Econ 3790: Business and Economic Statistics

1. Econ 3790: Business and Economic Statistics Instructor: Yogesh Uppal Email: yuppal@ysu.edu

2. Chapter 11Inferences About Population Variances • Inference about a Population Variance • Chi-Square Distribution • Interval Estimation of 2 • Hypothesis Testing

3. 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, Chi-squared value with 5 degrees of freedom (df) at a =0.05 is 11.07. Chi-Square Distribution

4. Interval Estimation of 2 .05 95% of the possible 2 values 2 0 = 11.07

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

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

7. Interval Estimation of 2 • Example: Buyer’s Digest (A): 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 68oF. The temperature readings of the ten thermostats are shown on the next slide.

8. Interval Estimation of 2 • Example: Buyer’s Digest (A) We will use the 10 readings below to develop a 95% confidence interval estimate of the population variance. Thermostat1 2 3 4 5 6 7 8 9 10 Temperature 67.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.2

9. Our value Interval Estimation of 2 For n - 1 = 10 - 1 = 9 d.f. and a = .05 Selected Values from the Chi-Square Distribution Table

10. Interval 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

11. where is the hypothesized value for the population variance Hypothesis Testing about a Population Variance • Left-Tailed Test • Hypotheses • Test Statistic

12. Reject H0 if where is based on a chi-square distribution with n - 1 d.f. Hypothesis TestingAbout a Population Variance • Left-Tailed Test (continued) • Rejection Rule Critical value approach: Reject H0 if p-value <a p-Value approach:

13. where is the hypothesized value for the population variance Hypothesis TestingAbout a Population Variance • Right-Tailed Test • Hypotheses • Test Statistic

14. Reject H0 if where is based on a chi-square distribution with n - 1 d.f. Hypothesis TestingAbout a Population Variance • Right-Tailed Test (continued) • Rejection Rule Critical value approach: Reject H0 if p-value <a p-Value approach:

15. where is the hypothesized value for the population variance Hypothesis TestingAbout a Population Variance • Two-Tailed Test • Hypotheses • Test Statistic

16. Reject H0 if where are based on a chi-square distribution with n - 1 d.f. Hypothesis TestingAbout a Population Variance • Two-Tailed Test (continued) • Rejection Rule Critical value approach: p-Value approach: Reject H0 if p-value <a

17. Hypothesis TestingAbout a Population Variance Example: Buyer’s Digest (B): 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. We will conduct a hypothesis test (with a = .10) to determine whether the ThermoRite thermostat’s temperature variance is “acceptable”.

18. Hypothesis TestingAbout a Population Variance • Example: Buyer’s Digest (B) Using the 10 readings, we will conduct a hypothesis test (with a = .10) to determine whether the ThermoRite thermostat’s temperature variance is “acceptable”. Thermostat1 2 3 4 5 6 7 8 9 10 Temperature 67.4 67.8 68.2 69.3 69.5 67.0 68.1 68.6 67.9 67.2

19. Hypothesis TestingAbout a Population Variance • Hypotheses • Rejection Rule Reject H0 if c 2> 14.684

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

21. Hypothesis TestingAbout a Population Variance • Rejection Region Area in Upper Tail = .10 2 14.684 0 Reject H0

22. Hypothesis TestingAbout a Population Variance The sample variance s 2 = 0.7 • Test Statistic • Conclusion Because c2 = 12.6 is less than 14.684, we cannot reject H0. The sample variance s2 = .7 is insufficient evidence to conclude that the temperature variance for ThermoRite thermostats is unacceptable.

23. Chapter 13, Part A: Analysis of Variance and Experimental Design • Introduction to Analysis of Variance • Analysis of Variance: Testing for the Equality of k Population Means

24. Introduction to Analysis of Variance Analysis of Variance (ANOVA) can be used to test for the equality of three or more population means. We want to use the sample results to test the following hypotheses: H0: 1=2=3=. . . = k Ha: Not all population means are equal

25. Introduction to Analysis of Variance H0: 1=2=3=. . . = k Ha: Not all population means are equal If H0 is rejected, we cannot conclude that all population means are different. Rejecting H0 means that at least two population means have different values.

