the chi-square distribution results when independent variables with standard normal distributions...

The chi-square distribution results when independent variables with standard normal distributions are squared and summed.

22 ( )

1ix x

sn

Sampling distribution of s2

22

2-stat ( 1)

sn


0 n – 1

2 ( 1)n

Sampling distribution ofc2


2.025

.025

22

2-stat ( 1)

sn



2.975

.975

22

2-stat ( 1)

sn


2 2.975 5

2

22 .0

( 1)n s

2

2.975

2( 1)n s

2

22.025

( 1)n s

Interval Estimation of s 2

To derive the interval estimate of 2, first substitute(n - 1)s2/ 2 for c2 into the following inequality

22 2.975 .025

Now write the above as two inequalities

2

22.0252

2( 1)n s

2

2

( 1)n s 2 2.025

2 2 2.025

2( 1)n s

Next, multiply the inequalities by 2

22

.975 22

2( 1)n s

2

2.9

275 2

( 1)n s 2 22 2.975 ( 1)n s

Divide both of the inequalities above by the respective chi-square critical value:

2 2

.2

2 2.025 .025

025( 1)n s

2

.02

2.0

5

225( 1)n s

2

2.025

22.025

2( 1)n s

2

2 2.975 .975

22.975 ( 1)n s 2.975

2

2.975

2

2.975

( 1)n s

22.97

2

5

( 1)n s


2 2

22 2.025 .975

( 1) ( 1)n s n s


2 2

22 2.025 .975

( 1) ( 1)n s n s

The 95% confidence interval for the population variance

2 2

22 2.025 .975

( 1) ( 1)n s n s


1 - a/2a/2

The 1 - a confidence interval for the population variance

Buyer’s Digest rates thermostats manufactured for home temperature control. In a recent test, ten thermostats manufactured by ThermoRite were selected at random and placed in a test room that was maintained at a temperature of 68oF.

Use the ten readings in the table below to develop a 95% confidence interval estimate of the population variance.

Example 1

Temperature 67.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.2

Thermostat 1 2 3 4 5 6 7 8 9 10


67.4 67.8 68.2 69.3 69.5 67.068.1 68.667.967.2

-0.7-0.30.11.21.4

-1.10.00.5

-0.2-0.9

0.490.090.011.441.961.210.000.250.040.81

22 ( )

1ix x

sn

2( )ix xix x

sum = 6.3s 2 = 0.7

268 )1( .ix 68.1ix ix


68.1x


2 22

2 2/ 2 1 / 2

( 1) ( 1)n s n s(10 -1)(0.7) (10 -1)(0.7)


2 22

2 2/ 2 1 / 2

( 1) ( 1)n s n s(10 -1)(0.7) (10 -1)(0.7)

290 .975

2

.025

.0252

.025

1-a = .95

0.9750.25

Selected Values from the Chi-Square Distribution Table

Degrees Area in Upper Tail

of Freedom .99.975 .95 .90 .10 .05 .025 .01

50.55

40.83

11.14

51.61

0 9.23611.07

012.83

215.08

6

60.87

21.23

71.63

52.20

410.64

512.59

214.44

916.81

2

71.23

91.69

02.16

72.83

312.01

714.06

716.01

318.47

5

81.64

72.18

02.73

33.49

013.36

215.50

717.53

520.09

0

92.08

82.70

03.32

54.16

814.68

416.91

919.02

321.66

6

102.55

83.24

73.94

04.86

515.98

718.30

720.48

323.20

9

2.025

2.975



2 22

2 2/ 2 1 / 2

( 1) ( 1)n s n s(10 -1)(0.7) (10 -1)(0.7)

0.9750.2519.023 2.700

20.331 2.333

We are 95% confident that the populationvariance is in this interval

Recall that Buyer’s Digest is rating ThermoRite thermostats. Buyer’s Digest gives an “acceptable” rating to a thermostat with a temperature variance of 0.5 or less.

Conduct a hypothesis test--at the 10% significance level--to

determine whether the ThermoRite thermostat’s temperature

variance is “acceptable”.

Hypotheses: 20 : 0.5H

2: 0.5aH

2 -stat

Recall that s2 = 0.7 and df = 9. With = 0.5, 20

2

20

( 1)sn

(0.(9) 7)

0.512.6

Example 2

Hypothesis Testing – One Variance

a = .10 (column)

Selected Values from the Chi-Square Distribution Table

Degrees Area in Upper Tail

of Freedom .99.975 .95 .90 .10 .05 .025 .01

50.55

40.83

11.14

51.61

0 9.23611.07

012.83

215.08

6

60.87

21.23

71.63

52.20

410.64

512.59

214.44

916.81

2

71.23

91.69

02.16

72.83

312.01

714.06

716.01

318.47

5

81.64

72.18

02.73

33.49

013.36

215.50

717.53

520.09

0

92.08

82.70

03.32

54.16

814.68

416.91

919.02

321.66

6

102.55

83.24

73.94

04.86

515.98

718.30

720.48

323.20

9

Our value 2

.10

and df = 10 – 1 = 9 (row)


214.684

.10

2-stat

Do not reject H0 Reject H0

9

There is insufficient evidence to conclude that the temperature variance for ThermoRite thermostats is unacceptable.

= .101 9n 20 : 0.5H

12.6


The F-distribution results from taking the ratio of variances of normally distributed variables.

Sampling distribution of F

2122

1

if s12 = s2

2



Bigger

2122

-stats

Fs

≈1 if s12 = s2

2

0 1

.025F

.025



2122

-stats

Fs

≈1

.975F



2122

-stats

Fs

≈1

.975

Buyer’s Digest has conducted the same test, but on 10 other thermostats. This time it test thermostats manufactured by TempKing. The temperature readings of the 10 thermostats are listed below.

We will conduct a hypothesis at a 10% level of significance to see if the variances are equal for both thermostats.

Example 3

ThermoRite Sample

TempKing Sample

Temperature 67.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.2

Temperature 67.7 66.4 69.2 70.1 69.5 69.7 68.1 66.6 67.3 67.5

s2 = 0.7 and df = 9

s2 = ? and df = 9

Hypothesis Testing – Two Variances

67.766.469.270.169.569.768.166.667.367.5

-0.51-1.810.991.891.291.49

-0.11-1.61-0.91-0.71

0.26013.27610.98013.57211.66412.22010.01212.59210.82810.5041

ix

22 ( )

1ix x

sn

2( )ix xix x

sum = 15.909s 2 = 1.768

68.21x

268 )1( .2ix 68.21ix TempKing

Since this is largerThan ThermoRite’s

21 1.768s

22 0.7s


n1 = 10 – 1 = 9 (column)

Selected Values from the F Distribution Table

Denominator Area in Numerator Degrees of Freedom

Degrees Upper

of Freedom Tail 7 8 9 10 15

.01 6.18 6.03 5.91 5.81 5.52

9 .10 2.51 2.47 2.44 2.42 2.34

.05 3.29 3.23 3.18 3.14 3.01

.025 4.20 4.10 4.03 3.96 3.77

.01 5.61 5.47 5.35 5.26 4.96.05F

& n 2 - 1 = 9 a/2 = .05 (row)


Hypotheses: 2 2

0 1 2:H 2 2

1 2:aH

F.95F

.05

Reject H0 Do not Reject H0 Reject H0

≈ 1

There is insufficient evidence to conclude that the population

variances differ for the two thermostat brands.

3.18

.05


-statF2122

ss

1.7680.70

2.53

2.53

http://www.halsnarr.com/ppt/data_unemp.xlsx

the chi-square distribution results when independent variables with standard normal distributions...

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