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Lecture 11

Lecture 11. One-way analysis of variance (Chapter 15.2). Review: Relat. between One-Sided Hypothesis Tests and CIs. Suppose we are given a CI for For the one-sided hypothesis test versus at significance level , we can conclude

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Lecture 11

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  1. Lecture 11 • One-way analysis of variance (Chapter 15.2)

  2. Review: Relat. between One-Sided Hypothesis Tests and CIs • Suppose we are given a CI for • For the one-sided hypothesis test versus at significance level , we can conclude • We reject the null hypothesis if and does not belong to the confidence interval • We do not reject the null hypothesis if either or belongs to the confidence interval.

  3. Review: CIs for Monotonic Functions of Parameters • A function f(x) is monotonic if it moves in one direction as its argument increases. • Suppose that we have a CI for a parameter and that we want to find a CI for the parameter . • If f is monotonically increasing, the CI is . If f is monotonically decreasing, the CI is

  4. Review of one-way ANOVA • Objective: Compare the means of K populations of interval data based on independent random samples from each. • H0: • H1: At least two means differ • Notation: xij – ith observation of jth sample; - mean of the jth sample; nj – number of observations in jth sample; - grand mean of all observations

  5. Example 15.1 • The marketing manager for an apple juice manufacturer needs to decide how to market a new product. Three strategies are considered, which emphasize the convenience, quality and low price of product respectively. • An experiment was conducted as follows: • In three cities an advertisement campaign was launched . • In each city only one of the three characteristics (convenience, quality, and price) was emphasized. • The weekly sales were recorded for twenty weeks following the beginning of the campaigns.

  6. Rationale Behind Test Statistic • Two types of variability are employed when testing for the equality of population means • Variability of the sample means • Variability within samples • Test statistic is essentially (Variability of the sample means)/(Variability within samples)

  7. The rationale behind the test statistic – I • If the null hypothesis is true, we would expect all the sample means to be close to one another (and as a result, close to the grand mean). • If the alternative hypothesis is true, at least some of the sample means would differ. • Thus, we measure variability between sample means.

  8. Variability between sample means • The variability between the sample means is measured as the sum of squared distances between each mean and the grand mean. This sum is called the Sum of Squares for Treatments SST In our example treatments are represented by the different advertising strategies.

  9. Sum of squares for treatments (SST) There are k treatments The mean of sample j The size of sample j Note: When the sample means are close toone another, their distance from the grand mean is small, leading to a small SST. Thus, large SST indicates large variation between sample means, which supports H1.

  10. Sum of squares for treatments (SST) • Solution – continuedCalculate SST = 20(577.55 - 613.07)2 + + 20(653.00 - 613.07)2 + + 20(608.65 - 613.07)2 = = 57,512.23 The grand mean is calculated by

  11. Sum of squares for treatments (SST) Is SST = 57,512.23 large enough to reject H0 in favor of H1?Large compared to what?

  12. 30 25 20 19 12 10 9 7 1 Treatment 3 Treatment 1 Treatment 2 20 16 15 14 11 10 9 A small variability within the samples makes it easier to draw a conclusion about the population means. The sample means are the same as before, but the larger within-sample variability makes it harder to draw a conclusion about the population means. Treatment 1 Treatment 2 Treatment 3

  13. The rationale behind test statistic – II • Large variability within the samples weakens the “ability” of the sample means to represent their corresponding population means. • Therefore, even though sample means may markedly differ from one another, SST must be judged relative to the “within samples variability”.

  14. Within samples variability • The variability within samples is measured by adding all the squared distances between observations and their sample means. This sum is called the Sum of Squares for Error SSE In our example this is the sum of all squared differences between sales in city j and the sample mean of city j (over all the three cities).

  15. Sum of squares for errors (SSE) • Solution – continuedCalculate SSE = (n1 - 1)s12 + (n2 -1)s22 + (n3 -1)s32 = (20 -1)10,774.44 + (20 -1)7,238.61+ (20-1)8,670.24 = 506,983.50

  16. Sum of squares for errors (SSE) Is SST = 57,512.23 large enough relative to SSE = 506,983.50 to reject the null hypothesis that specifies that all the means are equal?

  17. Calculation of MST - Mean Square for Treatments Calculation of MSE Mean Square for Error The mean sum of squares To perform the test we need to calculate the mean squaresas follows:

  18. H0: m1 = m2 = …=mk H1: At least two means differ Test statistic: R.R: F>Fa,k-1,n-k The F test rejection region And finally the hypothesis test:

  19. The F test Ho: m1 = m2= m3 H1: At least two means differ Test statistic F= MST/ MSE= 3.23 Since 3.23 > 3.15, there is sufficient evidence to reject Ho in favor of H1,and argue that at least one of the mean sales is different than the others.

  20. Required Conditions for Test • Independent simple random samples from each population • The populations are normally distributed (look for extreme skewness and outliers, probably okay regardless if each ). • The variances of all the populations are equal (Rule of thumb: Check if largest sample standard deviation is less than twice the smallest standard deviation)

  21. Source DF Sum of Squares Mean Square F Ratio Prob > F City 2 57512.23 28756.1 3.2330 0.0468 Error 57 506983.50 8894.4 C. Total 59 564495.73 ANOVA Table – Example 15.1 Analysis of Variance

  22. Model for ANOVA • = ith observation of jth sample • is the overall mean level, is the differential effect of the jth treatment and is the random error in the ith observation under the jth treatment. The errors are assumed to be independent, normally distributed with mean zero and variance The are normalized:

  23. Model for ANOVA Cont. • The expected response to the jth treatment is • Thus, if all treatments have the same expected response (i.e., H0 : all populations have same mean), . In general, is the difference between the means of population j and j’. • Sums of squares decomposition: SS(Total)=SST+SSE

  24. Relationship between F-test and t-test for two samples • For comparing two samples, the F-statistic equals the square of the t-statistic with equal variances. • For two samples, the ANOVA F-test is equivalent to testing versus .

  25. Practice Problems • 15.16, 15.22, 15.26 • Next Time: Chapter 15.7 (we will return to Chapters 15.3-15.5 after completing Chapter 15.7).

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