bandit thinkhamrop, phd. (statistics) department of biostatistics and demography faculty of public...
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Bandit Thinkhamrop, PhD. (Statistics)Department of Biostatistics and Demography
Faculty of Public HealthKhon Kaen University, THAILAND
Statistical inference revisited
Statistical inference use data from samples to make inferences about a population
1. Estimate the population parameter• Characterized by confidence interval of the
magnitude of effect of interest
2. Test the hypothesis being formulated before looking at the data
• Characterized by p-value
n = 25X = 52SD = 5
Sample
PopulationParameter estimation
[95%CI]
Hypothesis testing[P-value]
Parameter estimation[95%CI]
Hypothesis testing[P-value]
n = 25X = 52SD = 5SE = 1
Sample
PopulationParameter estimationParameter estimation
[95%CI] : 52-1.96(1) to 52+1.96(1) 50.04 to 53.96We are 95% confidence that the population mean would lie between 50.04 and 53.96
[95%CI] : 52-1.96(1) to 52+1.96(1) 50.04 to 53.96We are 95% confidence that the population mean would lie between 50.04 and 53.96
Z = 2.58Z = 1.96Z = 1.64
n = 25X = 52SD = 5SE = 1
Sample
Hypothesis testing
Hypothesis testing
Population
Z = 55 – 52 1
3H0 : = 55HA : 55
Hypothesis testing
H0 : = 55HA : 55If the true mean in the population is 55, chance to obtain a sample mean of 52 or more extreme is 0.0027.
Hypothesis testing
H0 : = 55HA : 55If the true mean in the population is 55, chance to obtain a sample mean of 52 or more extreme is 0.0027.
Z = 55 – 52 1
3 P-value = 1-0.9973 = 0.0027
5552
-3SE +3SE
Calculation of the previous example based on t-distribution
Stata command to find probability.di (ttail(24, 3))*2 .00620574
Stata command to find t value for 95%CL. di (invttail(24, 0.025))2.0638986
Web base stat table: http://vassarstats.net/tabs.html or www.stattrek.com
Revisit the example based on t-distribution (Stata output)
Variable | Obs Mean Std. Err. [95% Conf. Interval]-------------+--------------------------------------------------------------- | 25 52 1 49.9361 54.0639
1. Estimate the population parameter
2. Test the hypothesis being formulated before looking at the data
One-sample t test------------------------------------------------------------------------------ | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval]---------+-------------------------------------------------------------------- x | 25 52 1 5 49.9361 54.0639------------------------------------------------------------------------------ mean = mean(x) t = -3.0000Ho: mean = 55 degrees of freedom = 24
Ha: mean < 55 Ha: mean != 55 Ha: mean > 55 Pr(T < t) = 0.0031 Pr(|T| > |t|) = 0.0062 Pr(T > t) = 0.9969
Mean one group: T-test
a12255
1. Hypothesis H0: = 0Ha: 0
2. Data
3. Calculating for t-statistic
4. Obtain p-value based on t-distribution
5. Make a decision
P-value = 0.023
Reject the null hypothesis at level of significant of 0.05 The mean of y is statistically significantly different from zero.
Stata command .di (ttail(4, 3.59))*2 .02296182
Mean one group: T-test (cont.)
One-sample t test------------------------------------------------------------------------------Variable | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval]---------+-------------------------------------------------------------------- y | 5 3 .83666 1.870829 .6770594 5.322941------------------------------------------------------------------------------ mean = mean(y) t = 3.5857Ho: mean = 0 degrees of freedom = 4
Ha: mean < 0 Ha: mean != 0 Ha: mean > 0 Pr(T < t) = 0.9885 Pr(|T| > |t|) = 0.0231 Pr(T > t) = 0.0115
Comparing 2 means: T-test
a b1 52 92 95 85 9
1. Hypothesis H0: A = B
Ha: A B
2. Data
3. Calculating for t-statistic
4. Obtain p-value based on t-distribution
5. Make a decision
P-value = 0.002 (http://vassarstats.net/tabs.html)
Reject the null hypothesis at level of significant of 0.05 Mean of Group A is statistically significantly different from that of Group B.
