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Biostat 200 Lecture 6

Biostat 200 Lecture 6. Recap. We calculate confidence intervals to give the most plausible values for the population mean or proportion.

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Biostat 200 Lecture 6

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  1. Biostat 200Lecture 6

  2. Recap • We calculate confidence intervals to give the most plausible values for the population mean or proportion. • We conduct hypothesis tests of a mean or a proportion to make a conclusion about how our sample mean or proportion compares with some hypothesized value for the population mean or proportion. • You can use 95% confidence intervals to reach the same conclusions as hypothesis tests • If that value is null value of the mean is outside of the 95% confidence intervals, that would be the same as rejecting the null of a two sided test at significance level 0.05. Pagano and Gavreau, Chapter 10

  3. Recap • We make these conclusions based on what we observed in our sample -- we will never know the true population mean or proportion • If the data are very different from the hypothesized mean or proportion, we reject the null • Example: Phase I vaccine trial – does the candidate vaccine meet minimum thresholds for safety and efficacy? • Statistical significance can be driven by n, and does not equal clinical or biological significance • On the other hand, you might have a suggestive result but not statistical significance with a small sample that deserves a larger follow up study Pagano and Gavreau, Chapter 10

  4. Types of error • Type I error = significance level of the test =P(reject H0 | H0 is true) • Incorrectly reject the null • We take a sample from the population, calculate the statistics, and make inference about the true population. If we did this repeatedly, we would incorrectly reject the null 5% of the time that it is true if  is set to 0.05. Pagano and Gavreau, Chapter 10

  5. Types of error • Type II error –  = P(do not reject H0 | H0 is false) • Incorrectly fail to reject the null • This happens when the test statistic is not large enough, even if the underlying distribution is different

  6. Types of error • Remember, H0 is a statement about the population and is either true or false • We take a sample and use the information in the sample to try to determine the answer • Whether we make a Type I error or a Type II error depends on whether H0 is true or false • We set , the chance of a Type I error, and we can design our study to minimize the chance of a Type II error

  7. Chance of a type II error , chance of failing to reject the null if the alternative is true

  8. If the alternative is very different from the null, the chance of a Type II error is low

  9. If the alternative is not very different from the null, the chance of a Type II error is high

  10. Chance of a Type II error is lower if the SD is smaller This is relevant because the SD for the distribution of a sample mean is σ/n So increasing n decreases the SD of the sampling distribution

  11. What happens when you increase 

  12. Finding , P(Type II error) • Example: Mean age of walking • H0: μ<11.4 months (μ0) • Alternative hypothesis: HA: μ>11.4 months • Known SD=2 • Significance level=0.05 • Sample size=9 • We will reject the null if the zstat (assuming σ known) > 1.654 • So we will reject the null if • For our example, the null will be rejected if X> 1.654*2/3 + 11.4 = 13.9

  13. But if the true mean is really 16, what is the probability that the null will not be rejected? • The probability of a Type II error? • The null will be rejected if the sample mean is >13.9, not rejected if is ≤13.9 • What is the probability of getting a sample mean of ≤13.9 if the true mean is 16 in a sample with n=9? • P(Z<(13.9-16)/(2/9)) = 0.0008 • So if the true mean is 16 and the sample size is 9, the probability of rejecting the null incorrectly is 0.0008

  14. Note that this depended on: • The hypothesized alternative population mean (16) • What is the probability of failing to reject the null if the true population is 15? • P(Z<(13.9-15)/(2/9)) = P(Z< -1.6499)= 0.05 • What is the probability of failing to reject the null if the true population mean is 14? • P(Z<(13.9-14)/ (2/9)) =P(Z<-.1499) = 0.44 • The sample size n • What is the probability of failing to reject the null if the true population is 14 and n is 100? • P(Z<(13.9-14)/(2/100)) = P(Z<-.5)= 0.31 • What is the probability of failing to reject the null if the true population mean is 14 and n is 1000? • P(Z<(13.9-14)/(2/100)) = P(Z<-1.581)= 0.06

