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Basic Statistics II

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Significance/hypothesis tests

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RCT comparing drug A and drug B for the treatment of hypertension 50 patients allocated to A 50 patients allocated to B Outcome = systolic BP at 3 months

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Results Group A Mean = 145, sd = 9.9 Group B Mean = 135, sd = 10.0

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Null hypothesis : “μ (A) = μ (B)” [ie. difference equals 0] Alternative hypothesis : “μ (A) ≠ μ (B)” [ie. difference doesn’t equal zero] [where μ = population mean]

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Statistical problem When can we conclude that the observed difference mean(A) - mean(B) is large enough to suspect that μ (A) - μ (B) is not zero?

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P-value : “probability of obtaining observed data if the null hypothesis were true” [eg. if no difference in systolic BP between two groups]

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How do we evaluate the probability?

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Test Statistic Numerical value which can be compared with a known statistical distribution Expressed in terms of the observed data and the data expected if the null hypothesis were true

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Test statistic [mean (A) – mean (B)] / sd [mean(A)-mean(B)] Under null hypothesis this ratio will follow a Normal distribution with mean = 0 and sd = 1

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Hypertension example Test statistic = [mean (A) – mean (B)] / sd [mean(A)-mean(B)] = [ 145 – 135 ] / 1.99 = 5 → p <0.001

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Interpretation Drug B results in lower systolic blood pressure in patients with hypertension than does Drug A

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Two-sample t-test Compares two independent groups of Normally distributed data

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Significance test example I

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Null hypothesis : “μ (A) = μ (B)” [ie. difference equals 0] Alternative hypothesis : “μ (A) ≠ μ (B)” [ie. difference doesn’t equal zero] Two-sided test

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Null hypothesis : “μ (A) = μ (B) or μ (A) < μ (B) ” Alternative hypothesis : “μ (A) > μ (B)” One-sided test

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A one-sided test is only appropriate if a difference in the opposite direction would have the same meaning or result in the same action as no difference

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Paired-sample t-test Compares two dependent groups of Normally distributed data

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Paired-sample t-test Mean daily dietary intake of 11 women measured over 10 pre-menstrual and 10 post-menstrual days

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Dietary intake example Pre-menstrual (n=11): Mean=6753kJ, sd=1142 Post-menstrual (n=11): Mean=5433kJ, sd=1217 Difference Mean=1320, sd=367

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Dietary intake example Test statistic = 1320/[367/sqrt(11)] = 11.9 p<0.001

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Dietary intake example Dietary intake during the pre- menstrual period was significantly greater than that during the post- menstrual period

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The equivalent non-parametric tests Mann-Whitney U-test Wilcoxon matched pairs signed rank sum test

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Non-parametric tests Based on the ranks of the data Use complicated formula Hence computer package is recommended

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Significance test example II

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Type I error Significant result when null hypothesis is true (0.05) Type II error Non-significant result when null hypothesis is false [Power = 1 – Type II]

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The chi-square test Used to investigate the relationship between two qualitative variables The analysis of cross-tabulations

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The chi-square test Compares proportions in two independent samples

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Chi-square test example In an RCT comparing infra-red stimulation (IRS) with placebo on pain caused by osteoarthritis, 9/12 in IRS group ‘improved’ compared with 4/13 in placebo group

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Chi-square test example Improve? Yes No Placebo 4 9 13 IRS 9 3 12 13 12 25

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Placebo : 4/13 = 31% improve IRS: 9/12 = 75% improve

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Cross-tabulations The chi-square test tests the null hypothesis of no relationship between ‘group’ and ‘improvement’ by comparing the observed frequencies with those expected if the null hypothesis were true

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Cross-tabulations Expected frequency = row total x col total grand total

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Chi-square test example Improve? Yes No Placebo 4 9 13 IRS 9 3 12 13 12 25 Expected value for ‘4’ = 13 x 13 / 25 = 6.8

