STAT 205 Lecture Notes - Lecture 26: Minitab, Analysis Of Variance

66 views2 pages

Document Summary

Anova (analysis of variance) test analyzes all sample means at once. Completely randomized design: only one factor/independent variable with multiple treatment levels to which people are randomly assigned. P-value needs to be less than 0. 05 to reject h0. Cannot be performed until after normality and levene"s test are complete. I = 1,2,,k: all need to be simple random samples, all need normal distributions (test with qq plot, all need equal variances (levene"s test) Understand: assumptions, simple manual calculations (test score, critical value etc. , interpretation of minitab output. = family comparison error rate k = number of treatment levels. Mse = mean square error (obtained from minitab) n = sample size q ,k,n-k = critical value of the studentized range distribution nj = common sample size (n1 = n2 = = nj = = nk) | |> hsd, (cid:4666) (cid:4667)means there is a significant difference so you can reject h0 for that pair.

Get access

Grade+20% off
$8 USD/m$10 USD/m
Billed $96 USD annually
Grade+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
40 Verified Answers
Class+
$8 USD/m
Billed $96 USD annually
Class+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
30 Verified Answers

Related textbook solutions

Related Documents

Related Questions