  15. Power • The power of a test is the probability that you will reject the null hypothesis when it is not true and it is 1-  • Written P(reject H0 | HA ) • You can construct a power curve by plotting power versus different alternative hypotheses • You can also construct a power curve by plotting power versus different sample sizes (n’s) • This curve will allow you to see what gains you might have versus cost of adding participants • Power curves are not linear – you get to a point of diminishing returns

  16. Power curves

  17. The power of a statistical test is lower for alternative values that are closer to the null value (the chance of a Type II error is higher) and higher for more extreme alternative values. • The power of a statistical test can be increased by increasing n • You can fix =0.05 and =0.20 (for 80% power) and determine n for various alternative hypotheses • In practice, you often have n fixed by cost • Then you can calculate how big the alternative has to be to reject the null with 80% probability assuming the alternative is true • The difference between this alternative and the null is called the minimum detectable difference • In epidemiology when wanting to estimate an odds ratio it is call the minimum detectable odds ratio

  18. Comparison of two means: paired t-test • Paired samples, continuous variables • Two determinations on the same person (before and after) – e.g. before and after intervention • Matched samples – measurement on pairs of persons similar in some characteristics, i.e. identical twins (matching is on genetics) • Matching or pairing is performed to control for extraneous factors • Each person or pair has 2 data points, and we calculate the difference for each • Then we can use our one-sample methods to test hypotheses about the value of the difference Pagano and Gavreau, Chapter 11

  19. Two independent samples Paired data

  20. Comparison of two means: paired t-test • Step 1: The hypothesis (two sided) • Generically H0: μ1-μ2 =δ HA: μ1-μ2 ≠δ • Often δ=0, no difference So H0: μ1-μ2 =0, i.e. H0: μ1=μ2 HA: μ1-μ2 ≠0, i.e. HA: μ1≠μ2

  21. Comparison of two means: paired t-test • Step 1: The hypothesis (one sided) • Generically H0: μ1-μ2 ≥δ or H0: μ1-μ2 ≤δ HA: μ1-μ2 <δH0: μ1-μ2 <δ • Often δ=0, no difference So H0: μ1 ≥ μ2 or H0: μ1 ≤ μ2 HA: μ1 < μ2 HA: μ1 > μ2

  22. Comparison of two means: paired t-test • Step 2: Determine the compatibility with the null hypothesis The test statistic is

  23. Comparison of two means: paired t-test • Step 3: Reject or fail to reject the null • Is the p-value (the probability of observing a difference as large or larger, under the null hypothesis) greater than or less than the significance level, ?

  24. Paired t-test example list +-----------+ | wt1 wt2 | |-----------| 1. | 55 58 | 2. | 90 87 | 3. | 48 55 | 4. | 76 65 | 5. | 64 77 | |-----------| 6. | 84 92 | 7. | 56 57 | 8. | 81 76 | 9. | 92 93 | 10. | 65 72 | +-----------+

  25. . gen wtdiff=wt1-wt2 . summ wtdiff, detail wtdiff ------------------------------------------------------------- Percentiles Smallest 1% -13 -13 5% -13 -8 10% -10.5 -7 Obs 10 25% -7 -7 Sum of Wgt. 10 50% -2 Mean -2.1 Largest Std. Dev. 7.093816 75% 3 -1 90% 8 3 Variance 50.32222 95% 11 5 Skewness .3296938 99% 11 11 Kurtosis 2.396264 . di -2.1/(7.094/sqrt(10)) -.93611264 . di 2*ttail(9,.936113) .37365191

  26. . ttest wt1==wt2 Paired t test ------------------------------------------------------------------------------ Variable | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- wt1 | 10 71.1 4.933896 15.60235 59.93875 82.26125 wt2 | 10 73.2 4.535784 14.34341 62.93934 83.46066 ---------+-------------------------------------------------------------------- diff | 10 -2.1 2.243262 7.093816 -7.17461 2.97461 ------------------------------------------------------------------------------ mean(diff) = mean(wt1 - wt2) t = -0.9361 Ho: mean(diff) = 0 degrees of freedom = 9 Ha: mean(diff) < 0 Ha: mean(diff) != 0 Ha: mean(diff) > 0 Pr(T < t) = 0.1868 Pr(|T| > |t|) = 0.3736 Pr(T > t) = 0.8132 .

