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Krisztina Boda PhD Department of Medical Informatics, University of Szeged

Biostatistics, statistical software IV. Statistical estimation, confidence intervals. Hypothesis tests. One-and two sample t-tests. Krisztina Boda PhD Department of Medical Informatics, University of Szeged. Statistical estimation.

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Krisztina Boda PhD Department of Medical Informatics, University of Szeged

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  1. Biostatistics, statistical software IV.Statistical estimation, confidence intervals. Hypothesis tests. One-and two sample t-tests. Krisztina Boda PhD Department of Medical Informatics, University of Szeged

  2. Statistical estimation • A parameter is a number that describes the population (its value is not known). • For example: •  and  are parameters of the normal distribution N(,) • n, p are parameters of the binomial distribution •  is parameter of the Poisson distribution • Estimation: based on sample data, we can calculate a number that is an approximation of the corresponding parameter of the population. • A point estimate is a single numerical value used to approximate the corresponding population parameter. • For example, the sample mean is an estimation of the population’s mean, . approximates  approximates  INTERREG

  3. Interval estimate, confidence interval • Interval estimate: a range of values that we think includes the true value of the population parameter (with a given level of certainty) . • Confidence interval: an interval which contains the value of the (unknown) population parameter with high probability. • The higher the probability assigned, the more confident we are that the interval does, in fact, include the true value. • The probability assigned is the confidence level (generally: 0.90, 0.95, 0.99 ) INTERREG

  4. Interval estimate, confidence interval (cont.) • „high” probability: the probability assigned is the confidence level (generally: 0.90, 0.95, 0.99 ). • „small” probability: the „error” of the estimation (denoted by ) according to the confidence level is 1-0.90=0.1, 1-0.95=0.05, 1-0.99=0.01 • The most often used confidence level is 95% (0.95), • so the most often used value for  is =0.05 INTERREG

  5. The confidence interval is based on the concept of repetition of the study under consideration http://www.kuleuven.ac.be/ucs/java/index.htm • If the study were to be repeated 100 times, of the 100 resulting 95% confidence intervals, we would expect 95 of these to include the population parameter. INTERREG

  6. Formula of the confidence interval for the population’s mean when  is known • It can be shown that is a (1-)100% confidence interval for . • u/2 is the /2 critical value of the standard normal distribution, it can be found in standard normal distribution table for =0.05 u/2 =1.96 for =0.01 u/2 =2.58 • 95%CI for the population’s mean INTERREG

  7. The standard error of mean(SE or SEM) • is called the standard error of mean • Meaning: the dispersion of the sample means around the (unknown) population’s mean. • When  is unknown, the standard error of mean can be estimated from the sample by: INTERREG

  8. Formula of the confidence interval for the population’s mean when  is unknown • When  is unknown, it can be estimated by the sample SD (standard deviation). But, if we place the sample SD in the place of , u/2 is no longer valid, it also must be replace by t/2 . So is a (1-)100 confidence interval for . • t/2 is the /2 critical value of the Student's t statistic with n-1 degrees of freedom with n-1 degrees of freedom (see next slide) INTERREG

  9. t-distributions (Student’s t-distributions) df=19 df=200 INTERREG

  10. The Student’s t-distribution For =0.05 and df=12, the critical value is t/2=2.179 INTERREG

  11. Student’s t-distribution Degrees of freedom: 8 INTERREG

  12. Student’s t-distribution Degrees of freedom: 10 INTERREG

  13. Student’s t-distribution Degrees of freedom: 20 INTERREG

  14. Student’s t-distribution Degrees of freedom: 100 INTERREG

  15. Student’s t-distribution table INTERREG

  16. Student’s t-distribution table INTERREG

  17. Example 1. • We wish to estimate the average number of heartbeats per minute for a certain population. • The mean for a sample of 13 subjects was found to be 90, the standard deviation of the sample was SD=15.5. Supposed that the population is normally distributed the 95 % confidence interval for : • =0.05, SD=15.5 • Degrees of freedom: df=n-1=13 -1=12 • t/2 =2.179 • The lower limit is 90 – 2.179·15.5/√13=90-2.179 ·4.299=90-9.367=80.6326 • The upper limit is 90 + 2.179·15.5/√13=90+2.179 ·4.299=90+9.367=99.367 • The 95% confidence interval for the population mean is (80.63, 99.36) • It means that the true (but unknown) population means lies it the interval (80.63, 99.36) with 0.95 probability. We are 95% confident the true mean lies in that interval. INTERREG

