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Hypothesis test flow chart

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1 Hypothesis test flow chart
START HERE number of correlations number of variables χ2 test for independence (19.9) Table I Test H0: r=0 (17.2) Table G 1 correlation (r) frequency data 2 Measurement scale Means 2 1 basic χ2 test (19.5) Table I Test H0: r1= r2 (17.4) Tables H and A Do you know s? number of means number of factors z -test (13.1) Table A Yes 1 More than 2 2 2-way ANOVA Ch 21 Table E No 2 1 independent samples? 1-way ANOVA Ch 20 Table E t -test (13.14) Table D Yes No Test H0: m1= m2 (15.6) Table D Test H0: D=0 (16.4) Table D

2 Chapter 18: Testing for difference among three or more groups: One way Analysis of Variance (ANOVA)
B C Suppose you wanted to compare the results of three tests (A, B and C) to see if there was any differences difficulty. To test this, you randomly sample these ten scores from each of the three populations of test scores. How would you test to see if there was any difference across the mean scores for these three tests? The first thing is obvious – calculate the mean for each of the three samples of 10 scores. But then what? You could run three two-sample t-tests on each of the pairs (A vs. B, A vs. C and B vs. C). 84 85 62 78 62 79 93 62 83 76 74 79 92 71 80 68 81 74 79 69 84 76 87 67 81 68 67 87 61 75 81 72 75 Means

3 There are two problems with this:
C You could run an two-sample t-test on each of the pairs (A vs. B, A vs. C and B vs. C). There are two problems with this: The three tests wouldn’t be truly independent of each other, since they contain common values, and We run into the problem of making multiple comparisons: If we use an a value of .05, the probability of obtaining at least one significant comparison by chance is 1-(1-.05)3, or about .14 84 85 62 78 62 79 93 62 83 76 74 79 92 71 80 68 81 74 79 69 84 76 87 67 81 68 67 87 61 75 81 72 75 Means So how do we test the null hypothesis: H0: mA = mB = mC ?

4 So how do we test the null hypothesis: H0: mA = mB = mC ?
In the 1920’s Sir Ronald Fisher developed a method called ‘Analysis of Variance’ or ANOVA to test hypotheses like this. The trick is to look at the amount of variability between the means. So far in this class, we’ve usually talked about variability in terms of standard deviations. ANOVA’s focus on variances instead, which (of course) is the square of the standard deviation. The intuition is the same. The variance of these three mean scores (81, 72 and 75) is 22.5 Intuitively, you can see that if the variance of the means scores is ‘large’, then we should reject H0. But what do we compare this number 22.5 to? A B C 84 85 62 78 62 79 93 62 83 76 74 79 92 71 80 68 81 74 79 69 84 76 87 67 81 68 67 87 61 75 81 72 75 Means

5 So how do we test the null hypothesis: H0: mA = mB = mC ?
The variance of these three mean scores (81, 72 and 75) is 22.5 How ‘large’ is 22.5? Suppose we knew the standard deviation of the population of scores (s). If the null hypothesis is true, then all scores across all three columns are drawn from a population with standard deviation s. It follows that the mean of n scores should be drawn from a population with standard deviation: This means multiplying the variance of the means by n gives us an estimate of the variance of the population. A B C 84 85 62 78 62 79 93 62 83 76 74 79 92 71 80 68 81 74 79 69 84 76 87 67 81 68 67 87 61 75 81 72 75 Means With a little algebra:

6 The variance of these three mean scores (81, 72 and 75) is 22.5
Multiplying the variance of the means by n gives us an estimate of the variance of the population. For our example, A B C 84 85 62 78 62 79 We typically don’t know what s2 is. But like we do for t-tests, we can use the variance within our samples to estimate it. The variance of the 10 numbers in each column (61, 94, and 55) should each provide an estimate of s2. We can combine these three estimates of s2 by taking their average, which is 70. 93 62 83 76 74 79 92 71 80 68 81 74 79 69 84 76 87 67 81 68 67 87 61 75 81 72 75 Means n x Variance of means 225 61 94 55 Variances Mean of variances 70