  27. Comparison of two means: t-test • The goal is to compare means from two independent samples • Two different populations • E.g. vaccine versus placebo group • E.g. women with adequate versus in adequate micronutrient levels Pagano and Gavreau, Chapter 11

  28. Comparison of two means: t-test • Two sided hypothesis H0: μ1=μ2 ... μ1-μ2=0 HA: μ1≠μ2 μ1-μ2≠0 • One sided hypothesis H0: μ1≥μ2 μ1-μ2 ≥ 0 HA: μ1<μ2 • One sided hypothesis H0: μ1≤μ2 HA: μ1>μ2 Pagano and Gavreau, Chapter 11

  29. Comparison of two means: t-test • Even though the null and alternative hypotheses are the same as for the paired t-test, the test is different, it is wrong to use a paired t-test with independent samples and vice versa Pagano and Gavreau, Chapter 11

  30. Comparison of two means: t-test • By the CLT, is normally distributed with mean μ1-μ2 and standard deviation • In one version of the t-test, we assume that the population standard deviations are equal, so σ1 = σ2 = σ • Substituting, the standard deviation for the distribution of the difference of two sample means is • So we can calculate a z-score for the difference in the means and compare it to the standard normal distribution. The test statistic is Pagano and Gavreau, Chapter 11

  31. Comparison of two means: t-test • If the σ’s are unknown (pretty much always), we substitute with sample standard deviations, s, and compare the test statistic to the t-distribution • t-test test statistic  • The formula for the pooled SD is a weighted average of the individual sample SDs • The degrees of freedom for the test are n1+n2-2 Pagano and Gavreau, Chapter 11

  32. Comparison of two means: t-test • As in our other hypothesis tests, compare the t statistic to the t-distribution to determine the probability of obtaining a mean difference as large or larger as the observed difference • Reject the null if the probability, the p-value, is less than , the significance level • Fail to reject the null if p≥ 

  33. Comparison of two means, example • Study of non-pneumatic anti-shock garment (Miller et al) • Two groups – pre-intervention received usual treatment, intervention group received NASG • Comparison of hemorrhaging in the two groups • Null hypothesis: The hemorrhaging is the same in the two groups H0: μ1=μ2 HA: μ1≠μ2 • The data: • External blood loss: • Pre-intervention group (n=83) mean=340.4 SD=248.2 • Intervention group (n=83) mean=73.5 SD=93.9 Pagano and Gavreau, Chapter 11

  34. Calculating by hand • External blood loss: • Pre-intervention group (n=83) mean=340.4 SD=248.2 • Intervention group (n=83) mean=73.5 SD=93.9 • First calculate sp2 = (82*248.22 + 82*93.92)/(83+83-2) = 35210.2 tstat = (340.4-73.5)/sqrt(35210.2*(2/83)) = 9.16 df=83+83-2=164 . di 2*ttail(164,9.16) 2.041e-16

  35. Comparison of two means, example * ttesti n1 mean1 sd1 n2 mean2 sd2 ttesti 83 340.4 248.2 83 73.5 93.9 Two-sample t test with equal variances ------------------------------------------------------------------------------ | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- x | 83 340.4 27.24349 248.2 286.204 394.596 y | 83 73.5 10.30686 93.9 52.99636 94.00364 ---------+-------------------------------------------------------------------- combined | 166 206.95 17.85377 230.0297 171.6987 242.2013 ---------+-------------------------------------------------------------------- diff | 266.9 29.12798 209.3858 324.4142 ------------------------------------------------------------------------------ diff = mean(x) - mean(y) t = 9.1630 Ho: diff = 0 degrees of freedom = 164 Ha: diff < 0 Ha: diff != 0 Ha: diff > 0 Pr(T < t) = 1.0000 Pr(|T| > |t|) = 0.0000 Pr(T > t) = 0.0000 Pagano and Gavreau, Chapter 11