  18. Example 2. • We wish to estimate the average number of heartbeats per minute for a certain population. • The mean for a sample of 36 subjects was found to be 90, the standard deviation of the sample was SD=15.5. Supposed that the population is normally distributed the 95 % confidence interval for : • =0.05, SD=15.5 • Degrees of freedom: df=n-1=36-1=35 • t /2=2.0301 • The lower limit is 90 – 2.0301·15.5/√36=90-2.0301 ·2.5833=90-5.2444=84.755 • The upper limit is 90 + 2.0301·15.5/√36=90+2.0301 ·2.5833=90+5.2444=95.24 • The 95% confidence interval for the population mean is (84.76, 95.24) • It means that the true (but unknown) population means lies it the interval (84.76, 95.24) with 0.95 probability. We are 95% confident the true mean lies in that interval. INTERREG

  19. Example INTERREG

  20. Presentation of results INTERREG

  21. Hypothesis testing • Hypothesis: a statement about the population • Based on our data (sample) we conclude to the whole phenomenon (population) • We examine whether our result (difference in samples) is greater then the difference caused only by chance. INTERREG

  22. Hypothesis • Hypothesis: a statement about the population • Examples • H1: =16 (the population mean is 16) • H2: ≠16 (the population mean is not 16) • H3: B=G (boys and girls score the same on mathematics exams) • H4: B≠G (boys and girls score differently on mathematics exams) • Statisticians usually test the hypothesis which tells them what to expect by giving a specific value to work with. They refer to this hypothesis as the null hypothesis and symbolize it as H0. The null hypothesis is often the one that assumes fairness, honesty or equality. • The opposite hypothesis is called alternative hypothesis and is symbolized by Ha. This hypothesis, however, is often the one that is of interest. INTERREG

  23. Steps of hypothesis-testing • Step 1. State the motivated (alternative) hypothesis Ha. • Step 2. State the null hypothesis H0. • Step 3. You select the probability of „error”, or the α significance level. α =0.05 or α =0.01. • Step 4. You choose the size n of the random sample • Step 5.Select a random sample from the appropriate population and obtain your data. • Step 6. Calculate the decision rule. • Step 7. Decision. • a) Reject the null hypothesis and claim that your alternative hypothesis was correct the difference is significant at α100% level. • b) Fail to reject the null hypothesis correct the difference is not significant at α 100% level . INTERREG

  24. Testing the mean  of a sample drawn from a normal population: one sample t-test • Problem, data. • The normal value of the systolic blood pressure is 120 mm Hg. • The following are the systolic blood pressures (mm Hg) of n=9 patients undergoing drug therapy for hypertension. • 182.00 152.00 178.00 157.00 194.00 163.00 144.00 114.00 174.00 • Summary statistics • The mean=162 mmHg, the standard deviation SD=23.92 . • Question • Can we conclude with 95% confidence on the basis of these data that the population mean is =120? • HO: the population mean is 120, =120 • Ha: the population mean is not 120 , 120 • Assumption: the sample is drawn from a normally distributed population INTERREG

  25. Decision rule based on confidence interval • Find the 95% CI for the above data! • α=0.05 • mean=162 • SD=23.92 • The standard error, SE=SD/ n=7.97 • t8,0.05=2.306) • The confidence interval: • (mean - t*SE, mean + t * SE )=(162-2.306*23.92/9, 162+2.306*7.97)=(143.61,180.386) • Can we conclude with 95% confidence on the basis of these data that the population mean is =120? • No, because the confidence interval does not contain 120. • Decision rule based on confidence interval: is the given number (the number in the null hypothesis) in the confidence interval? • If yes: the difference is not significant at α level • If not: the difference is significant at α level • In our case 120 is not in the 95%Ci, the difference is significant at 5% level. INTERREG