T-testTwo-sample t test with equal variances------------------------------------------------------------------------------ Group | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval]---------+-------------------------------------------------------------------- a | 5 3 .83666 1.870829 .6770594 5.322941 b | 5 8 .7745967 1.732051 5.849375 10.15063---------+--------------------------------------------------------------------combined | 10 5.5 .9916317 3.135815 3.256773 7.743227---------+-------------------------------------------------------------------- diff | -5 1.140175 -7.629249 -2.370751------------------------------------------------------------------------------ diff = mean(1) - mean(2) t = -4.3853Ho: diff = 0 degrees of freedom = 8
Ha: diff < 0 Ha: diff != 0 Ha: diff > 0 Pr(T < t) = 0.0012 Pr(|T| > |t|) = 0.0023 Pr(T > t) = 0.9988
Two independent sample t-test using SPSS
Please do the data analysis using SPSS and paste the results here.
Mann-Whitney U test Wilcoxon rank-sum test
Two-sample Wilcoxon rank-sum (Mann-Whitney) test
group | obs rank sum expected-------------+--------------------------------- 1 | 5 16 27.5 2 | 5 39 27.5-------------+--------------------------------- combined | 10 55 55
unadjusted variance 22.92adjustment for ties -1.25 ----------adjusted variance 21.67
Ho: y(group==1) = y(group==2) z = -2.471 Prob > |z| = 0.0135
Mann-Whitney U test Wilcoxon rank-sum test
using SPSS
Please do the data analysis using SPSS and paste the results here.
Comparing 2 means : ANOVAMathematical model of ANOVA X = + +
X = Grand mean + Treatment effect + ErrorX = M + T + E
3. Calculating for F-statistic
4. Obtain p-value based on F-distribution
5. Make a decision
P-value = 0.002 (http://vassarstats.net/tabs.html)
Reject the null hypothesis at level of significant of 0.05 Mean of Group A is statistically significantly different from that of Group B.
X M T
Mean: 3 8
= + +
E
[3-5.5] [8-5.5]
SST
SSE
Degree of freedom 1 1 8
Between groups
Within groups
ANOVA 2 groups
Analysis of Variance Source SS df MS F Prob > F------------------------------------------------------------------------Between groups 62.5 1 62.5 19.23 0.0023 Within groups 26 8 3.25------------------------------------------------------------------------ Total 88.5 9 9.83333333
Bartlett's test for equal variances: chi2(1) = 0.0211 Prob>chi2 = 0.885
Comparing 3 means: ANOVA1. Hypothesis H0: A = B = C
Ha: At least one mean is difference
2. Data a b c1 5 42 9 42 9 65 8 85 9 4
ANOVA 3 groups (cont.)Mathematical model of ANOVA X = + +
X = Grand mean + Treatment effect + ErrorX = M + T + E
3. Calculating for F-statistic
4. Obtain p-value based on F-distribution
5. Make a decision
P-value = 0.003 (http://vassarstats.net/tabs.html)
Reject the null hypothesis at level of significant of 0.05 At least one mean of the three groups is statistically significantly different from the others.
X M T
Mean: 3 8 5.2
= + +
E
[3-5.4] [8-5.4] [5.2-5.4]
SST
SSE
Df: 15 1 2 12
Between groups
Within groups
ANOVA 3 groups
Analysis of Variance Source SS df MS F Prob > F------------------------------------------------------------------------Between groups 62.8 2 31.4 9.71 0.0031 Within groups 38.8 12 3.23333333------------------------------------------------------------------------ Total 101.6 14 7.25714286
Bartlett's test for equal variances: chi2(2) = 0.0217 Prob>chi2 = 0.989
Kruskal-Wallis testKruskal-Wallis equality-of-populations rank test
+------------------------+ | group | Obs | Rank Sum | |-------+-----+----------| | 1 | 5 | 22.00 | | 2 | 5 | 61.50 | | 3 | 5 | 36.50 | +------------------------+
chi-squared = 7.985 with 2 d.f.probability = 0.0185
chi-squared with ties = 8.190 with 2 d.f.probability = 0.0167
Comparing 2 means: Regressiona b1 52 92 95 85 9
1. Data x y (x-x) (x-x)2 (y-y) (x-x)(y-y)1 1 -0.5 0.25 -4.5 2.251 2 -0.5 0.25 -3.5 1.751 2 -0.5 0.25 -3.5 1.751 5 -0.5 0.25 -0.5 0.251 5 -0.5 0.25 -0.5 0.252 5 0.5 0.25 -0.5 -0.302 9 0.5 0.25 3.5 1.752 9 0.5 0.25 3.5 1.752 8 0.5 0.25 2.5 1.252 9 0.5 0.25 3.5 1.75