  36. You can calculate a 95% confidence interval for the difference in the means • If the confidence interval for the difference does not include 0, then you can reject the null hypothesis of no difference • This is NOT equivalent to calculating separate confidence intervals for each mean and determining whether they overlap

  37. Comparison of two means: t-test • This t-test assumes equal variances in the two underlying populations • Assumes that both samples come from populations, each with variance σ2 • If we do not assume equal variances we use a slightly different test statistic • Variances not assumed to be equal, so you do not use a pooled estimate • There is another formula for degrees of freedom • Often the two different t-tests yield the same answer, but it’s important to check if you don’t have good reason to assume equivalence Pagano and Gavreau, Chapter 11

  38. The t statistic is Round up to the nearest integer to get the degrees of freedom

  39. Comparison of two means, example ttesti 83 340.4 248.2 83 73.5 93.9, unequal Two-sample t test with unequal variances ------------------------------------------------------------------------------ | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- x | 83 340.4 27.24349 248.2 286.204 394.596 y | 83 73.5 10.30686 93.9 52.99636 94.00364 ---------+-------------------------------------------------------------------- combined | 166 206.95 17.85377 230.0297 171.6987 242.2013 ---------+-------------------------------------------------------------------- diff | 266.9 29.12798 209.1446 324.6554 ------------------------------------------------------------------------------ diff = mean(x) - mean(y) t = 9.1630 Ho: diff = 0 Satterthwaite's degrees of freedom = 105.002 Ha: diff < 0 Ha: diff != 0 Ha: diff > 0 Pr(T < t) = 1.0000 Pr(|T| > |t|) = 0.0000 Pr(T > t) = 0.0000 • Note that the t-statistic stayed the same. This is because the sample sizes in each group are equal. When the sample sizes are not equal this will not be the case • The degrees of freedom are decreased, so if the sample sizes are equal in the two groups this is a more conservative test

  40. Test of the means of independent samples • When you have the data in Stata, with the different groups in different columns, use ttest var1==var2, unpaired or ttest var1==var2, unpaired unequal • More often, you will have the data all in one variable, and the grouping in another variable. Then use ttest var, by(groupvar) or ttest var, by(groupvar) unequal

  41. Testing whether BMI in our class data set differs by sex Null hypothesis: BMI of females = BMI of males . ttest bmi, by(sex) Two-sample t test with equal variances ------------------------------------------------------------------------------ Group | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- male | 292 22.84854 .2370304 4.050377 22.38203 23.31506 female | 241 24.76146 .1948815 3.025374 24.37757 25.14536 ---------+-------------------------------------------------------------------- combined | 533 23.71348 .1621327 3.743124 23.39499 24.03198 ---------+-------------------------------------------------------------------- diff | -1.912917 .3153224 -2.53235 -1.293485 ------------------------------------------------------------------------------ diff = mean(male) - mean(female) t = -6.0665 Ho: diff = 0 degrees of freedom = 531 Ha: diff < 0 Ha: diff != 0 Ha: diff > 0 Pr(T < t) = 0.0000 Pr(|T| > |t|) = 0.0000 Pr(T > t) = 1.0000

  42. . ttest bmi, by(sex) unequal Two-sample t test with unequal variances ------------------------------------------------------------------------------ Group | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- male | 292 22.84854 .2370304 4.050377 22.38203 23.31506 female | 241 24.76146 .1948815 3.025374 24.37757 25.14536 ---------+-------------------------------------------------------------------- combined | 533 23.71348 .1621327 3.743124 23.39499 24.03198 ---------+-------------------------------------------------------------------- diff | -1.912917 .3068586 -2.515736 -1.310099 ------------------------------------------------------------------------------ diff = mean(male) - mean(female) t = -6.2339 Ho: diff = 0 Satterthwaite's degrees of freedom = 525.975 Ha: diff < 0 Ha: diff != 0 Ha: diff > 0 Pr(T < t) = 0.0000 Pr(|T| > |t|) = 0.0000 Pr(T > t) = 1.0000

