Difference of Means and ANOVA Problems

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1 Difference of Means and Problems Dr. Tom Ilvento FREC 408 Accounting Firm Study An accounting firm specializes in auditing the financial records of large firm It is interested in evaluating its fee structure,particularly in relation to charges by the size of the firm It takes a random sample of 0 companies from three size classes Sales over $50 million Sales of $00 to $50 million Sales of Less than $00 million Stem and Leaf Plot STEM Leaf Descriptive Statistics with 95% C.I. $50+ $00-$50 < $00 Total Mean Standard Error Median Mode 50 #N/A #N/A 50 Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count Confidence Level(95.0%) Things to note The mean cost for the largest firms is much larger than the other two firm classes While there is a difference in the means for the lower two classes, when we look at the BOE for the 95% C.I. there is overlap The variances of the lower two firms is very similar but both are considerably lower than that for the largest firms Difference of Means Test Let s test to see if there is a significant difference between firms of sales between $00 to $50 million Less than $00 million Is this reasonable? Ratio of variances is 35.7/ =.00

2 Decision Tree for Two Means Target Assumptions Test Statistic Independent random samples z, using Large sample size (n, n >30) sample variance H 0 : µ -µ =D Independent random samples Small sample size t, using Populations appr. normal pooled variance Equal variances S p POOLED ESTIMATE OF THE VARIANCE The Our formula will be a weighted average of s and s ( n ) s + ( n ) s s p = ( n + n ) s p (0 ) (0 ) = (0 + 0 ) s p = s 8 = p = NOTE Since the sample sizes were equal, we could have simply taken the average of the two variances ( )/ = Use the Pooled Estimate of the Variance to calculate the standard error ˆ σ ( x x ) = s p + n n ˆ ( x ) = = x σ Accounting Firm Problem EXCEL Output Null hypothesis Alternative Assumptions Test Statistic Rejection Region Calculation Conclusion H 0 : (: -: ) = 0 H a : (: -: ) 0 two-tailed test Small independent samples, approx normal, variances are equal t* = ( )/5.493 t.05/, 8 d.f. =.0 t* =.9 t* < -t.05/, 8 d.f..9 <.0 We cannot reject H 0 : (: -: ) = 0 t-test: Two-Sample Assuming Equal Variances $00-$50 < $00 Mean Variance Observations 0 0 Pooled Variance Hypothesized Mean Difference 0 df 8 t Stat 0.9 P(T<=t) one-tail 0.84 t Critical one-tail.734 P(T<=t) two-tail t Critical two-tail.0

3 Let s shift to would allow us to test the difference of means across all classes of firm size We need to assume equal variances is this reasonable? Review Degrees of freedom k = 3 n = 30 SS(Total) df = n- = 30 - = 9 SST df = k- = 3 = SSE df = n-k = 30 3 = 7 Excel Output $ $00-$ < $ Between Groups Within Groups Hypothesis Test Null hypothesis H 0 : : =: = : 3 Alternative H a : At least two means differ Assumptions F* = 8.44 Test Statistic F Rejection Region.05,, 7 d.f. = 3.35 F* > F Conclusion 8.44 > 3.35 Equal variances, normal distribution Reject H 0 : : =: = : 3 Total Brake Test A firm makes disc brakes for the automobile industry The R&D department tested four different brake systems In the test, they used 40 identical mid-sized cars, 0 each for the four brake systems The cars were driven on a test track and stopped electronically They measured the distance in feet to bring the car to a stop Are there differences in stopping distance across the different brakes? The data Brake A Brake B Brake C Brake D Total Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Confidence Level(95.0% )

4 The Output Brake A Brake B Brake C Brake D Between Groups Within Groups Hypothesis Test Null hypothesis H 0 : : =: = : 3 = : 4 Alternative H a : At least two means differ Assumptions F* = Test Statistic F Rejection Region.05, 3, 36 d.f. =.866 F* > F Conclusion >.866 Equal variances, normal distribution Reject H 0 : : =: = : 3 = : 4 Total What is R-square? Cock Roach Data / =.4 On page 660 there is a Statistics in the Real World problem on roaches. This is a study to see if the navigation of cockroaches is random or not. The researcher hypothesized that cockroaches do follow trails, much like bees, ants, and termites, using chemical trails. She used a chemical trail with pheromones as the main treatment (EXTRACT - from cockroach feces), but also included a control trail using methanol (TRAIL). Today, we will focus on a Two-Way of the Trail Type (Extract versus Control group) and Roach Type (Gravid, Male, Female, Nymph) Roach Two-Way The researcher released German cockroaches of different age, sex, and reproductive stage to see if these factors influenced trail following ability Factor = TYPE: The levels = Female; Gravid; Male; and Nymph Factor = TRAIL The levels = Extract and Control She measured the movement pattern of the cockroaches and calculated an average perpendicular distance Response Variable = MOVE 4

5 The is how the procedure arranges the Data Extract Control Male Female Gravid Nymph Total Mean Standard Error Median Mode.40 #N/A.40 #N/A #N/A #N/A.40 Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count Male Female Gravid Nymphs Total Extract Sum Average Variance Control Sum Average Variance Total Count Sum Average Variance I can graph it to try to make sense of it R = 464/ = EXTRACT CONTROL Male Female Nymph Gravid Female Gravid Male Nymph Between Groups Within Groups Total Conduct a Test to see if there is a mean difference in MOVE by the levels for Type (Male, Female, Nymph, and Gravid). Use an F-test with α=.0. Null Hypothesis H0: : = : = : 3 = : 4 Alternative Hypothesis Assumptions of Test Test Statistic F* =.605 Ha: At least one mean is different Large sample, equal variances, normal distribution Rejection Region F.0, 3 and 76 d.f. = 4.05 Comparison of Test Statistics with Rejection Region F* > F.0, 3 and 76 d.f..605 > 4.05 We can reject H0: : = : = : 3 = : 4 p-value <.000 What if I focused on just the TRAIL Factor? Extract Control Between Groups Within Groups Total R = 30,984.9/3, =.78 or 7% of the variability in MOVE 5

6 What if I focused on just the TRAIL Factor? Male Female Gravid Nymph Between Groups Within Groups Total R = 0,37.89/3, =.0905 or 9% of the variability in MOVE Two Factor Male Female Gravid Nymphs Total Extract Sum Average Variance Control Sum Average Variance Total Count Sum Average Variance Two-Factor How to Solve for R? Sample Columns Interaction Within Total Sample refers to TRAIL (Extract versus Control) Columns refers to TYPE (Male, Female, Gravid, Nymph) Interaction refers to the interaction between TRAIL and TYPE Use this formula: SSE/SST R = 70,89./3,993.0 = -.69 = % of the variability in MOVE is accounted for by considering the trail type and the Roach Type 6

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