  26. Decision rule based on t-value • Calculatet= (mean - c)/SE=(162-120)/7.97=5.26. • Degrees of freedom: n-1=9-1=8 • Compare the absolute value of the calculated t to the critical t-value in the table: t8,0.05=2.306 • 5.26>2.306 • Decision rule: • if |t|>ttable, the difference is significant at α level • if |t|<ttable, the difference is not significant at α level • The acceptance (non-rejection) region is the set of values for which we accept the null hypothesis (- ttable, ttable) • The critical region (rejection region) is the set of values for which the null hypothesis is rejected. Acceptance region t=5.26 INTERREG

  27. Decision rule based on p-value • The probability, computed assuming that H0 is true, that the test statistic would take a value as extreme or more extreme than that actually observed. • Decision: • If p<, then the difference is significant at  level • If p>, then the difference is not significant at  level • In our case the difference is significant at 5% level, because p=0.001<0.05 INTERREG

  28. One-sample t-test, example • A company produces a 16 ml bottle of some drug (solution). The bottles are filled by an automated bottle-filling process. If this process is substantially overfilling or under filling bottles, then this process must be shut down and readjusted. Overfilling results in lost profits for the company, while under filling is unfair to consumers. For a given adjustment of the bottles consider the infinite population of all the bottle fills that could potentially be produced. We let denote the mean of the infinite population of all the bottle fills. • The company has decided that it will shut down and readjust the process if it can be very certain that the mean fill is above or below the desired 16 ml. • Now suppose that the company observes the following sample of n=6 bottle fills: • 15.68, 16.00, 15.61, 15.93, 15.86, 15.72 • It can be verified that this sample has mean=15.8 and standard deviation s=0.156. • Question: Is it true that the mean bottle fill in the population is 16? INTERREG

  29. INTERREG

  30. INTERREG

  31. A one-sample t test for paired differences (paired t-test) INTERREG

  32. Comparison of the means of two related samples: paired t-test • Self-control experiment (measure the data before and after the treatment on the same patient) or • Related data: • Before treatment – after treatment • Left side – right side • Matched pairs • Null hypothesis: the two sample means are approximations of the same population mean (there is no treatment-effect, the difference is only by chance) • HO: before= after or difference= 0 • Alternative hypothesis: there is a treatment effect • HA: before≠ after or difference≠ 0 INTERREG

  33. Comparison of the means of two related samples: paired t-test (cont.) • Calculation: take the difference of the two samples, calculate the mean and SE of the differences • Fix  • The paired t-test is a one-sample t-test for the differences. The decision rule is based on: • Confidence interval for the mean difference • Calculation of t-value and comparison its absolute value with the ttable • p-value (software) INTERREG

  34. A study was conducted to determine weight loss, body composition, etc. in obese women before and after 12 weeks of treatment with a very-low-calorie diet . We wish to know if these data provide sufficient evidence to allow us to conclude that the treatment is effective in causing weight reduction in obese women. The mean difference is actually 4. Is it a real difference? Big or small? If the study were to be repeated, would we get the same result or less, even 0? Before After Difference 85 86 -1 95 90 5 75 72 3 110 100 10 81 75 6 92 88 4 83 83 0 94 93 1 88 82 6 105 99 6 Mean 90.8 86.8 4. SD 10.79 9.25 3.333 Paired t-test, example INTERREG

  35. Paired t-test, example (cont). • Idea: if the treatment is not effective, the mean sample difference is small (close to O), if it is effective, the mean difference is big. • HO: before= after or difference= 0 • HA: before≠ after or difference≠ 0 • Let =0. • Degrees of freedom=10-1=9, t0.05,9=2.262 • Mean=4, SD=3.333 • SE=3.333/10=1.054 INTERREG

