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The chi-square distribution results when independent variables with standard normal distributions are squared and summed. Sampling distribution ofs2s2.

Presentation on theme: "The chi-square distribution results when independent variables with standard normal distributions are squared and summed. Sampling distribution ofs2s2."— Presentation transcript:

The chi-square distribution results when independent variables with standard normal distributions are squared and summed. Sampling distribution ofs2s2

The chi-square distribution results when independent variables with standard normal distributions are squared and summed. 0 n – 1 Sampling distribution of 22

The chi-square distribution results when independent variables with standard normal distributions are squared and summed..025 Sampling distribution of 22

The chi-square distribution results when independent variables with standard normal distributions are squared and summed..975 Sampling distribution of 22

Interval Estimation of  2 To derive the interval estimate of  2, first substitute (n  1)s 2 /  2 for  2 into the following inequality Now write the above as two inequalities

Next, multiply the inequalities by  2 Divide both of the inequalities above by the respective chi- square critical value: Interval Estimation of  2

The 95% confidence interval for the population variance

Interval Estimation of  2   The 1 -  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 68 o F. 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  2

67.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.2 -0.7 -0.3 0.1 1.2 1.4 -1.1 0.0 0.5 -0.2 -0.9 0.49 0.09 0.01 1.44 1.96 1.21 0.00 0.25 0.04 0.81 sum = 6.3 s 2 = 0.7 Interval Estimation of  2

(10 -1)(0.7)

Interval Estimation of  2 (10 -1)(0.7) 22 9 0.025   

Selected Values from the Chi-Square Distribution Table DegreesArea in Upper Tail of Freedom.99.975.95.90.10.05.025.01 50.5540.8311.1451.6109.23611.07012.83215.086 60.8721.2371.6352.20410.64512.59214.44916.812 71.2391.6902.1672.83312.01714.06716.01318.475 81.6472.1802.7333.49013.36215.50717.53520.090 92.0882.7003.3254.16814.68416.91919.02321.666 102.5583.2473.9404.86515.98718.30720.48323.209 Interval Estimation of  2

(10 -1)(0.7)   We are 95% confident that the population variance 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: Recall that s 2 = 0.7 and df = 9. With = 0.5, Example 2 Hypothesis Testing – One Variance

 =.10 (column) Selected Values from the Chi-Square Distribution Table DegreesArea in Upper Tail of Freedom.99.975.95.90.10.05.025.01 50.5540.8311.1451.6109.23611.07012.83215.086 60.8721.2371.6352.20410.64512.59214.44916.812 71.2391.6902.1672.83312.01714.06716.01318.475 81.6472.1802.7333.49013.36215.50717.53520.090 92.0882.7003.3254.16814.68416.91919.02321.666 102.5583.2473.9404.86515.98718.30720.48323.209 Our value and df = 10 – 1 = 9 (row) Hypothesis Testing – One Variance

 Do not reject H 0 Reject H 0 9 There is insufficient evidence to conclude that the temperature variance for ThermoRite thermostats is unacceptable.  =.10 Hypothesis Testing – One Variance

The F-distribution results from taking the ratio of variances of normally distributed variables. Sampling distribution of F if  1 2 =  2 2

The F-distribution results from taking the ratio of variances of normally distributed variables. Sampling distribution of F Bigger ≈1≈1 if  1 2 =  2 2 0 1

.025 The F-distribution results from taking the ratio of variances of normally distributed variables. Sampling distribution of F ≈1≈1

The F-distribution results from taking the ratio of variances of normally distributed variables. Sampling distribution of F ≈1≈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 s 2 = 0.7 and df = 9 s 2 = ? and df = 9 Hypothesis Testing – Two Variances

67.7 66.4 69.2 70.1 69.5 69.7 68.1 66.6 67.3 67.5 -0.51 -1.81 0.99 1.89 1.29 1.49 -0.11 -1.61 -0.91 -0.71 0.2601 3.2761 0.9801 3.5721 1.6641 2.2201 0.0121 2.5921 0.8281 0.5041 sum = 15.909 s 2 = 1.768 TempKing Since this is larger Than ThermoRite’s Hypothesis Testing – Two Variances

n 1 = 10 – 1 = 9 (column) Selected Values from the F Distribution Table DenominatorArea in Numerator Degrees of Freedom DegreesUpper of FreedomTail 7891015.016.186.035.915.815.52 9.102.512.472.442.422.34.053.293.233.183.143.01.0254.204.104.033.963.77.015.615.475.355.264.96 & n    =.05 (row) Hypothesis Testing – Two Variances Hypotheses:

.05 Reject H 0 Do not Reject H 0 Reject H 0 ≈ 1 There is insufficient evidence to conclude that the population variances differ for the two thermostat brands..05 Hypothesis Testing – Two Variances

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