Exam 4 Flashcards
define Analysis of Variance
Analysis of variance is a technique that allows us to compare two or more populations of interval data.
Analysis of variance is: three things….
Analysis of variance is:
an extremely powerful and widely used procedure
a procedure which determines whether differences exist between population means
a procedure which works by analyzing sample variance
example of a one-way analysis of variance
Examples: Accident rates for 1st, 2nd, and 3rd shift Expected mileage for five brands of tires
assumptions of a one-way analysis of variance
Populations are normally distributed
Populations have equal variances
Samples are randomly and independently drawn
x is the ________ variable, and its values are __________.
x is the response variable, and its values are responses.
xij refers to the _________, ___________
xij refers to the observation, treatment
Each population is a _______ ________.
Each population is a factor level.
Population classification criterion is called a _______
Population classification criterion is called a factor
hypothesis of One-way ANOVA H0
H0: All population means are equal
i.e., no factor effect (no variation in means among groups)
hypothesis of One-way ANOVA H1
H1: At least one population mean is different
i.e., there is a factor effect
Does not mean that all population means are different (some pairs may be the same)
Since µ1 = µ2 = µ3 = µ4 is of interest to us, a statistic that measures the proximity of the sample means to each other
between-treatments variation. It is denoted SST, short for “sum of squares for treatments”
SSE (_____ ___ _______ ___ ______) measures the ______-_______ _________.
SSE (Sum of Squares for Error) measures the within-treatments variation. measure of the amount of variation we can expect from the random variable we’ve observed.
MST stands for
mean square for treatments
MSE stands for
Mean square for errors
ANOVA: in the F table….
numerator degrees of freedom determine the column
denominator degrees of freedom determine the row
ANOVA: Degrees of freedom (for F crit)
df1=k-1 (MST), df2=n-k (MSE)
ANOVA: Treatments df
treatments df: k-1 (k=treatments)
ANOVA: Error df
Error df: n-k (n= #of obs, k=treatments)
ANOVA: Total df
n-1
What is a multinomial experiment?
Unlike a binomial experiment which only has two possible outcomes (e.g. heads or tails), a multinomial experiment:
•Consists of a fixed number, n, of trials. • Each trial can have one of k outcomes, called cells. • Each probability pi remains constant. • Our usual notion of probabilities holds, namely: p1 + p2 + … + pk = 1 • Each trial is independent of the other trials.
what is a Chi-squared goodness of fit test used for
How “close” are the observed values to those which would be expected under the fitted model
how are degrees of freedom calculated for a X^2 test of a contingency table?
r-1 x c-1
what is ei=npi
expected frequency
which test statistic measures the similarity of the expected and observed frequencies?
chi-squared goodness of fit test
when performing a test of a contingency table, what should you do if the expected frequency isn’t given?
row total x column total / n
CSGOF: observed frequency, fi, comes from where
actual number from the problem
CSGOF: expected frequency is calculated as:
n x pi
CSGOF: Delta (difference) is calculated as:
( fi - ei ) Observed minus expected
CSGOF: Summation Component is calculated as:
( fi - ei )^2 / ei (observed minus expected) / expected
The Chi-squared test of a contingency table is used to:
The Chi-squared test of a contingency table is used to:
• determine whether there is enough evidence to infer that two nominal variables are related, and
• to infer that differences exist among two or more populations of nominal variables.
In order to use these techniques, we need to classify the data according to two different criteria.
Chi-Squared test of a contingency table expected value calculation
eij = (row total x column total) / n n=number of obs
chi-squared distribution with ________________ degrees of freedom
chi-squared distribution with (r – 1)(c – 1) degrees of freedom
degrees of freedom chi-squared test of a contingency table
( r - 1 ) ( c - 1 ) r=rows c=columns
chi-squared test of a contingency table Rule of 5
In a contingency table where one or more cells have expected values of less than 5, we need to combine rows or columns to satisfy the rule of five.
chi-squared test of a contingency table example
120 females, 12 were left handed, 108 were right handed
180 males, 24 were left handed, 156 were right handed
chi-squared test of a contingency table expected value
Expected value = (row total x column total) / grand total