  36. Paired t-test, example (cont.) • Decision based on confidence interval: • 95%CI:(4-2.262*1.054, 4+2.262*1.054)=(1.615, 6.384) • If H0 were true, 0 were inside the confidence interval • Now 0 is outside the confidence interval, the difference is significant at 5% level, the treatment was effective. • The mean loss of body weight was 4 kg, which could be even 6.36 but minimum 1.615, with 95% probability. INTERREG

  37. Decision based on test statistic (t-value): • t= (mean - 0)/SE=mean/SE=4/1.054=3.795. This t has to be compared to the critical t-value in the table. • |t|=3.795>2.262(=t0.05,9), the difference is significant at 5% level • Decision based on p-value: • p=0.004, p<0.05, the difference is significant at 5% level Acceptance region tcomputed, test statistic ttable, critical value INTERREG

  38. Testing the mean of two independent samples from normal populations: two-sample t-test • Independent samples: • Control group, treatment group • Male, female • Ill, healthy • Young, old • etc. • Assumptions: • Independent samples : x1, x2, …, xn and y1, y2, …, ym • the xi-s are distributed as N(µ1, 1) and the yi-s are distributed as N µ2, 2 ). • H0: 1=2, Ha: 12 INTERREG

  39. The case when the standard deviations are equal • Assumptions: • 1. Both populations are approximately normal. • 2. The variances of the two populations are equal( 1= 1 = ). • That is the xi-s are distributed as N(µ1,) and the yi-s are distributed asN(µ2,) • H0: 1=2, Ha: 12 • IfH0 is true, then has Student’s t distribution with n+m-2 degrees of freedom. • Decision: • If |t|>tα,n+m-2, the difference is significant at α level, we reject H0 • If |t|<tα,n+m-2, the difference is not significant at α level, we do not reject H0 . INTERREG

  40. The case when the standard deviations are not equal • Both populations are approximately normal. • 2. The variances of the two populations are not equal ( 1 1 ). • That is the xi-s are distributed as N(µ1, 1) and the yi-s are distributed asN(µ2, 2) • H0: 1=2, Ha: 12 • : IfH0 is true, then has Student t distribution with df degrees of freedom. • Decision: • If |t|>tα,n+m-2, the difference is significant at α level, we reject H0 • If |t|<tα,n+m-2, the difference is not significant at α level, we do not reject H0 . INTERREG

  41. Comparison of the variances of two normal populations: F-test • H0: 1=2 • Ha:1 > 2 (one sided test) • F: the higher variance divided by the smaller variance: • Degrees of freedom: • 1. Sample size of the nominator-1 • 2. Sample size of the denominator-1 • Decision based on F-table • If F>Fα,table, the twp variances are significantly different at α level INTERREG

  42. Table of the F-distribution α=0.05 Nominator-> Denominator| INTERREG

  43. Example INTERREG

  44. Result of SPSS INTERREG

  45. Two sample t-test, example. • A study was conducted to determine weight loss, body composition, etc. in obese women before and after 12 weeks in two groups: • Group I. treatment with a very-low-calorie diet . • Group II. no diet • Volunteers were randomly assigned to one of these groups. • We wish to know if these data provide sufficient evidence to allow us to conclude that the treatment is effective in causing weight reduction in obese women compared to no treatment. INTERREG

  46. Two sample t-test, cont. Data INTERREG

  47. Two sample t-test, example, cont. • HO: diet=control, (the mean change in body weights are the same in populations) • Ha: diet control (the mean change in body weights are different in the populations) • Assumptions: • normality (now it cannot be checked because of small sample size) • Equality of variances (check: visually compare the two standard deviations) INTERREG

  48. Two sample t-test, example, cont. • Assuming equal variances, compute the t test- statistic: t==2.477 • Degrees of freedom: 10+11-2=19 • Critical t-value: t0.05,19=2.093 • Comparison and decision: • |t|=2.477>2.093(=t0.05,19), the difference is significant at 5% level • p=0.023<0.05 the difference is significant at 5% level INTERREG

  49. Example from the literature Circulation,2004;109:1212-1214. INTERREG

  50. INTERREG

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