Econ 3790: Business and Economic Statistics Instructor: Yogesh Uppal

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

Chi-Square Distribution 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. For example, Chi-squared value with 5 degrees of freedom (df) at  =0.05 is

95% of the possible  2 values 95% of the possible  2 values 22 2 Interval Estimation of  2 = 11.07

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 1 -  is the confidence coefficient.

Interval Estimation of  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.

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 68 o F. The temperature readings of the ten thermostats are shown on the next 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. Example: Buyer’s Digest (A) Temperature Thermostat

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

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

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

n Left-Tailed Test (continued) Hypothesis Testing About a Population Variance Reject H 0 if p -value <  p -Value approach: Critical value approach: Rejection Rule Rejection Rule Reject H 0 if where is based on a chi-square distribution with n - 1 d.f.

Right-Tailed Test Hypothesis Testing About a Population Variance where is the hypothesized value for the population variance Test Statistic Test Statistic Hypotheses Hypotheses

n Right-Tailed Test (continued) Hypothesis Testing About a Population Variance Reject H 0 if Reject H 0 if p -value <  where is based on a chi-square distribution with n - 1 d.f. p -Value approach: Critical value approach: Rejection Rule Rejection Rule

Two-Tailed Test Hypothesis Testing About a Population Variance where is the hypothesized value for the population variance Test Statistic Test Statistic Hypotheses Hypotheses

n Two-Tailed Test (continued) Hypothesis Testing About a Population Variance Reject H 0 if p -value <  p -Value approach: Critical value approach: Rejection Rule Rejection Rule Reject H 0 if where are based on a chi-square distribution with n - 1 d.f.

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. Hypothesis Testing About a Population Variance We will conduct a hypothesis test (with  =.10) to determine whether the ThermoRite thermostat’s temperature variance is “acceptable”.

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

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

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

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

Test Statistic Hypothesis Testing About a Population Variance 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. Conclusion The sample variance s 2 = 0.7

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

Introduction to Analysis of Variance Analysis of Variance (ANOVA) can be used to test Analysis of Variance (ANOVA) can be used to test for the equality of three or more population means. for the equality of three or more population means. Analysis of Variance (ANOVA) can be used to test Analysis of Variance (ANOVA) can be used to test for the equality of three or more population means. for the equality of three or more population means. We want to use the sample results to test the We want to use the sample results to test the following hypotheses: following hypotheses: We want to use the sample results to test the We want to use the sample results to test the following hypotheses: following hypotheses: H 0 :  1  =  2  =  3  = ... =  k H a : Not all population means are equal

Introduction to Analysis of Variance H 0 :  1  =  2  =  3  = ... =  k H a : Not all population means are equal If H 0 is rejected, we cannot conclude that all If H 0 is rejected, we cannot conclude that all population means are different. population means are different. If H 0 is rejected, we cannot conclude that all If H 0 is rejected, we cannot conclude that all population means are different. population means are different. Rejecting H 0 means that at least two population Rejecting H 0 means that at least two population means have different values. means have different values. Rejecting H 0 means that at least two population Rejecting H 0 means that at least two population means have different values. means have different values.

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

Test for the Equality of k Population Means F = MSTR/MSE H 0 :  1  =  2  =  3  = ... =  k  H a : Not all population means are equal n Hypotheses n Test Statistic

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

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. Within-Samples Estimate of Population Variance Denominator represents the degrees of freedom the degrees of freedom associated with SSE associated with SSE Numerator is the sum of squares sum of squares due to error and is denoted SSE

Test for the Equality of k Population Means n Rejection Rule where the value of F  is based on an F distribution with k - 1 numerator d.f. and n T - k denominator d.f. Reject H 0 if F > F  n k: # of subpopulations you are comparing. n n T : Total number of observations.

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

Comparing the Variance Estimates: The F Test If the null hypothesis is true and the ANOVA If the null hypothesis is true and the ANOVA assumptions are valid, the sampling distribution of assumptions are valid, the sampling distribution of MSTR/MSE is an F distribution with MSTR d.f. MSTR/MSE is an F distribution with MSTR d.f. equal to k - 1 and MSE d.f. equal to n T - k. equal to k - 1 and MSE d.f. equal to n T - k. If the means of the k populations are not equal, the If the means of the k populations are not equal, the value of MSTR/MSE will be inflated because MSTR value of MSTR/MSE will be inflated because MSTR overestimates  2. overestimates  2. Hence, we will reject H 0 if the resulting value of Hence, we will reject H 0 if the resulting value of MSTR/MSE appears to be too large to have been MSTR/MSE appears to be too large to have been selected at random from the appropriate F selected at random from the appropriate F distribution. distribution.

ANOVA Table SST is partitioned into SSTR and SSE. SST’s degrees of freedom (d.f.) are partitioned into SSTR’s d.f. and SSE’s d.f. Treatment Error Total SSTR SSE SST k– 1 n T n T – k nT nT nT nT - 1 MSTR MSE Source of Variation Sum of Squares Degrees of Freedom MeanSquares MSTR/MSE F

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

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

Example: Reed Manufacturing Test for the Equality of k Population Means Janet Reed would like to know if 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).

Example: Reed Manufacturing Test for the Equality of k Population Means A simple random sample of five 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  =.05. Conduct an F test using  =.05.

Plant 1 Buffalo Plant 2 Pittsburgh Plant 3 Detroit Observation Sample Mean Sample Variance Test for the Equality of k Population Means

H 0 :  1  =  2  =  3  H a : Not all the means are equal where:  1 = mean number of hours worked per week by the managers at Plant 1 week by the managers at Plant 1  2 = mean number of hours worked per  2 = mean number of hours worked per week by the managers at Plant 2 week by the managers at Plant 2  3 = mean number of hours worked per week by the managers at Plant 3 week by the managers at Plant 3 1. Develop the hypotheses. p -Value and Critical Value Approaches p -Value and Critical Value Approaches

Treatment Error Total Source of Variation Sum of Squares Degrees of Freedom MeanSquares 9.5 F Test for the Equality of k Population Means Compute the test statistic using ANOVA Table Compute the test statistic using ANOVA Table

Test for the Equality of k Population Means 5. Determine whether to reject H 0. 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. The F > F , so we reject H 0. With 2 numerator d.f. and 12 denominator d.f., F  = p –Value Approach p –Value Approach 4. Compute the critical value.