n−1=6−1=5

Test Statistic

t=

d¯ − d0

5.333 − 0

√ =

√ = 1.754.

sd / n

7.448/ 6

14

• Define the critical region. α = 0.05. We have

−tα/2 (n − 1) = −t0.025 (5) = −2.571;

tα/2 (n − 1) = t0.025 (5) = 2.571.

Our critical region is:

Reject H0 if t < −2.571 or t > 2.571.

• Make your decision and state your conclusion.

Since 1.754 6> 2.571, Do Not Reject H0 . There is not enough evidence that the

average difference in student test scorers from June to August are different.

(b) Perform a hypothesis test. Use the p-value method to make your decision, with the cutoff

value being the level of significance.

Solution:

• State your hypotheses.

H0 : µD = 0

versus

µD 6= 0.

• Is the data paired?

Yes.

• Sample Information

Student

1

2

3

4

5

6

June Percentage

54

49

68

66

62

62

August Percentage August – June

50

−4

65

16

74

6

64

−2

68

6

72

10

d¯

5.333

sd

7.448

• Calculate your test statistic.

Degrees of Freedom

v =n−1=6−1=5

Test Statistic

t=

5.333 − 0

d¯ − d0

√ =

√ = 1.754.

sd / n

7.448/ 6

• Calculate the p-value.

p-value = 2 · P (T > |1.754|)

(

= 2 · tcdf(1.754, 999, 5) = 0.1398

= 2 · P (T > 1.754)

∈ 2 · (0.05, 0.10) = (0.10, 0.20)

15

via calculator

via table

• Make your decision and state your conclusion.

Since our p-value is large (even larger than our cutoff), we Do Not Reject H0 . There

is not enough evidence that the average difference in student test scores from June

to August are different.

(c) It is possible to find our 95% CI for µD to be

−2.484 < µD < 13.15.
Use this confidence interval to perform your hypothesis test.
Solution:
Our hypotheses are
H0 : µD = 0
versus
µD 6= 0.
Our hypothesis is a 2-sided hypothesis. We have α = 0.05. This corresponds directly to
a 95% confidence interval. In our case, we have d0 = 0. Therefore, we look to see if 0 is
within our interval. Since 0 is within our interval, we Do Not Reject H0 . There is not
enough evidence that the average difference in student test scores from June to August
are different.
16
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