D2 = a • b (r – 1) = 3 • 2 (5 – 1) = 24

At ∞ = 0.05, QA = 3.53
▪
CRA = 3.53 √ 1879.8 / (2 • 5) = 48.4
▪
Determining Sample Means that Are Different

Means are different if abs difference of smeans > CRA
o
Formula – Critical Range for Factor B
▪
CRB = QB √ MSB / ar

a = number of levels in factor A

r = number of replications in each cell

QB – critical value for factor B
▪
Critical value for Factor B (QB)

D1 = a = 2 (cities)

D2 = a • b (r – 1) = 3 • 2 (5 – 1) = 24

At ∞ = 0.05, QB = 2.92
▪
CRB = 2.92 √ 1879.8 / (3 • 5) = 32.67
▪
Determining Sample Means that Are Different

Means are different if abs difference of smeans > CRA
Chapter 12: ChiSquare Tests
●
Quadrophobia: Fear of the Number 4
o
Financial Statements Analyzed
▪
For thousands of US firms
▪
Found the EPS rarely ended in a 0.04
▪
Statistics underrepresented the 4 digit in the sample
▪
Some companies cooking the books (financial statements)

b/c 0.04 rounds down, while 0.05 rounds up
o
ChiSquare Distribution
– a distribution that allows us to:
▪
Compare 2+ population proportions

Det. if diff in proportions of populations
▪
Perform a goodnessoffit test to det. if data follow a particular
probability distribution
▪
Test 2 categorical variables for independence

Det. if variable 1 has impact on variable 2
➢
12.1 COMPARING TWO OR MORE POPULATION PROPORTIONS
o
Example: Monday is best day on Wall Street
▪
Calculated number of “up” and “downs” in the market for each
weekday
▪
Proportion of “ups” in each day are calculated, and Monday has
the highest proportion
o
Observed Frequencies
(f0) – the sample data collected from a population
of interest
▪
Do we have enough evidence from this sample to support claim
that Mondays have more “ups” in the week?
▪
Is this a real pattern that investors can rely on to make decisions?
●
Step 1: Identify the null and alternative hypotheses
o
Similar to OneWay ANOVA
▪
H0:π1 = π2 = π3 = π4 = π5
▪
H1: Not all πs are equal
o
π = proportion of “up” days for the allocated day of the week (15)
o
Set significance level (∞) = 0.05
●
Step 2: Calculate the expected frequencies
o
Expected frequencies
(Fe) – the frequencies that are most likely to occur if
the null hypothesis is true
▪
Assume proportions are equal – successes are random
▪
= (P) • (total data points in segment – days)
o
Calculate overall proportion of successes
▪
Total “up days” ÷ total days
▪
P = 78 / 125 = 0.624
▪
→ 62.4% of the days during the period were “up”
days
o
Calculating Fe
▪
Calculating the Expected Frequencies of “Up” Days

= P (0.624) • (total days for each week day) →
Fe
▪
Calculating the Expected Frequencies of “Down” Days

= (total days for each week day) – (Up day Fe)
▪
Lowest expected frequency must be = or > 5

b/c must follow chisquare distribution
●
Step 3: Calculate the chisquare test statistic, x²
o
Formula – ChiSquare Test Statistic
▪
= ∑ (f0 – Fe)² / (Fe)
o
Logic of the Test Statistic
▪
If observed frequencies are close to expected frequencies

→ smaller chisquare test statistic

Increased likelihood that there is support for the null
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 Fall '12
 Donnelly
 Normal Distribution, Null hypothesis, Hypothesis testing, Statistical hypothesis testing