the chi-square distribution results when independent variables with standard normal distributions...
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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
The chi-square distribution results when independent variables with standard normal distributions are squared and summed.
0 n – 1
2 ( 1)n
Sampling distribution ofc2
The chi-square distribution results when independent variables with standard normal distributions are squared and summed.
2.025
.025
22
2-stat ( 1)
sn
Sampling distribution ofc2
The chi-square distribution results when independent variables with standard normal distributions are squared and summed.
2.975
.975
22
2-stat ( 1)
sn
Sampling distribution ofc2
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
Interval Estimation of s 2
2 2
22 2.025 .975
( 1) ( 1)n s n s
Interval Estimation of s 2
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
Interval Estimation of s 2
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
Interval Estimation of s 2
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
Interval Estimation of s 2
68.1x
Interval Estimation of s 2
2 22
2 2/ 2 1 / 2
( 1) ( 1)n s n s(10 -1)(0.7) (10 -1)(0.7)
Interval Estimation of s 2
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
Interval Estimation of s 2
Interval Estimation of s 2
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)
Hypothesis Testing – One Variance
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
Hypothesis Testing – One Variance
The F-distribution results from taking the ratio of variances of normally distributed variables.
Sampling distribution of F
2122
1
if s12 = s2
2
The F-distribution results from taking the ratio of variances of normally distributed variables.
Sampling distribution of F
Bigger
2122
-stats
Fs
≈1 if s12 = s2
2
0 1
.025F
.025
The F-distribution results from taking the ratio of variances of normally distributed variables.
Sampling distribution of F
2122
-stats
Fs
≈1
.975F
The F-distribution results from taking the ratio of variances of normally distributed variables.
Sampling distribution of F
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
Hypothesis Testing – Two Variances
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)
Hypothesis Testing – Two Variances
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
Hypothesis Testing – Two Variances
-statF2122
ss
1.7680.70
2.53
2.53
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