Exam 4

Question 1

define Analysis of Variance

Answer 1

Analysis of variance is a technique that allows us to compare two or more populations of interval data.

Question 2

Analysis of variance is: three things….

Answer 2

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

Question 3

example of a one-way analysis of variance

Answer 3

Examples: Accident rates for 1st, 2nd, and 3rd shift Expected mileage for five brands of tires

Question 4

assumptions of a one-way analysis of variance

Answer 4

Populations are normally distributed
Populations have equal variances
Samples are randomly and independently drawn

Question 5

x is the ________ variable, and its values are __________.

Answer 5

x is the response variable, and its values are responses.

Question 6

xij refers to the _________, ___________

Answer 6

xij refers to the observation, treatment

Question 7

Each population is a _______ ________.

Answer 7

Each population is a factor level.

Question 8

Population classification criterion is called a _______

Answer 8

Population classification criterion is called a factor

Question 9

hypothesis of One-way ANOVA H0

Answer 9

H0: All population means are equal

i.e., no factor effect (no variation in means among groups)

Question 10

hypothesis of One-way ANOVA H1

Answer 10

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)

Question 11

Since µ1 = µ2 = µ3 = µ4 is of interest to us, a statistic that measures the proximity of the sample means to each other

Answer 11

between-treatments variation. It is denoted SST, short for “sum of squares for treatments”

Question 12

SSE (_____ ___ _______ ___ ______) measures the ______-_______ _________.

Answer 12

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.

Question 13

MST stands for

Answer 13

mean square for treatments

Question 14

MSE stands for

Answer 14

Mean square for errors

Question 15

ANOVA: in the F table….

Answer 15

numerator degrees of freedom determine the column

denominator degrees of freedom determine the row

Question 16

ANOVA: Degrees of freedom (for F crit)

Answer 16

df1=k-1 (MST), df2=n-k (MSE)

Question 17

ANOVA: Treatments df

Answer 17

treatments df: k-1 (k=treatments)

Question 18

ANOVA: Error df

Answer 18

Error df: n-k (n= #of obs, k=treatments)

Question 19

ANOVA: Total df

Answer 19

n-1

Question 20

What is a multinomial experiment?

Answer 20

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.

Question 21

what is a Chi-squared goodness of fit test used for

Answer 21

How “close” are the observed values to those which would be expected under the fitted model

Question 22

how are degrees of freedom calculated for a X^2 test of a contingency table?

Answer 22

r-1 x c-1

Question 23

what is ei=npi

Answer 23

expected frequency

Question 24

which test statistic measures the similarity of the expected and observed frequencies?

Answer 24

chi-squared goodness of fit test

Question 25

when performing a test of a contingency table, what should you do if the expected frequency isn’t given?

Answer 25

row total x column total / n

Question 26

CSGOF: observed frequency, fi, comes from where

Answer 26

actual number from the problem

Question 27

CSGOF: expected frequency is calculated as:

Answer 27

n x pi

Question 28

CSGOF: Delta (difference) is calculated as:

Answer 28

( fi - ei ) Observed minus expected

Question 29

CSGOF: Summation Component is calculated as:

Answer 29

( fi - ei )^2 / ei (observed minus expected) / expected

Question 30

The Chi-squared test of a contingency table is used to:

Answer 30

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.

Question 31

Chi-Squared test of a contingency table expected value calculation

Answer 31

eij = (row total x column total) / n n=number of obs

Question 32

chi-squared distribution with ________________ degrees of freedom

Answer 32

chi-squared distribution with (r – 1)(c – 1) degrees of freedom

Question 33

degrees of freedom chi-squared test of a contingency table

Answer 33

( r - 1 ) ( c - 1 ) r=rows c=columns

Question 34

chi-squared test of a contingency table Rule of 5

Answer 34

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.

Question 35

chi-squared test of a contingency table example

Answer 35

120 females, 12 were left handed, 108 were right handed

180 males, 24 were left handed, 156 were right handed

Question 36

chi-squared test of a contingency table expected value

Answer 36

Expected value = (row total x column total) / grand total