1. Biostatistics, statistical software IV
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 
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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 )
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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, =0.05, =0.01
The most often used confidence level is
95% (0.95),
so the most often used value for  is
=0.05
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5. The confidence interval is based on the concept of repetition of the study under consideration
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.
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6. is a (1-)100% confidence interval for .
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
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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:
A measure of how much the value of the mean may vary from sample to sample taken from the same distribution. It can be used to roughly compare the observed mean to a hypothesized value (that is, you can conclude the two values are different if the ratio of the difference to the standard error is less than -2 or greater than +2).
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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)
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9. t-distributions (Student’s t-distributions)
df=19
df=200
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10. The Student’s t-distribution
For =0.05 and df=12,
the critical value is t/2 =2.179
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11. Student’s t-distribution
Degrees of freedom: 8
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12. Student’s t-distribution
Degrees of freedom: 10
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13. Student’s t-distribution
Degrees of freedom: 20
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14. Student’s t-distribution
Degrees of freedom: 100
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15. Student’s t-distribution table
Two sided alfa
Degrees of freedom
0.2
0.1
0.05
0.02
0.01 1 2 3 4 5 6 7 8 9 10 11
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16. Student’s t-distribution table
two sided alfa
degrees of freedom
0.2
0.1
0.05
0.02
0.01
0.001 1 2 3 4 5 6 7
... …
100
500
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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= ·4.299= =
The upper limit is
·15.5/√13= ·4.299= =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.
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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 – ·15.5/√36= ·2.5833= =84.755
The upper limit is
·15.5/√36= ·2.5833= =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.
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19. Example
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20. Presentation of results
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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.
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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.
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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 .
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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.
Summary statistics
The mean=162 mmHg, the standard deviation SD=
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
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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 )=( *23.92/9, *7.97)=(143.61, )
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.
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26. Decision rule based on t-value
Calculate t= (mean - c)/SE=( )/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
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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
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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?
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31. A one-sample t test for paired differences (paired t-test)
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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
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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)
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34. Paired t-test, example
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
Mean SD
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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
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36. Paired t-test, example (cont.)
Decision based on confidence interval:
95%CI:( *1.054, *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.
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37. tcomputed, test statistic
Decision based on test statistic (t-value):
t= (mean - 0)/SE=mean/SE=4/1.054= 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
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38. Testing the mean of two independent samples from normal populations: two-sample t-test
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
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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 as N(µ2,)
H0: 1=2, Ha: 12
If H0 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 .
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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 as N(µ2, 2)
H0: 1=2, Ha: 12
: If H0 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 .
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41. Comparison of the variances of two normal populations: F-test
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
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42. Table of the F-distribution α=0.05
Nominator->
Denominator|
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43. Example
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44. Result of SPSS
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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.
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46. Two sample t-test, cont. Data
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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)
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48. Two sample t-test, example, cont.
Assuming equal variances, compute the t test- statistic: t==2.477
Degrees of freedom: =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
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49. Example from the literature
Circulation,2004;109:
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51. Step1. H0, HA Basic Question is: Are these two groups different?
Formalized as
H0: p1 = p2 (null, no difference)
HA: p1 ≠ p2 (alternative, it is what we want to prove) or
H0: 1 = 2 (null, no difference)
HA: 1 ≠ 2 (alternative, it is what we want to prove)
The formalization of H0 and HA depends on the data type and on the measure of the endpoint.
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52. Step 2. Fix   is generally chosen to be 0.05
It represents the mistake of our later decision when reject H0.
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53. Step 3. Decide the sample size n
The sample size depends on…
The objective(s) of the research (estimation, hypothesis test, …)
Main outcome: categorical or continuous; one or more, primary or secondary, and the estimation of the distribution of the outcome – based on earlier trials.
Probability of type I. error, - see later
The power of the test (1-) (1- probability of type II. error) –see later
The research design
effect size of clinical importance …
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54. Step 4. Collect data (perform the experiment)
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55. Step 5. Decision rule
Test statistic: the summary statistics (signal/noise) of the data used in decision making between H0 and Ha.
Compute the t-statistic.
When the null hypothesis is true, the probability distribution of the test statistic (null distribution) is known .
Based on the null distribution, the critical values can be found according to the given . .
the p-value can be computed (probability of the observed test statistic as is or more extreme in either direction when the null hypothesis is true).
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56. Step 4. Decision. Decision based on p-value
If p<α: statistical significance at α level. H0 is rejected in favour of Ha.
If p≥ α : non-significance at α level. H0 cannot be rejected.
Decision based on test statistic
compare the computed t-value to the critical t-value. The difference is significant, if the absolute value of the test statistics > critical t-value
Decision based on confidence interval
the confidence interval does not contain 0
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57. Compare the mean age in the two groups!
The sample means are „similar”. Is this small difference really
caused by chance?
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58. Step 1. Step 2. Step 3. Step 4. Decision.
H0: the means in the two populations are equal: 1=2
HA: the means in the two populations are not equal: 1≠2
Step 2.
Let α=0.05
Step 3.
Decision rule: two-sample t-test.
Step 4. Decision.
Decision based on test statistic:
Compute the test statistics: t=-1.059, the degrees of freedom is =25
ttable=2.059
|t|=1.059<2.059, the difference is not significant at 5% level.
p=0.28, p>0.05, the difference is not significant at 5% level.
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59. How to get the p-value?
If H0 is true, the computed test statistic has a t-distribution with 25 degrees of freedom.
Then with 95% probability, the t-value lies in the „acceptance region”
Check it: now t=-1.059
0.025
0.95
0.025
ttable, critical value
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60. tcomputed, test statistic
How to get the p-value?
If H0 is true, the computed test statistic has a t-distribution with 25 degrees of freedom
Then with 95% probability, the t-value lies in the „acceptance region”
Check it: now t=-1.059
The p-value is the shaded area, p=0.28. The probability of the observed test statistic as is or more extreme in either direction when the null hypothesis is true.
0.025
0.95
0.025
tcomputed, test statistic
ttable, critical value
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61. Review questions and exercises
Problems to be solved by hand-calculations
..\Handouts\Problems hand IV.DOC,
..\Handouts\Practice_t-test.doc
Solutions
..\Handouts\Problems hand IV solutions.DOC,
..\Handouts\Practice_t-test, solutions.doc
Problems to be solved using computer
..\Handouts\PRACT4M.DOC
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62. Main tests to compare two groups
Comparing means, normality supposed (t-tests)
Paired samples (before-after treatment, left side-right side, etc.): paired t-test
Independent samples (control-treatment, treatment 1 – treatment 2, etc): two-sample t-test
Comparing means medians, normality not supposed : nonparametric tests, e.g. tests based on ranks
Paired samples : Wilcoxon test
Independent samples Mann-Whitney U-test
Comparing „percentages” (frequencies):
Paired samples : chi-squared test
Independent samples: McNemar test
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