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1 1 Slide Chapter 11 Inferences About Population Variances n Inference about a Population Variance n Inferences about Two Population Variances.

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Presentation on theme: "1 1 Slide Chapter 11 Inferences About Population Variances n Inference about a Population Variance n Inferences about Two Population Variances."— Presentation transcript:

1 1 1 Slide Chapter 11 Inferences About Population Variances n Inference about a Population Variance n Inferences about Two Population Variances

2 2 2 Slide Inferences About a Population Variance n Chi-Square Distribution n Interval Estimation n Hypothesis Testing

3 3 3 Slide Chi-Square Distribution We can use the chi-square distribution to develop We can use the chi-square distribution to develop interval estimates and conduct hypothesis tests interval estimates and conduct hypothesis tests about a population variance. about a population variance. The sampling distribution of ( n - 1) s 2 /  2 has a chi- The sampling distribution of ( n - 1) s 2 /  2 has a chi- square distribution whenever a simple random sample square distribution whenever a simple random sample of size n is selected from a normal population. of size n is selected from a normal population. The chi-square distribution is based on sampling The chi-square distribution is based on sampling from a normal population. from a normal population. n The chi-square distribution is the sum of squared standardized normal random variables such as standardized normal random variables such as ( z 1 ) 2 +( z 2 ) 2 +( z 3 ) 2 and so on. ( z 1 ) 2 +( z 2 ) 2 +( z 3 ) 2 and so on.

4 4 4 Slide Examples of Sampling Distribution of ( n - 1) s 2 /  With 2 degrees of freedom of freedom With 2 degrees of freedom of freedom With 5 degrees of freedom of freedom With 5 degrees of freedom of freedom With 10 degrees of freedom of freedom With 10 degrees of freedom of freedom

5 5 5 Slide

6 6 6 Chi-Square Distribution For example, there is a.95 probability of obtaining a  2 (chi-square) value such that For example, there is a.95 probability of obtaining a  2 (chi-square) value such that We will use the notation to denote the value for the chi-square distribution that provides an area of  to the right of the stated value. We will use the notation to denote the value for the chi-square distribution that provides an area of  to the right of the stated value.

7 7 7 Slide 95% of the possible  2 values 95% of the possible  2 values 22 2 Interval Estimation of  2

8 8 8 Slide Interval Estimation of  2 Substituting ( n – 1) s 2 /  2 for the  2 we get Substituting ( n – 1) s 2 /  2 for the  2 we get n Performing algebraic manipulation we get There is a (1 –  ) probability of obtaining a  2 value There is a (1 –  ) probability of obtaining a  2 value such that such that

9 9 9 Slide n Interval Estimate of a Population Variance Interval Estimation of  2 where the    values are based on a chi-square distribution with n - 1 degrees of freedom and where 1 -  is the confidence coefficient.

10 10 Slide Interval Estimation of  n Interval Estimate of a Population Standard Deviation Taking the square root of the upper and lower Taking the square root of the upper and lower limits of the variance interval provides the confidence interval for the population standard deviation.

11 11 Slide Buyer’s Digest rates thermostats manufactured for home temperature control. In a recent test, 10 thermostats manufactured by ThermoRite were selected and placed in a test room that was maintained at a temperature of 68 o F. The temperature readings of the ten thermostats are The temperature readings of the ten thermostats are shown on the next slide. Interval Estimation of  2 n Example: Buyer’s Digest (A)

12 12 Slide Interval Estimation of  2 We will use the 10 readings below to We will use the 10 readings below to develop a 95% confidence interval estimate of the population variance. n Example: Buyer’s Digest (A) Temperature Thermostat

13 13 Slide Interval Estimation of  2 Selected Values from the Chi-Square Distribution Table Our value For n - 1 = = 9 d.f. and  =.05 For n - 1 = = 9 d.f. and  =.05

14 14 Slide Interval Estimation of  2 22 2 Area in Upper Tail = For n - 1 = = 9 d.f. and  =.05 For n - 1 = = 9 d.f. and  =.05

15 15 Slide Interval Estimation of  2 Selected Values from the Chi-Square Distribution Table For n - 1 = = 9 d.f. and  =.05 For n - 1 = = 9 d.f. and  =.05 Our value

16 16 Slide 22 2 Interval Estimation of  2 n - 1 = = 9 degrees of freedom and  =.05 n - 1 = = 9 degrees of freedom and  = Area in Upper Tail =.025 Area in Upper Tail =.025

17 17 Slide Sample variance s 2 provides a point estimate of  2. Sample variance s 2 provides a point estimate of  2. Interval Estimation of  2.33 <  2 < 2.33 n A 95% confidence interval for the population variance is given by:

18 18 Slide Hypothesis Testing About a Population Variance Lower-tail test: H 0 : σ 2  σ 0 2 H 1 : σ 2 < σ 0 2 Upper-tail test: H 0 : σ 2 ≤ σ 0 2 H 1 : σ 2 > σ 0 2 Two-tail test: H 0 : σ 2 = σ 0 2 H 1 : σ 2 ≠ σ 0 2  /2  Reject H 0 if or

19 19 Slide Recall that Buyer’s Digest is rating Recall that Buyer’s Digest is rating ThermoRite thermostats. Buyer’s Digest gives an “acceptable” rating to a thermo- stat with a temperature variance of 0.5 or less. Hypothesis Testing About a Population Variance n Example: Buyer’s Digest (B) We will conduct a hypothesis test (with We will conduct a hypothesis test (with  =.10) to determine whether the ThermoRite thermostat’s temperature variance is “acceptable”.

20 20 Slide Hypothesis Testing About a Population Variance Using the 10 readings, we will Using the 10 readings, we will conduct a hypothesis test (with  =.10) to determine whether the ThermoRite thermostat’s temperature variance is “acceptable”. n Example: Buyer’s Digest (B) Temperature Thermostat

21 21 Slide n Hypotheses Hypothesis Testing About a Population Variance Reject H 0 if  2 > n Rejection Rule

22 22 Slide Selected Values from the Chi-Square Distribution Table For n - 1 = = 9 d.f. and  =.10 For n - 1 = = 9 d.f. and  =.10 Hypothesis Testing About a Population Variance Our value

23 23 Slide 22 2 Area in Upper Tail =.10 Area in Upper Tail =.10 Hypothesis Testing About a Population Variance n Rejection Region Reject H 0

24 24 Slide n Test Statistic Hypothesis Testing About a Population Variance Because  2 = 12.6 is less than , we cannot Because  2 = 12.6 is less than , we cannot reject H 0. The sample variance s 2 =.7 is insufficient evidence to conclude that the temperature variance for ThermoRite thermostats is unacceptable. n Conclusion The sample variance s 2 = 0.7

25 25 Slide n Using the p -Value The sample variance of s 2 =.7 is The sample variance of s 2 =.7 is insufficient evidence to conclude that the insufficient evidence to conclude that the temperature variance is unacceptable (>.5). temperature variance is unacceptable (>.5). Because the p –value >  =.10, we Because the p –value >  =.10, we cannot reject the null hypothesis. cannot reject the null hypothesis. The rejection region for the ThermoRite The rejection region for the ThermoRite thermostat example is in the upper tail; thus, the thermostat example is in the upper tail; thus, the appropriate p -value is less than.90 (  2 = 4.168) appropriate p -value is less than.90 (  2 = 4.168) and greater than.10 (  2 = ). and greater than.10 (  2 = ). Hypothesis Testing About a Population Variance A precise p -value can be found using Minitab or Excel. A precise p -value can be found using Minitab or Excel.

26 26 Slide

27 27 Slide

28 28 Slide Hypothesis Tests for Two Variances Tests for Two Population Variances F test statistic H 0 : σ x 2 = σ y 2 H 1 : σ x 2 ≠ σ y 2 Two-tail test Lower-tail test Upper-tail test H 0 : σ x 2  σ y 2 H 1 : σ x 2 < σ y 2 H 0 : σ x 2 ≤ σ y 2 H 1 : σ x 2 > σ y 2  Goal: Test hypotheses about two population variances The two populations are assumed to be independent and normally distributed

29 29 Slide F 分布 F 分布定义为 : 设 X 、 Y 为两个独立的随机变量, X 服从自由度为 m 的卡方分布, Y 服从自由度为 n 的 卡方分布,这 2 个独立的卡方分布被各自的自由 度除以后的比率这一统计量的分布即: F= ( x/m ) /(y/n) F= ( x/m ) /(y/n) 服从自由度为( m,n) 的 F- 分布, 上式 F 服从第一 自由度为 m ,第二自由度为 n 的 F 分布 服从自由度为( m,n) 的 F- 分布, 上式 F 服从第一 自由度为 m ,第二自由度为 n 的 F 分布

30 30 Slide F 分布 n F 分布的性质 1 、它是一种非对称分布; 2 、它有两个自由度,即 n1 -1 和 n2-1 ,相应的分布 记为 F ( n1 –1 , n2-1 ), n1 –1 通常称为分子自 由度, n2-1 通常称为分母自由度; 3 、 F 分布是一个以自由度 n1 –1 和 n2-1 为参数的分布 族,不同的自由度决定了 F 分布的形状。 4 、 F 分布的倒数性质: F α,df1,df2 =1/F 1-α,df2,df1

31 31 Slide F 分布

32 32 Slide Hypothesis Tests for Two Variances Tests for Two Population Variances F test statistic The random variable Has an F distribution with (n x – 1) numerator degrees of freedom and (n y – 1) denominator degrees of freedom Denote an F value with 1 numerator and 2 denominator degrees of freedom by (continued)

33 33 Slide Test Statistic Tests for Two Population Variances F test statistic The critical value for a hypothesis test about two population variances is where F has (n x – 1) numerator degrees of freedom and (n y – 1) denominator degrees of freedom

34 34 Slide n One-Tailed Test Test Statistic Test Statistic Hypotheses Hypotheses Hypothesis Testing About the Variances of Two Populations Denote the population providing the larger sample variance as population 1.