7 One is n times the variance of the means of each column.
If H0: mA = mB = mC is true, we now have two separate estimates of the variance of the population (s2). One is n times the variance of the means of each column. The other is the mean of the variances of each column. If H0 is true, then these two numbers should be, on average, the same, since they’re both estimates of the same thing (s2). For our example, these two numbers (225 and 70) seem quite different. Remember our intuition that a large variance of the means should be evidence against H0. Now we have something to compare it to seems large compared to 70. A B C 84 85 62 78 62 79 93 62 83 76 74 79 92 71 80 68 81 74 79 69 84 76 87 67 81 68 67 87 61 75 81 72 75 Means n x Variance of means 225 Variances Mean of variances 61 94 55 70

8 Fcrit for a = .05 and df’s of 2 and 27 is 3.35.
When conducting an ANOVA, we compute the ratio of these two estimates of s2. This ratio is called the ‘F statistic’. For our example, 225/70 = If H0 is true, then the value of F should be around 1. If H0 is not true, then F should be significantly greater than 1. We determine how large F should be for rejecting H0 by looking up Fcrit in Table E. F distributions depend on two separate degrees of freedom – one for the numerator and one for the denominator. df for the numerator is k-1, where k is the number of columns or ‘treatments’. For our example, df is 3-1 =2. df for the denominator is N-k, where N is the total number of scores. In our case, df is 30-3 = 27. A B C 84 85 62 78 62 79 93 62 83 76 74 79 92 71 80 68 81 74 79 69 84 76 87 67 81 68 67 87 61 75 81 72 75 Means n x Variance of means Fcrit for a = .05 and df’s of 2 and 27 is 3.35. Since Fobs = 3.23 is less than Fcrit, we fail to reject H0. We cannot conclude that the exam scores come from populations with different means. 225 Ratio (F) 3.23 Variances Mean of variances 61 94 55 70

9 Fcrit for a = .05 and df’s of 2 and 27 is 3.35.
Since Fobs = 3.23 is less than Fcrit, we fail to reject H0. We cannot conclude that the exam scores come from populations with different means. Instead of finding Fcrit in Table E, we could have calculated the p-value using our F-calculator. Reporting p-values is standard. Our p-value for F=3.23 with 2 and 27 degrees of freedom is p=.0552 Since our p-value is greater then .05, we fail to reject H0

10 Our resulting F statistic is 15.32.
Example: Consider the following n=12 samples drawn from k=5 groups. Use an ANOVA to test the hypothesis that the means of the populations that these 5 groups were drawn from are different. Answer: The 5 means and variances are calculated below, along with n x variance of means, and the mean of variances. Our resulting F statistic is Our two dfs are k-1=4 (numerator) and 60-5 = 55(denominator). Table E shows that Fcrit for 4 and 55 is 2.54. Fobs > Fcrit so we reject H0. A B C D E 87 99 78 91 81 69 105 70 51 76 104 75 64 79 71 108 86 68 74 83 92 62 72 66 93 85 111 65 107 84 67 61 97 90 82 78 96 74 100 97 76 46 75 149 Means Variances Mean of variances 93 n x Variance of means 1429 Ratio (F) 15.32

11 What does the probability distribution F(dfbet,dfw) look like?
1 2 3 4 5 F(2,5) F(2,10) F(2,50) F(2,100) F(10,5) F(10,10) F(10,50) F(10,100) F(50,5) F(50,10) F(50,50) F(50,100)

12 For a typical ANOVA, the number of samples in each group may be different, but the intuition is the same - compute F which is the ratio of the variance of the means over the mean of the variances. Formally, the variance is divided up the following way: Given a table of k groups, each containing ni scores (i= 1,2, …, k), we can represent the deviation of a given score, X from the mean of all scores, called the grand mean as: Deviation of X from the grand mean Deviation of X from the mean of the group Deviation of the mean of the group from the grand mean

13 The total sums of squares can be partitioned into two numbers:
Total sum of squares: SStotal Within-groups sum of squares: SSw Between-groups sum of squares: SSbet SSbet is a measure of the variability between groups. It is used as the numerator in our F-tests The variance between groups, called ‘mean squared error’, or MSbet, is calculated by dividing SSbet by its degrees of freedom dfbet = k-1 MSbet=SSbet/dfbet and is another estimate of s2 if H0 is true. This is essentially n times the variance of the means. If H0 is not true, then s2bet is an estimate of s2 plus any ‘treatment effect’ that would add to a difference between the means. .

