Presentation on theme: "Chapter 11 Inferences About Population Variances"— Presentation transcript:
1Chapter 11 Inferences About Population Variances Inference about a Population VarianceInferences about Two Population Variances
2Inferences About a Population Variance Chi-Square DistributionInterval EstimationHypothesis Testing卡方分布
3Chi-Square Distribution The chi-square distribution is the sum of squaredstandardized normal random variables such as(z1)2+(z2)2+(z3)2 and so on.The chi-square distribution is based on samplingfrom a normal population.The sampling distribution of (n - 1)s2/ 2 has a chi-square distribution whenever a simple random sampleof size n is selected from a normal population.卡方分布 (χ2分布)是概率论与统计学中常用的一种概率分布。k 个独立的标准正态分布变量的平方和服从自由度为k 的卡方分布。卡方分布常用于假设检验和置信区间的计算。We can use the chi-square distribution to developinterval estimates and conduct hypothesis testsabout a population variance.
4Examples of Sampling Distribution of (n - 1)s2/ 2 With 2 degreesof freedomWith 5 degreesof freedomWith 10 degreesof freedom
6Chi-Square Distribution We will use the notation to denote the value for the chi-square distribution that provides an area of a to the right of the stated value.For example, there is a .95 probability of obtaining a c2 (chi-square) value such that
7Interval Estimation of 2 .025.02595% of thepossible 2 values2
8Interval Estimation of 2 There is a (1 – a) probability of obtaining a c2 valuesuch thatSubstituting (n – 1)s2/s 2 for the c2 we getPerforming algebraic manipulation we get
9Interval Estimation of 2 Interval Estimate of a Population Variancewhere the values are based on a chi-squaredistribution with n - 1 degrees of freedom andwhere 1 - is the confidence coefficient.
10Interval Estimation of Interval Estimate of a Population Standard DeviationTaking the square root of the upper and lowerlimits of the variance interval provides the confidenceinterval for the population standard deviation.
11Interval Estimation of 2 Example: Buyer’s Digest (A)Buyer’s Digest rates thermostatsmanufactured for home temperaturecontrol. In a recent test, 10 thermostatsmanufactured by ThermoRite wereselected and placed in a test room thatwas maintained at a temperature of 68oF.The temperature readings of the ten thermostats areshown on the next slide.
12Interval Estimation of 2 Example: Buyer’s Digest (A)We will use the 10 readings below todevelop a 95% confidence intervalestimate of the population variance.ThermostatTemperature
13Interval Estimation of 2 For n - 1 = = 9 d.f. and a = .05Selected Values from the Chi-Square Distribution TableOur value
14Interval Estimation of 2 For n - 1 = = 9 d.f. and a = .05.025Area inUpper Tail= .97522.700
15Interval Estimation of 2 For n - 1 = = 9 d.f. and a = .05Selected Values from the Chi-Square Distribution TableOur value
16Interval Estimation of 2 n - 1 = = 9 degrees of freedom and a = .05.025Area in UpperTail = .02522.70019.023
17Interval Estimation of 2 Sample variance s2 provides a point estimate of 2.A 95% confidence interval for the population variance is given by:.33 < 2 < 2.33
19Hypothesis Testing About a Population Variance Example: Buyer’s Digest (B)Recall that Buyer’s Digest is ratingThermoRite thermostats. Buyer’s Digestgives an “acceptable” rating to a thermo-stat with a temperature variance of 0.5or less.We will conduct a hypothesis test (witha = .10) to determine whether the ThermoRitethermostat’s temperature variance is “acceptable”.
20Hypothesis Testing About a Population Variance Example: Buyer’s Digest (B)Using the 10 readings, we willconduct a hypothesis test (with a = .10)to determine whether the ThermoRitethermostat’s temperature variance is“acceptable”.ThermostatTemperature
21Hypothesis Testing About a Population Variance HypothesesRejection RuleReject H0 if c 2 >
22Hypothesis Testing About a Population Variance For n - 1 = = 9 d.f. and a = .10Selected Values from the Chi-Square Distribution TableOur value
23Hypothesis Testing About a Population Variance Rejection RegionArea in UpperTail = .10214.684Reject H0
24Hypothesis Testing About a Population Variance Test StatisticThe sample variance s 2 = 0.7ConclusionBecause c2 = 12.6 is less than , we cannotreject H0. The sample variance s2 = .7 is insufficientevidence to conclude that the temperature variancefor ThermoRite thermostats is unacceptable.
