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Problem Set 7 Notes

# Problem Set 7 Notes - Some housekeeping Formulae will be...

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Some housekeeping Formulae will be provided in the mid-term test “Simple” calculations “Short” answers No admission after 4:35 pm

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Standard error of a proportion ( 29 = - - = n i i n x x x Variance 1 2 1 ) ( For continuous data: And SE = SD / √n If we score a ‘success’ as 1, and a ‘failure’ as 0, then applying the formula above (with n, rather than n-1, in the denominator of the variance formula) gives SE ( π ) = SD = Standard Deviation = Variance n ) π π - (1
Approximation of binomial distribution to the normal distribution It was noted that the approximation could be applied when np > 5 or nq > 5 This is because the approximation is not very good otherwise Similar to the idea that the distribution of sample means is a good approximation to the normal distribution when the size of a sample is 30 or more

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Effect size We have looked at the general principles for computing sample sizes For comparing two groups with the t-test, the formula is: But often we don’t know σ in advance One way round this is to look at ‘effect size’, which is essentially ∆/σ This can be used directly in the formula

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Click to edit Master subtitle style Problem set 7: Comparing more than two groups Harry Shannon
What this problem will cover How do we avoid ‘multiple testing’ when comparing several groups One-way ANOVA ANOVA tables F-tests Degrees of freedom

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Do Sun-Screens Save your Skin? In the previous problem set, we analysed the data as both an unmatched and matched design using t- tests. However, there is more to sun-screens than addressed by the simple question of "Does it work?" All screens prominently display a SPF code (Sun Protection Factor) which is defined as the ratio of time to burn with sun-screen on to time to burn with no sun-screen
Is there any difference among sunscreens? Can we establish whether there really is a dose-response relationship? We might proceed by taking three samples of Bronzetone off the shelf, with SPF ratings of 2, 8, and 15 for the sake of argument (staying within one brand name for consistency). For six subjects in each group, the degree of burning scores might look like this:

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SPF-2 SPF-8 SPF-15 Subject Score Subject Score Subject Score 1 10 7 7 13 4 2 11 8 5 14 5 3 12 9 12 15 6 4 4 10 2 16 3 5 5 11 5 17 2 6 6 12 5 18 4 Mean 8.0 Mean 6.0 Mean 4.0 Once again, we ask the statistical question: Are these SPF-specific results statistically different from each other?
How do we compare the means of more than two groups? In our example we have 3 groups. We could simply do 3 t-tests between each pair of means: 1 vs 2; 1 vs 3; 2 vs 3 Or more generally compare the means of all pairs of groups

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Number of comparisons = n(n-1)/2 No. Groups No. Comparisons 3 3 4 6 6 15 9 36
This messes with the p-value: What’s the probability of getting at least one pair significantly different under all the possible hypotheses like H0 : μr = μs Prob (at least 1 diff) = 1 - Prob(no diff) = 1 - (1 - ° )k No groups Prob (1 or more sig) 3 .15 4 .27

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Pascal’s triangle 1 1 1 1 2 1 1 3 3 1
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Problem Set 7 Notes - Some housekeeping Formulae will be...

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