  43. Confidence interval for the difference of two means from independent samples, when unequal variances are assumed

  44. Comparison of two proportions • Similar to comparing two means • Null hypothesis about two proportions, p1 and p2, H0: p1= p2 HA: p1≠ p2 • If n1 and n2 are sufficiently large, the difference between the two proportions follows a normal distribution. So we can use the z statistic to find the probability of observing a difference as large as we do, under the null hypothesis of no difference Pagano and Gavreau, Chapter 14

  45. Comparison of two proportions • Example: HIV prevalence among the men and women testing at Mulago Hospital • N=466 males; 107 (23.0%) tested HIV+ • N=467 females; 162 (34.7%) tested HIV+ • Null hypothesis: The HIV prevalence in males and females is the same? • H0: p1= p2 • Z statistic is calculated: • p̂ = (107+162)/(466+467) = 0.288 • zstat = (.230-.347)/sqrt[ .288*(1-.288)*(1/466+1/467) • =-.117/.030 = -3.95 • 2*p(Z<-.395) < 0.001 Pagano and Gavreau, Chapter 14

  46. Comparison of two proportions prtesti n1 p1 n2 p2 prtesti 466 .23 467 .347 Two-sample test of proportion x: Number of obs = 466 y: Number of obs = 467 ------------------------------------------------------------------------------ Variable | Mean Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------- x | .23 .0194947 .1917911 .2682089 y | .347 .0220274 .3038271 .3901729 -------------+---------------------------------------------------------------- diff | -.117 .0294151 -.1746525 -.0593475 | under Ho: .0296673 -3.94 0.000 ------------------------------------------------------------------------------ diff = prop(x) - prop(y) z = -3.9437 Ho: diff = 0 Ha: diff < 0 Ha: diff != 0 Ha: diff > 0 Pr(Z < z) = 0.0000 Pr(|Z| < |z|) = 0.0001 Pr(Z > z) = 1.0000 Pagano and Gavreau, Chapter 14

  47. Comparison of two proportions . prtest hiv, by(sex) Two-sample test of proportion 0: Number of obs = 466 1: Number of obs = 467 ------------------------------------------------------------------------------ Variable | Mean Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------- 0 | .2296137 .0194832 .1914274 .2678001 1 | .3468951 .0220258 .3037253 .3900649 -------------+---------------------------------------------------------------- diff | -.1172813 .0294063 -.1749167 -.059646 | under Ho: .0296598 -3.95 0.000 ------------------------------------------------------------------------------ diff = prop(0) - prop(1) z = -3.9542 Ho: diff = 0 Ha: diff < 0 Ha: diff != 0 Ha: diff > 0 Pr(Z < z) = 0.0000 Pr(|Z| < |z|) = 0.0001 Pr(Z > z) = 1.0000 Pagano and Gavreau, Chapter 14

  48. Comparison of several means: analysis of variance • A different test is needed when we want to test for differences among more than two independent samples • This test is called the Analysis of Variance, or ANOVA • Null hypothesis H0: all equal means μ1=μ2=μ3=… • The alternative HA is that at least one of the means differs from the others Pagano and Gavreau, Chapter 12

  49. Testing each pair of means would increase the chances of making a type I error (rejecting the null when it is true) • Say the null of no difference is true • Then P(Type I error) = P(reject H0 | H0 is true) =  for each test • P( no Type I error on 1 test) = 1-  • P( no Type I error on all 3 tests) = (1- )3 • If =0.05, then (1- )3 = .857 • The P(at least 1 type I error on 3 tests) = 1-.875=.143 • So there is a reasonable chance (.143)that the null will be rejected even if it is true • This is known as the multiple comparison problem

  50. Comparison of several means: analysis of variance • Why is it called analysis of variance? • The test compares the between-group variability (how different are the groups from the overall mean) to the within-group variability Pagano and Gavreau, Chapter 12

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