25Hypothesis Testing About a Population Variance Using the p-ValueThe rejection region for the ThermoRitethermostat example is in the upper tail; thus, theappropriate p-value is less than .90 (c 2 = 4.168)and greater than .10 (c 2 = ).A precise p-valuecan be found usingMinitab or Excel.Because the p –value > a = .10, wecannot reject the null hypothesis.The sample variance of s 2 = .7 isinsufficient evidence to conclude that thetemperature variance is unacceptable (>.5).
28Hypothesis Tests for Two Variances PopulationVariancesGoal: Test hypotheses about twopopulation variancesH0: σx2 σy2H1: σx2 < σy2Lower-tail testF test statisticH0: σx2 ≤ σy2H1: σx2 > σy2Upper-tail test在统计学的一些应用中，我们或许想比较两种生产线生产出来的产品质量的方差、两种装配方法所需装配时间的方差或者两种加热装置的温度的方差。对两个总体的方差进行比较时，我们可以使用两个独立的随机抽取的样本，设它们分别取自总体1和总体2。两个样本方法可以作为推断两个总体方差的基础。F分布表是以两个正态分布的抽样分布为基础的。H0: σx2 = σy2H1: σx2 ≠ σy2Two-tail testThe two populations are assumed to be independent and normally distributed
32Hypothesis Tests for Two Variances (continued)The random variableTests for TwoPopulationVariancesF test statisticHas an F distribution with (nx – 1) numerator degrees of freedom and (ny – 1) denominator degrees of freedomDenote an F value with 1 numerator and 2 denominator degrees of freedom by
33Test Statistic Tests for Two PopulationVariancesThe critical value for a hypothesis test about two population variances isF test statisticwhere F has (nx – 1) numerator degrees of freedom and (ny – 1) denominator degrees of freedom
34Hypothesis Testing About the Variances of Two Populations One-Tailed TestHypothesesDenote the population providing thelarger sample variance as population 1.Test Statistic
35Hypothesis Testing About the Variances of Two Populations One-Tailed Test (continued)Rejection RuleCritical value approach:Reject H0 if F > Fwhere the value of F is based on anF distribution with n1 - 1 (numerator)and n2 - 1 (denominator) d.f.p-Value approach:Reject H0 if p-value < a
36Hypothesis Testing About the Variances of Two Populations Two-Tailed TestHypothesesDenote the population providing thelarger sample variance as population 1.Test Statistic
38Hypothesis Testing About the Variances of Two Populations Two-Tailed Test (continued)Rejection RuleCritical value approach:Reject H0 if F > F/2where the value of F/2 is based on anF distribution with n1 - 1 (numerator)and n2 - 1 (denominator) d.f.p-Value approach:Reject H0 if p-value < a
39Decision Rules: Two Variances Use sx2 to denote the larger variance.H0: σx2 = σy2H1: σx2 ≠ σy2H0: σx2 ≤ σy2H1: σx2 > σy2/2FFDo notreject H0Reject H0Do notreject H0Reject H0rejection region for a two-tail test is:where sx2 is the larger of the two sample variances
40Hypothesis Testing About the Variances of Two Populations Example: Buyer’s Digest (C)Buyer’s Digest has conducted thesame test, as was described earlier, onanother 10 thermostats, this timemanufactured by TempKing. Thetemperature readings of the tenthermostats are listed on the next slide.We will conduct a hypothesis test with = .10 to seeif the variances are equal for ThermoRite’s thermostatsand TempKing’s thermostats.
41Hypothesis Testing About the Variances of Two Populations Example: Buyer’s Digest (C)ThermoRite SampleThermostatTemperatureTempKing SampleThermostatTemperature
42Hypothesis Testing About the Variances of Two Populations Hypotheses(TempKing and ThermoRite thermostatshave the same temperature variance)(Their variances are not equal)Rejection RuleThe F distribution table (on next slide) shows that withwith = .10, 9 d.f. (numerator), and 9 d.f. (denominator),F.05 = 3.18.Reject H0 if F > 3.18
43Hypothesis Testing About the Variances of Two Populations Selected Values from the F Distribution Table
44Hypothesis Testing About the Variances of Two Populations Test StatisticTempKing’s sample variance is 1.768ThermoRite’s sample variance is .700= 1.768/.700 = 2.53ConclusionWe cannot reject H0. F = 2.53 < F.05 = 3.18.There is insufficient evidence to conclude thatthe population variances differ for the twothermostat brands.
45Hypothesis Testing About the Variances of Two Populations Determining and Using the p-ValueArea in Upper TailF Value (df1 = 9, df2 = 9)Because F = 2.53 is between 2.44 and 3.18, the areain the upper tail of the distribution is between .10and .05.But this is a two-tailed test; after doubling the upper-tail area, the p-value is between .20 and .10. (A precisep-value can be found using Minitab or Excel.)Because a = .10, we have p-value > a and thereforewe cannot reject the null hypothesis.