D2 a b r 1 3 2 5 1 24 At 005 QA 353 CRA 353 18798 2 5 484 Determining Sample

D2 a b r 1 3 2 5 1 24 at 005 qa 353 cra 353 18798 2 5

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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 s-means > 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
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- 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 s-means > CRA Chapter 12: Chi-Square 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 Chi-Square Distribution – a distribution that allows us to: Compare 2+ population proportions - Det. if diff in proportions of populations Perform a goodness-of-fit 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?
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Step 1: Identify the null and alternative hypotheses o Similar to One-Way 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 (1-5) 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 chi-square distribution Step 3: Calculate the chi-square test statistic, x² o Formula – Chi-Square Test Statistic = ∑ (f0 – Fe)² / (Fe) o Logic of the Test Statistic If observed frequencies are close to expected frequencies - → smaller chi-square 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

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