14 The total sums of squares can be partitioned into two numbers:
Total sum of squares: SStotal Within-groups sum of squares: SSw Between-groups sum of squares: SSbet SSw is a measure of the variability within each group. It is used as the denominator in all F-tests. The variance within each group, MSw is calculated by dividing SSw by its degrees of freedom dfw = ntotal – k MSw=SSw/dfw This is an estimate of s2 This is essentially the mean of the variances within each group. (It is exactly the mean of variances if our sample sizes are all the same.)

15 The F ratio is calculated by dividing up the sums of squares and df into ‘between’ and ‘within’
SStotal = SSw + SSbet Variances are then calculated by dividing SS by df MSbet=SSbet/dfbet SStotal MSw=SSw/dfw SSw SSbet F is the ratio of variances between and within F= 𝑀𝑆 𝑏𝑒𝑡 𝑀𝑆 𝑤 dftotal = dfw + dfbet dftotal =ntotal-1 dfw =ntotal-k dfbet =k-1

16 F= 𝑀𝑆 𝑏𝑒𝑡 𝑀𝑆 𝑤 Finally, the F ratio is the ratio of MSbet and MSw
We can write all these calculated values in a summary table like this: Source SS df MS F Between k-1 MSbet=SSbet/dfbet Within ntotal-k MSw=SSw/dfw Total ntotal-1 𝑀𝑆 𝑏𝑒𝑡 𝑀𝑆 𝑤 (k is the number of groups)

17 Calculating SStotal grand mean: Source SS df MS F Between Within Total
87 99 78 91 81 69 105 70 51 76 104 75 64 79 71 108 86 68 74 83 92 62 72 66 93 85 111 65 107 84 67 61 97 90 82 Calculating SStotal grand mean: 78 96 74 100 97 76 46 75 149 Means Variances Mean of variances 93 n x Variance of means 1429 Ratio (F) 15.32 Source SS df MS F Between Within Total 10847 59

18 𝑀𝑆 𝑏𝑒𝑡 = 𝑆𝑆 𝑏𝑒𝑡 𝑑𝑓 𝑏𝑒𝑡 = 5717 4 =1429 Calculating SSbet and MSbet
87 99 78 91 81 69 105 70 51 76 104 75 64 79 71 108 86 68 74 83 92 62 72 66 93 85 111 65 107 84 67 61 97 90 82 Calculating SSbet and MSbet 𝑀𝑆 𝑏𝑒𝑡 = 𝑆𝑆 𝑏𝑒𝑡 𝑑𝑓 𝑏𝑒𝑡 = =1429 78 96 74 100 97 76 46 75 149 Means Variances Mean of variances 93 n x Variance of means 1429 Ratio (F) 15.32 Source SS df MS F Between 5717 5-1=4 1429 Within Total 10847 59

19 𝑀𝑆 𝑤 = 𝑆𝑆 𝑤 𝑑𝑓 𝑤 = 5130 55 =93 Calculating SSw and MSw Source SS df MS
B C D E 87 99 78 91 81 69 105 70 51 76 104 75 64 79 71 108 86 68 74 83 92 62 72 66 93 85 111 65 107 84 67 61 97 90 82 Calculating SSw and MSw 𝑀𝑆 𝑤 = 𝑆𝑆 𝑤 𝑑𝑓 𝑤 = =93 78 96 74 100 97 76 46 75 149 Means Variances Mean of variances 93 n x Variance of means 1429 Ratio (F) 15.32 Source SS df MS F Between 5717 5-1=4 1429 Within 5130 12x5-5=55 93 Total 10847 59

20 A B C D E 87 99 78 91 81 69 105 70 51 76 104 75 64 79 71 108 86 68 74 83 92 62 72 66 93 85 111 65 107 84 67 61 97 90 82 Fcrit with dfs of 4 and 55 and a = .05 is 2.54 Our decision is to reject H0 since > 2.54 Calculating F F= 𝑀𝑆 𝑏𝑒𝑡 𝑀𝑆 𝑤 = =15.32 78 96 74 100 97 76 46 75 149 Means Variances Mean of variances 93 n x Variance of means 1429 Ratio (F) 15.32 Source SS df MS F Between 5717 5-1=4 1429 15.32 Within 5130 12x5-5=55 93 Total 10847 59

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