Mean 1.5 5.5Sum 2.5 12.5
y = a + bxwhere b = 12.5/2.5 = 5, then5.5 = a + 5(1.5) Thus a = 5.5-7.5 = -2
Comparing 2 means: Regression (cont.)Y
x
10
0
2
4
6
8
-2
1 2
y = a + bx
b difference of y between x=1 vs. x=2
a
y = 3 if x = 1
y = 8 if x = 2
y = -2 if x = 0
y = -2 + 5x
y = 5.5; x = 1.5
Regression model (2 means)
Source | SS df MS Number of obs = 10-------------+------------------------------ F( 1, 8) = 19.23 Model | 62.5 1 62.5 Prob > F = 0.0023 Residual | 26 8 3.25 R-squared = 0.7062-------------+------------------------------ Adj R-squared = 0.6695 Total | 88.5 9 9.83333333 Root MSE = 1.8028
------------------------------------------------------------------------------ y | Coef. Std. Err. t P>|t| [95% Conf. Interval]-------------+---------------------------------------------------------------- group | 5 1.140175 4.39 0.002 2.370751 7.629249 _cons | -2 1.802776 -1.11 0.299 -6.157208 2.157208------------------------------------------------------------------------------
i.group _Igroup_1-3 (naturally coded; _Igroup_1 omitted)
Source | SS df MS Number of obs = 15-------------+------------------------------ F( 2, 12) = 9.71 Model | 62.8 2 31.4 Prob > F = 0.0031 Residual | 38.8 12 3.23333333 R-squared = 0.6181-------------+------------------------------ Adj R-squared = 0.5545 Total | 101.6 14 7.25714286 Root MSE = 1.7981
------------------------------------------------------------------------------ y | Coef. Std. Err. t P>|t| [95% Conf. Interval]-------------+---------------------------------------------------------------- _Igroup_2 | 5 1.137248 4.40 0.001 2.522149 7.477851 _Igroup_3 | 2.2 1.137248 1.93 0.077 -.2778508 4.677851 _cons | 3 .8041559 3.73 0.003 1.247895 4.752105------------------------------------------------------------------------------
Regression model (3 means)
Correlation coefficient
• Pearson product moment correlation – Denoted by r (for the sample) or (for the
population)– Require bivariate normal distribution assumption– Require linear relationship
• Spearman rank correlation– For small sample, not require bivariate normal
distribution assumption
Calculation for correlation coefficient(r)
[1]x
[2]y
[3](x-x)/SD
[4](y-y)/SD [3] x [4]
1 5 -1.07 -1.73 1.852 9 -0.53 0.58 -0.312 9 -0.53 0.58 -0.315 8 1.07 0.00 0.005 9 1.07 0.58 0.62
Sum 1.85Mean 3 8SD 1.87 1.73
Interpretation of correlation coefficient
Correlation Negative Positive
None −0.09 to 0.00 0.00 to 0.09
Small −0.30 to −0.10 0.10 to 0.30
Medium −0.50 to −0.30 0.30 to 0.50
Strong −1.00 to −0.50 0.50 to 1.00
These serve as a guide, not a strict rule. In fact, the interpretation of a correlation coefficient depends on the context and purposes.
From Wikipedia, the free encyclopedia
The correlation coefficient reflects the non-linearity and direction of a linear relationship (top row), but not the slope of that relationship (middle), nor many aspects of nonlinear relationships (bottom). The figure in the center has a slope of 0 but in that case the correlation coefficient is undefined because the variance of Y is zero.
This is a file from the Wikimedia Commons.
Inference on correlation coefficient
Stata commands:.di tanh(-0.885)-.70891534.di tanh(1.887).95511058
Stata command
• ci2 x y, corr spearman
Confidence interval for Spearman's rank correlation of x and y, based on Fisher's transformation.
Correlation = 0.354 on 5 observations (95% CI: -0.768 to 0.942)
Warning: This method may not give valid results with small samples (n<= 10) for rank correlations.
Inference on correlation coefficientusing SPSS
Please do the data analysis using SPSS and paste the results here.
One proportion: Z-test
y101
. . .0
1. Hypothesis H0: 1 = 0Ha: 1 0
2. Data
3. Calculating for z-statistic
ny = 50, py = 0.1
4. Obtain p-value based on Z-distribution
5. Make a decision
P-value = 0.018 (http://vassarstats.net/tabs.html)
Reject the null hypothesis at a level of significant of 0.05 Proportion of Y is statistically significantly different from zero.