35 35 Slide n One-Tailed Test (continued) Reject H 0 if p -value <  where the value of F  is based on an F distribution with n (numerator) and n (denominator) d.f. p -Value approach: Critical value approach: Rejection Rule Rejection Rule Hypothesis Testing About the Variances of Two Populations Reject H 0 if F > F 

36 36 Slide n Two-Tailed Test Test Statistic Test Statistic Hypotheses Hypotheses Hypothesis Testing About the Variances of Two Populations Denote the population providing the larger sample variance as population 1.

37 37 Slide n 虽然我们可以通过以上公式求左侧 F 值,但通常在进行假设检验 计算时,只需知道右侧 F 值。进行 假设检验时,可以随意 规定哪个总体是总体 1 或总体 2 ,我们用总体 1 来代表方差较大的总 体时,只有在右侧才有可能出现 H 0 被拒绝的情况。虽然左侧临界 值仍存在,但我们不需要知道它值,因为使用方差较大的总体作 为总体 1 , s 1 2 /s 2 2 往往出现在右侧。

38 38 Slide n Two-Tailed Test (continued) Reject H 0 if p -value <  p -Value approach: Critical value approach: Rejection Rule Rejection Rule Hypothesis Testing About the Variances of Two Populations Reject H 0 if F > F  /2 where the value of F  /2 is based on an F distribution with n (numerator) and n (denominator) d.f.

39 39 Slide Decision Rules: Two Variances rejection region for a two- tail test is: F 0  Reject H 0 Do not reject H 0 F0  /2 Reject H 0 Do not reject H 0 H 0 : σ x 2 = σ y 2 H 1 : σ x 2 ≠ σ y 2 H 0 : σ x 2 ≤ σ y 2 H 1 : σ x 2 > σ y 2 Use s x 2 to denote the larger variance. where s x 2 is the larger of the two sample variances

40 40 Slide Buyer’s Digest has conducted the same test, as was described earlier, on another 10 thermostats, this time manufactured by TempKing. The temperature readings of the ten thermostats are listed on the next slide. Hypothesis Testing About the Variances of Two Populations n Example: Buyer’s Digest (C) We will conduct a hypothesis test with  =.10 to see We will conduct a hypothesis test with  =.10 to see if the variances are equal for ThermoRite’s thermostats and TempKing’s thermostats.

41 41 Slide Hypothesis Testing About the Variances of Two Populations n Example: Buyer’s Digest (C) ThermoRite Sample TempKing Sample Temperature Thermostat Temperature Thermostat

42 42 Slide n Hypotheses Hypothesis Testing About the Variances of Two Populations Reject H 0 if F > 3.18 The F distribution table (on next slide) shows that with with  =.10, 9 d.f. (numerator), and 9 d.f. (denominator), F.05 = (Their variances are not equal) (TempKing and ThermoRite thermostats have the same temperature variance) n Rejection Rule

43 43 Slide Selected Values from the F Distribution Table Hypothesis Testing About the Variances of Two Populations

44 44 Slide n Test Statistic Hypothesis Testing About the Variances of Two Populations We cannot reject H 0. F = 2.53 < F.05 = There is insufficient evidence to conclude that the population variances differ for the two thermostat brands. Conclusion Conclusion = 1.768/.700 = 2.53 TempKing’s sample variance is ThermoRite’s sample variance is.700

45 45 Slide n Determining and Using the p -Value Hypothesis Testing About the Variances of Two Populations Because  =.10, we have p -value >  and therefore Because  =.10, we have p -value >  and therefore we cannot reject the null hypothesis. we cannot reject the null hypothesis. But this is a two-tailed test; after doubling the upper- But this is a two-tailed test; after doubling the upper- tail area, the p -value is between.20 and.10. (A precise tail area, the p -value is between.20 and.10. (A precise p -value can be found using Minitab or Excel.) p -value can be found using Minitab or Excel.) Because F = 2.53 is between 2.44 and 3.18, the area Because F = 2.53 is between 2.44 and 3.18, the area in the upper tail of the distribution is between.10 in the upper tail of the distribution is between.10 and.05. and.05. Area in Upper Tail F Value (df 1 = 9, df 2 = 9)

46 46 Slide

47 47 Slide

48 48 Slide End of Chapter 11


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