Statistics for Business and Economics Chapter 11 Inferences About Population Variances

2 Understand the role of statistical inference in developing conclusions about the variance of a single population.. Know that the sampling distribution of s s has an F distribution and

Trang 1

Inferences About Population Variances

Learning Objectives

1 Understand the importance of variance in a decision-making situation

2 Understand the role of statistical inference in developing conclusions about the variance of a single

population

3 Know that the sampling distribution of (n - 1) s2/σ2 has a chi-square distribution and be able to use

this result to develop a confidence interval estimate of σ2

4 Be able to compute p-values using the chi-square distribution.

5 Know how to test hypotheses involving σ2

6 Understand the role of statistical inference in developing conclusions about two population

variances

7 Know that the sampling distribution of s s has an F distribution and be able to use this result to 12/ 22

test hypotheses involving two population variances

8 Be able to compute p-values using the F distribution.

Solutions:

11 - 1

Trang 2

1 a 11.070

b 27.488

c 9.591

d 23.209

e 9.390

2 s2 = 25

a With 19 degrees of freedom χ.052 = 30.144 and χ.952 = 10.117

2

15.76 ≤σ2 ≤ 46.95

b With 19 degrees of freedom χ.0252 = 32.852 and χ.9752 = 8.907

2

14.46 ≤σ2 ≤ 53.33

c 3.8 ≤σ ≤ 7.3

3

2

2 0

27.08 50

χ

σ

Degrees of Freedom = (16 - 1) = 15

Using χ2table, p-value is between 025 and 05

Exact p-value usingχ2=27.08 is 0281

p-value ≤ 05, reject H0

Critical value approach

2

.05

χ = 24.996

Reject H0 if χ ≥2 24.996

27.08 > 24.996, reject H0

4 a n = 18

11 - 2

Trang 3

s2 = 36

2

.05

χ = 27.587

2

.95

χ = 8.672 (17 degrees of freedom)

2

.22 ≤σ2 ≤ 71

b .47 ≤σ ≤ 84

5 a

2

31.07 1

i

x x s

n

−

b χ.0252 = 16.013 χ.9752 = 1.690

2

13.6 ≤σ2 ≤ 128.7

c 3.7 ≤σ ≤ 11.3

.2205 12

i x x

n

∑

2

47.9500 1

i

x x s

n

−

b χ.0252 = 32.852 χ.9752 = 8.907

The 95% confidence interval for the population standard deviation ranges from a quarterly standard deviation of 5.27% to 10.11%

7 a

2

57.7230 1

i

x x s

n

−

Trang 4

57.7230 7.60

b With 11 df, χ.0252 = 21.920 χ.9752 = 3.816

2

c 5.38 ≤σ ≤ 12.90

8 a x=.78

2

.4748

i

x x s

n

−

∑

b s= 4748 6891=

c 11 degrees of freedom

2

.025

χ = 21.920 χ.9752 = 3.816

2 (12 1).4748 (12 1).4748

21.920− ≤σ ≤ 3.816−

.2383 ≤σ2 ≤ 1.3687

.4882 ≤σ ≤ 1.1699

9 H0: σ2 ≤ 0004

Ha: σ2 > 0004

2 2

2 0

.0004

χ

σ

Degrees of freedom = n - 1 = 29

Using χ2table, p-value is greater than 10

Exact p-value usingχ2= 36.25 is 1664

p-value > 05, do not reject H0 The product specification does not appear to be violated.