26. Assumptions for Analysis of Variance For each population, the response variable is normally distributed. The variance of the response variable, denoted  2, is the same for all of the populations. The observations must be independent.

27. Test for the Equality of k Population Means • Hypotheses H0: 1=2=3=. . . = k Ha: Not all population means are equal • Test Statistic F = MSTR/MSE

28. Between-Treatments Estimateof Population Variance • A between-treatment estimate of  2 is called the mean square treatment and is denoted MSTR. Numerator is the sum of squares due to treatments and is denoted SSTR Denominator represents the degrees of freedom

29. Within-Samples Estimateof Population Variance • The estimate of  2 based on the variation of the sample observations within each sample is called the mean square error and is denoted by MSE. Numerator is the sum of squares due to error and is denoted SSE Denominator represents the degrees of freedom associated with SSE

30. Test for the Equality of k Population Means • k: # of subpopulations you are comparing. • nT: Total number of observations. • Rejection Rule Reject H0 if F>Fa where the value of F is based on an F distribution with k - 1 numerator d.f. and nT - k denominator d.f.

31. Hypothesis Testing About theVariances of Two Populations Selected Values from the F Distribution Table

32. Comparing the Variance Estimates: The F Test • If the null hypothesis is true and the ANOVA • assumptions are valid, the sampling distribution of • MSTR/MSE is an F distribution with MSTR d.f. • equal to k - 1 and MSE d.f. equal to nT - k. • If the means of the k populations are not equal, the • value of MSTR/MSE will be inflated because MSTR • overestimates  2. • Hence, we will reject H0 if the resulting value of • MSTR/MSE appears to be too large to have been • selected at random from the appropriate F • distribution.

33. ANOVA Table Source of Variation Sum of Squares Mean Squares Degrees of Freedom F SSTR SSE SST k – 1 nT – k nT - 1 Treatment Error Total MSTR MSE MSTR/MSE SST’s degrees of freedom (d.f.) are partitioned into SSTR’s d.f. and SSE’s d.f. SST is partitioned into SSTR and SSE.

34. ANOVA Table SST divided by its degrees of freedom nT – 1 is the overall sample variance that would be obtained if we treated the entire set of observations as one data set. With the entire data set as one sample, the formula for computing the total sum of squares, SST, is:

35. ANOVA Table ANOVA can be viewed as the process of partitioning the total sum of squares and the degrees of freedom into their corresponding sources: treatments and error. Dividing the sum of squares by the appropriate degrees of freedom provides the variance estimates and the F value used to test the hypothesis of equal population means.

36. Test for the Equality of k Population Means • Example: Reed Manufacturing Janet Reed would like to know if there is any significant difference in the mean number of hours worked per week for the department managers at her three manufacturing plants (in Buffalo, Pittsburgh, and Detroit).

37. Test for the Equality of k Population Means • Example: Reed Manufacturing A simple random sample of five managers from each of the three plants was taken and the number of hours worked by each manager for the previous week is shown on the next slide. Conduct an F test using a = .05.

38. Test for the Equality of k Population Means Plant 3 Detroit Plant 2 Pittsburgh Plant 1 Buffalo Observation 1 2 3 4 5 48 54 57 54 62 51 63 61 54 56 73 63 66 64 74 Sample Mean 55 68 57 Sample Variance 26.0 26.5 24.5

39. Test for the Equality of k Population Means • p -Value and Critical Value Approaches 1. Develop the hypotheses. H0:  1= 2= 3 Ha: Not all the means are equal where:  1 = mean number of hours worked per week by the managers at Plant 1  2 = mean number of hours worked per week by the managers at Plant 2   3 = mean number of hours worked per week by the managers at Plant 3

40. Test for the Equality of k Population Means • Compute the test statistic using ANOVA Table Source of Variation Sum of Squares Mean Squares Degrees of Freedom F 490 308 798 2 12 14 Treatment Error Total 245 25.67 9.5

41. Test for the Equality of k Population Means • p –Value Approach 4. Compute the critical value. With 2 numerator d.f. and 12 denominator d.f., Fa = 3.89. 5. Determine whether to reject H0. The F > Fa,so we reject H0. We have sufficient evidence to conclude that the mean number of hours worked per week by department managers is not the same at all 3 plant.