Stata command to get the p-vale. di (1-normal(2.357))*2.01842325
Comparing 2 proportions: Z-test
x y1 11 00 1
. . . . . .1 0
1. Hypothesis H0: 1 = 0
Ha: 1 0
2. Data
3. Calculating for z-statistic
n0 = 50, p0 = 0.1
n1 = 50, p1 = 0.4
4. Obtain p-value based on t-distribution
5. Make a decision
P-value = 0.0005 (http://vassarstats.net/tabs.html)
Reject the null hypothesis at level of significant of 0.05 Proportion of Y between group of x is statistically significantly different from each other.
xy
0 1 Total
0 45 5 50
1 30 20 50
Total 75 25 100
Z-test for two proportionsTwo-sample test of proportions 0: Number of obs = 50 1: Number of obs = 50------------------------------------------------------------------------------ Variable | Mean Std. Err. z P>|z| [95% Conf. Interval]-------------+---------------------------------------------------------------- 0 | .1 .0424264 .0168458 .1831542 1 | .4 .069282 .2642097 .5357903-------------+---------------------------------------------------------------- diff | -.3 .0812404 -.4592282 -.1407718 | under Ho: .0866025 -3.46 0.001------------------------------------------------------------------------------ diff = prop(0) - prop(1) z = -3.4641 Ho: diff = 0
Ha: diff < 0 Ha: diff != 0 Ha: diff > 0 Pr(Z < z) = 0.0003 Pr(|Z| < |z|) = 0.0005 Pr(Z > z) = 0.9997
Comparing 2 proportions: Chi-square-test1. Hypothesis H0: ij = i+ +j where I = 0, 1; j = 0, 1
Ha: ร่� i+ +j
2. Data
3. Calculating for 2-statistic
4. Obtain p-value based on t-distribution
5. Make a decision
P-value = 0.001 (http://vassarstats.net/tabs.html)
Reject the null hypothesis at level of significant of 0.05 There is statistically significantly association between x and y.
xy
0 1 Total
0 45 5 50
1 30 20 50
Total 75 25 100 O E (O-E) (O-E)2 (O-E)2/E
45 (75/100)50 = 37.50 7.50 56.25 1.50
5 (25/100)50 =12.50 -7.50 56.25 4.50
30 (75/100)50 =37.50 -7.50 56.25 1.50
20 (25/100)50 =12.50 7.50 56.25 4.50
Chi-square (df = 1) 12.00
Comparing 2 proportions: Chi-square-test
| y x | 0 1 | Total-----------+----------------------+---------- 0 | 45 5 | 50 1 | 30 20 | 50 -----------+----------------------+---------- Total | 75 25 | 100
Pearson chi2(1) = 12.0000 Pr = 0.001
csi 20 5 30 45, or exact
| Exposed Unexposed | Total-----------------+------------------------+------------ Cases | 20 5 | 25 Noncases | 30 45 | 75-----------------+------------------------+------------ Total | 50 50 | 100 | | Risk | .4 .1 | .25 | | | Point estimate | [95% Conf. Interval] |------------------------+------------------------ Risk difference | .3 | .1407718 .4592282 Risk ratio | 4 | 1.62926 9.820408 Attr. frac. ex. | .75 | .3862245 .8981712 Attr. frac. pop | .6 | Odds ratio | 6 | 2.086602 17.09265 (Cornfield) +------------------------------------------------- 1-sided Fisher's exact P = 0.0005 2-sided Fisher's exact P = 0.0010
Binomial regression. binreg y x, rr
Generalized linear models No. of obs = 100Optimization : MQL Fisher scoring Residual df = 98 (IRLS EIM) Scale parameter = 1Deviance = 99.80946404 (1/df) Deviance = 1.018464Pearson = 99.99966753 (1/df) Pearson = 1.020405
Variance function: V(u) = u*(1-u) [Bernoulli]Link function : g(u) = ln(u) [Log]
BIC = -351.4972
------------------------------------------------------------------------------ | EIM y | Risk Ratio Std. Err. z P>|z| [95% Conf. Interval]-------------+---------------------------------------------------------------- x | 4 1.833024 3.03 0.002 1.629265 9.820377 _cons | .1 .0424262 -5.43 0.000 .0435379 .2296851------------------------------------------------------------------------------
Logistic regression
. logistic y x
Logistic regression Number of obs = 100 LR chi2(1) = 12.66 Prob > chi2 = 0.0004Log likelihood = -49.904732 Pseudo R2 = 0.1125
------------------------------------------------------------------------------ y | Odds Ratio Std. Err. z P>|z| [95% Conf. Interval]-------------+---------------------------------------------------------------- x | 6 3.316625 3.24 0.001 2.030635 17.72844 _cons | .1111111 .0523783 -4.66 0.000 .044106 .2799096------------------------------------------------------------------------------