10 σ2 = (18.2)2 = 331.24

H0: σ2 ≤ 331.24

Ha: σ2 > 331.24

11 - 4

Trang 5

2 2 2

0

52.075 (18.2)

χ

σ

Usingχ2table, p-value is between 025 and 05

Exact p-value corresponding toχ2= 52.075 is 0317

p-value ≤ 05, reject H0 The standard deviation for Vanguard PRIMECAP fund is greater than the

average standard deviation for large cap mutual funds

3.3567 12

i

x x

n

=∑ = =

2

.6899

i

s

n

−

∑

.6899 8306

b H0: σ2 =.70

Ha: σ2 ≠.70

c

2 2

2 0

( 1) (12 1)(.6899)

10.84 70

χ

σ

Usingχ2table, area in upper tail is greater than 10

Two-tail p-value is greater than 20

Exact two-tailed p-value corresponding toχ2= 10.84 is 2(.4568) = 9136

p-value > 05, do not reject H0 We cannot conclude the variance in bond yields has changed.

12 a

2

1

i

x x s

n

−

b H0: σ2 = 94

Ha: σ2 ≠ 94

2 2

2 0

9.49 94

χ

σ

Trang 6

Usingχ2table, area in tail is greater than 10

p-value > 05, cannot reject H0.

13 a F.05 = 3.33

b F.025 = 2.76

c F.01 = 4.50

d F.10 =1.94

14 a

2

1

2

2.4

s

F

s

Degrees of freedom 15 and 20

Using F table, p-value is between 025 and 05

Exact p-value corresponding to F = 2.4 is 0345

p-value ≤ 05, reject H0 Conclude σ12 >σ22

b F.05 = 2.20

Reject H0 if F ≥ 2.20

2.4 ≥ 2.20, reject H0 Conclude σ12 >σ22

15 a Larger sample variance is s12

2

1

2

4

s

F

s

Using F table, area in tail is between 025 and 05

Two-tail p-value is between 05 and 10

p-value > 05, do not reject H0

b Since we have a two-tailed test

/ 2 025 2.30

Fα =F =

11 - 6

Trang 7

2.05 < 2.30, do not reject H0

16 For this type of hypothesis test, we place the larger variance in the numerator So the Fidelity

variance is given the subscript of 1

0: 1 2

:

a

1

2

18.9 1.59 15.0

s

F

s

= = =

Degrees of freedom in the numerator and denominator are both 60

Using the F table, p-value is less than 05

p-value ≤ 05, reject H0 We conclude that the Fidelity fund has a greater variance than the

American Century fund

17 a Population 1 is 4 year old automobiles

b

1

2

100

s

F

s

Using F table, p-value is less than 01

p-value ≤ 01, reject H0 Conclude that 4 year old automobiles have a larger variance in annual

repair costs compared to 2 year old automobiles This is expected due to the fact that older

automobiles are more likely to have some more expensive repairs which lead to greater variance in the annual repair costs

18 We place the larger sample variance in the numerator So, the Merrill Lynch variance is given the

subscript of 1

0: 1 2

Trang 8

2 2

:

a

1

2

587 1.44 489

s

F

s

= = =

Using F table, area in tail is greater than 10

p-value > 10, do not reject H0 We cannot conclude there is a statistically significant difference

between the variances for the two companies

19 H :σ0 12 =σ22

2

1

s =.0489

2

s =.0059

2

1

2

.0059

s

F

s

Using F table, area in tail is less than 01

Two-tail p-value is less than 02

Exact p-value ≈ 0

p-value≤ 05, reject H0 The process variances are significantly different Machine 1 offers the best

opportunity for process quality improvements

Note that the sample means are similar with the mean bag weights of approximately 3.3 grams However, the process variances are significantly different

20 H :σ0 12 =σ22

2

1

2

11.1 5.29 2.1

s

F

s

Using F table, area in tail is less than 01

11 - 8

Trang 9

Two-tail p-value is less than 02

Exact p-value ≈ 0

p-value ≤ 05, reject H0 The population variances are not equal for seniors and managers

21 Consider the Small cap fund as population 1 and the large cap fund as population 2

1

2

(13.03)

2.15 (8.89)

s

F

s

Upper-tail p-value is the area to the right of the test statistic.

