# Lecture 11 One-way analysis of variance (Chapter 15.2)

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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 –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.

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

Review of one-way ANOVA Objective: Compare the means of K populations of interval data based on independent random samples from each. H 0 : H 1 : At least two means differ Notation: x ij – ith observation of jth sample; - mean of the jth sample; n j – number of observations in jth sample; - grand mean of all observations

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. Example 15.1

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)

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.

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. Variability between sample means

There are k treatments The size of sample j The mean of sample j Sum of squares for treatments (SST) Note: When the sample means are close to one 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 H 1.

Solution – continued Calculate 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 Sum of squares for treatments (SST)

Is SST = 57,512.23 large enough to reject H 0 in favor of H 1 ? Large compared to what? Sum of squares for treatments (SST)

20 25 30 1 7 Treatment 1Treatment 2 Treatment 3 10 12 19 9 Treatment 1Treatment 2Treatment 3 20 16 15 14 11 10 9 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. A small variability within the samples makes it easier to draw a conclusion about the population means.

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”. The rationale behind test statistic – II

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). Within samples variability

Solution – continued Calculate SSE Sum of squares for errors (SSE)  (n 1 - 1)s 1 2 + (n 2 -1)s 2 2 + (n 3 -1)s 3 2 = (20 -1)10,774.44 + (20 -1)7,238.61+ (20-1)8,670.24 = 506,983.50

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? Sum of squares for errors (SSE)

mean squares To perform the test we need to calculate the mean squares as follows: The mean sum of squares Calculation of MST - M ean S quare for T reatments Calculation of MSE M ean S quare for E rror

And finally the hypothesis test: H 0 :  1 =  2 = …=  k H 1 : At least two means differ Test statistic: R.R: F>F ,k-1,n-k The F test rejection region

The F test H o :  1 =  2 =  3 H 1 : At least two means differ Test statistic F= MST  MSE= 3.23 Since 3.23 > 3.15, there is sufficient evidence to reject H o in favor of H 1, and argue that at least one of the mean sales is different than the others.

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)

ANOVA Table – Example 15.1 Analysis of Variance SourceDFSum of Squares Mean Square F RatioProb > F City257512.2328756.13.23300.0468 Error57506983.508894.4 C. Total 59564495.73

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:

Model for ANOVA Cont. The expected response to the jth treatment is Thus, if all treatments have the same expected response (i.e., H 0 : 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

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.

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).