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Expected values Improve? Yes No Placebo 6.8 6.2 13 IRS 6.2 5.8 12 13 12 25

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Test Statistic = (observed freq – expected freq) 2 expected freq

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Test Statistic = (O – E) 2 E = (4 - 6.8) 2 /6.8 + (9 – 6.2) 2 /6.2 + (4 - 6.8) 2 /6.8 + (9 – 6.2) 2 /6.2 = 4.9 → p=0.027

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Chi-square test example Statistically significant difference in improvement between the IRS and placebo groups

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Small samples The chi-square test is valid if: at least 80% of the expected frequencies exceed 5 and all the expected frequencies exceed 1

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Small samples If criterion not satisfied then combine or delete rows and columns to give bigger expected values

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Small samples Alternatively: Use Fisher’s Exact Test [calculates probability of observed table of frequencies - or more extreme tables-under null hypothesis]

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Yates’ Correction Improves the estimation of the discrete distribution of the test statistic by the continuous chi-square distribution

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Chi-square test with Yates’ correction Subtract ½ from the O-E difference (|O – E|-½) 2 E

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Significance test example III

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McNemar’s test Compares proportions in two matched samples

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McNemar’s test example Severe cold age 14 Yes No Severe Yes 212 144 356 cold No 256 707 963 age 468 851 1319 12

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McNemar’s test example Null hypothesis = proportions saying ‘yes’ on the 1 st and 2 nd occasions are the same the frequencies for ‘yes,no’ and ‘no,yes’ are equal

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McNemar’s test Test statistic based on observed and expected ‘discordant’ frequencies Similar to that for simple chi-square test

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McNemar’s test example Test statistic = 31.4 => p <0.001 Significant difference between the two ages

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Significance test example IV

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Comparison of means 2 groups 2-sample t-test 3 or more groups ANOVA

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One-way analysis of variance Example: Assessing the effect of treatment on the stress levels of a cohort of 60 subjects. 3 age-groups: 15-25, 26-45, 46-65 Stress measured on scale 0-100

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Stress levels GroupMean (SD) 15-25 (n=20)52.8 (11.2) 26-45 (n=20)33.4 (15.0) 46-65 (n=20)35.6 (11.7)

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Graph of stress levels

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ANOVA Sum of squares DfMean square FSig Between groups 4513.622256.813.8<0.001 Within groups 9294.857163.1 Total13808.459

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Interpretation Significant difference between the three age-groups with respect to stress levels But what about the specific (pairwise) differences?

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Stress levels GroupMean (SD) 15-25 (n=20)52.8 (11.2) 26-45 (n=20)33.4 (15.0) 46-65 (n=20)35.6 (11.7)

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Multiple comparisons Comparing each pair of means in turn gives a high probability of finding a significant result by chance A multiple comparison method (eg. Scheffé, Duncan, Newman-Keuls) makes appropriate adjustment

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Scheffés test Comparison 15-25 vs. 26-45p<0.001 15-25 vs. 46-65p<0.001 26-45 vs. 46-65p=0.86

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Stress levels GroupMean (SD) 15-25 (n=20)52.8 (11.2) 26-45 (n=20)33.4 (15.0) 46-65 (n=20)35.6 (11.7)

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Comparison of medians 2 groups Mann-Whitney 3 or more groups Kruskal-Wallis

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Kruskal-Wallis Example: Stress levels Overall comparison of 3 groups: p<0.001

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Multiple comparisons There are no non-parametric equivalents to the multiple comparison tests such as Scheffés Need to apply Bonferroni’s correction to multiple Mann-Whitney U-tests

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Bonferroni’s correction For k comparisons between means: multiply each p value by k

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Mann-Whitney U-test Comparison 15-25 vs. 26-45p<0.001 15-25 vs. 46-65p<0.001 26-45 vs. 46-65p=0.68 Need to multiple each p-value by 3

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Significance test example V

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