From the F table, the p-value is between 025 and 05

p-value .05, reject ≤ H0 The population variance and standard deviation for the small cap growth

fund are larger Financial analysts would say the small cap growth fund is riskier

22 a Population 1 - Wet pavement

1

2

32 4.00 16

s

F

s

Using F table, p-value is less than 01

p-value≤ 05, reject H0 Conclude that there is greater variability in stopping distances on wet

pavement

b Drive carefully on wet pavement because of the uncertainty in stopping distances

23 a s2 = (30) 2 = 900

b χ.052 = 30.144 and χ.952 = 10.117 (19 degrees of freedom)

2

Trang 10

567 ≤σ2≤ 1690

c 23.8 ≤σ ≤ 41.1

24 With 12 degrees of freedom,

2

.025

χ = 23.337 χ.9752 = 4.404

2

114.9 ≤σ2≤ 609

10.72 ≤σ ≤ 24.68

25 a x i 260.16

x

n

Σ

b

2

4996.8 1

i

x x s

n

−

c χ.0252 = 32.852 χ.9752 = 8.907 (19 degrees of freedom)

2

2890 ≤σ2≤ 10,659

53.76 ≤σ ≤ 103.24

26 a H0: σ2 ≤ 0001

H a: σ2 > 0001

2

.0001

χ

σ

Usingχ2table, p- value is between 01 and 025

p-value ≤ 10, reject H0 Variance exceeds maximum variance requirement.

b χ.052 = 23.685

11 - 10

Trang 11

.95

2

.00012 ≤σ2≤ 00042

27 H0: σ2 ≤ 02

Ha: σ2 > 02

2

2 0

.02

χ

σ

Usingχ2table, p- value is greater than 10

p-value > 05, do not reject H0 The population variance does not appear to be exceeding the

standard

28 H0: σ2 ≤ 1

Ha: σ2 > 1

2 2

2 0

31.50 1

χ

σ

Usingχ2table, p-value is between 05 and 10

p-value ≤ 10, reject H0 Conclude that σ2 > 1

29

2

12.69

i

x x s

n

H0: σ2 = 10

Ha: σ2≠ 10

2 2

2

10.16 10

χ

σ

Trang 12

Usingχ2table, area in tail is greater than 10

Two-tail p- value is greater than 20

p-value > 10, do not reject H0

30 a Try n = 15

2

.025

χ = 26.119

2

.975

2

34.3 ≤σ2≤ 159.2

5.86 ≤σ ≤ 12.62

∴ A sample size of 15 was used

b n = 25; expect the width of the interval to be smaller.

2

.05

χ = 39.364

2

.975

2

39.02 ≤σ2≤ 126.86

6.25 ≤σ ≤ 11.13

31 H0:σ12 =σ22

:

a

Population 1 is women’s scores

1

2

2.4623

1.24 2.2118

s

F

s

= = =

11 - 12

Trang 13

Using Excel of Minitab, we find the exact two-tail p-value corresponding to F = 1.24 is 5876 p-value > 10, do not reject H0 There is not a statistically significant difference in the variances

We cannot conclude that there is a difference in the variability of golf scores for male and female professional golfers

32 H0:σ12 =σ22

:

a

H σ ≠σ

Use critical value approach since F tables do not have 351 and 72 degrees of freedom.

F.025 = 1.47

1

2

.797

s

F

s

F < 1.47, do not reject H0 We are not able to conclude students who complete the course and

students who drop out have different variances of grade point averages

33 H0:σ12 =σ22

:

a

H σ ≠σ

Population 1 has the larger sample variance

2

1

2

2.3

s

F

s

p-value > 10, do not reject H0 Cannot conclude that there is a difference between the population

variances

34 H0:σ12 =σ22

:

a

H σ ≠σ

2

1

2

25 2.08 12

s

F

s

Trang 14

p-value ≤ 10, reject H0 Conclude that the population variances are not equal.

11 - 14

Định dạng
Số trang	14
Dung lượng	399 KB