AP Statistics Curriculum 2007 Infer 2Means Dep
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(→Paired vs. Independent Testing) 
(→Paired vs. Independent Testing) 

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====Paired vs. Independent Testing====  ====Paired vs. Independent Testing====  
Suppose we accidentally analyzed the groups independently (using the [[AP_Statistics_Curriculum_2007_Infer_2Means_Indep independent Ttest]]) rather than using this paired test (this would be an incorrect way of analyzing this ''beforeafter'' data). How would this change our results and findings?  Suppose we accidentally analyzed the groups independently (using the [[AP_Statistics_Curriculum_2007_Infer_2Means_Indep independent Ttest]]) rather than using this paired test (this would be an incorrect way of analyzing this ''beforeafter'' data). How would this change our results and findings?  
  :  +  : \(T_o = {\overline{x}\overline{y}  \mu_o \over SE(\overline{x}+\overline{y})} \sim T(df=17)\) 
  :  +  : \(T_o = {\overline{x}\overline{y}  \mu_o \over SE(\overline{x}+\overline{y})} = {0.682 0.637 0 \over \sqrt{SE^2(\overline{x})+SE^2(\overline{y})}}= \) \({0.682 0.637\over \sqrt{{0.0742^2\over 10}+ {0.0709^2\over 10}}}={0.682 0.637\over 0.0325}=1.38\) 
  :  +  : \(pvalue=P(T>1.38)= 0.100449\) and we would have failed to reject the nullhypothesis ('''incorrect!''') 
Similarly, had we incorrectly used the [[AP_Statistics_Curriculum_2007_Infer_2Means_Indep independent design]] and constructed a corresponding Confidence interval, we would obtain an incorrect inference:  Similarly, had we incorrectly used the [[AP_Statistics_Curriculum_2007_Infer_2Means_Indep independent design]] and constructed a corresponding Confidence interval, we would obtain an incorrect inference:  
  :  +  : \(CI: {\overline{x}\overline{y}  \mu_o \pm t_{(df=17, \alpha/2)} \times SE(\overline{x}+\overline{y})} = \) \(0.045 \pm 1.740\times 0.0325 = [0.0116 ; 0.1016].\) 
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Revision as of 19:56, 17 June 2013
Contents

General AdvancePlacement (AP) Statistics Curriculum  Inferences about Two Means: Dependent Samples
In the previous chapter we saw how to do significance testing in the case of a single random sample. Now, we show how to do hypothesis testing comparing two samples and we begin with the simple case of paired samples.
Inferences About Two Means: Dependent Samples
In all study designs, it is always critical to clearly identify whether samples we compare come from dependent or independent populations. There is a general formulation for the significance testing when the samples are independent. The fact that there may be uncountable many different types of dependencies that prevents us from having a similar analysis protocol for all dependent sample cases. However, in one specific case  paired samples  we have a theory to generalize the significance testing analysis protocol. Two populations (or samples) are dependent because of pairing (or paired) if they are linked in some way, usually by a direct relationship. For example, measure the weight of subjects before and after a six month diet.
Paired Designs
These are the most common Paired Designs, in which the idea of pairing is that members of a pair are similar to each other with respect to extraneous variables.
 Randomized block experiments with two units per block
 Observational studies with individually matched controls (e.g., clinical trials of drug efficacy  patient pre vs. post treatment results are compared)
 Repeated (time or treatment affected) measurements on the same individual
 Blocking by time – formed implicitly when replicate measurements are made at different times.
Background
 Recall that for a random sample {} of the process, the population mean may be estimated by the sample average, .
 The standard error of is given by .
Analysis Protocol for Paired Designs
To study paired data, we would like to examine the differences between each pair. Suppose {} and {} represent the 2 paired samples. Then we want to study the difference sample {}. Notice the effect of the pairings of each X_{i} and Y_{i}.
Now we can clearly see that the group effect (group differences) is directly represented in the {d_{i}} sequence. The onesample T test is the proper strategy to analyze the difference sample {d_{i}}, if the X_{i} and Y_{i} samples come from Normal distributions.
Since we are focusing on the differences, we can use the same reasoning as we did in the single sample case to calculate the standard error (i.e., the standard deviation of the sampling distribution of ) of .
Thus, the standard error of is given by , where .
Confidence Interval of the Difference of Means
The interval estimation of the difference of two means (or Confidence intervals) is constructed as follows. Choose a confidence level (1 − α)100%, where α is small (e.g., 0.1, 0.05, 0.025, 0.01, 0.001, etc.). Then a (1 − α)100% confidence interval for μ_{1} − μ_{2} is defined in terms of the Tdistribution:
Both the confidence intervals and the hypothesis testing methods in the paired design require Normality of both samples. If these parametric assumptions are invalid we must use a notparametric (distribution free test), even if the latter is less powerful.
Hypothesis Testing about the Difference of Means
 Null Hypothesis: H_{o}:μ_{1} − μ_{2} = μ_{o} (e.g., μ_{1} − μ_{2} = 0)
 Alternative Research Hypotheses:
 One sided (unidirectional): H_{1}:μ_{1} − μ_{2} > μ_{o}, or H_{1}:μ_{1} − μ_{2} < μ_{o}
 Double sided:
Test Statistics
 If the two populations that the {X_{i}} and {Y_{i}} samples were drawn from are approximately Normal, then the Test Statistics is:
 .
Effects of Ignoring the Pairing
The SE estimate will be smaller for correctly paired data. If we look within each sample at the data, we notice variation from one subject to the next. This information gets incorporated into the SE for the independent ttest via s_{1} and s_{2}. The original reason we paired was to try to control for some of this intersubject variation, which is not of interest in the paired design. Notice that the intersubject variation has no influence on the SE for the paired test, because only the differences were used in the calculation. The price of pairing is smaller degrees of freedom of the Ttest. However, this can be compensated with a smaller SE if we had paired correctly.
Pairing is used to reduce bias and increase precision in our inference. By matching/blocking we can control variation due to extraneous variables.
For example, if two groups are matched on age, then a comparison between the groups is free of any bias due to a difference in age distribution.
Pairing is a strategy of design, not an analysis tool. Pairing needs to be carried out before the data are observed. It is not correct to use the observations to make pairs after the data has been collected.
Example
Suppose we measure the thickness of plaque (mm) in the carotid artery of 10 randomly selected patients with mild atherosclerotic disease. Two measurements are taken, thickness before treatment with Vitamin E (baseline) and after two years of taking Vitamin E daily. Formulate testable hypothesis and make inference about the effect of the treatment at α = 0.05.
 What makes this paired data rather than independent data?
 Why would we want to use pairing in this example?
Data in row format
Before  0.66,0.72,0.85,0.62,0.59,0.63,0.64,0.7,0.73,0.68 
After  0.6,0.65,0.79,0.63,0.54,0.55,0.62,0.67,0.68,0.64 
Data in column format
Subject  Before  After  Difference 

1  0.66  0.60  0.06 
2  0.72  0.65  0.07 
3  0.85  0.79  0.06 
4  0.62  0.63  0.01 
5  0.59  0.54  0.05 
6  0.63  0.55  0.08 
7  0.64  0.62  0.02 
8  0.70  0.67  0.03 
9  0.73  0.68  0.05 
10  0.68  0.64  0.04 
Mean  0.682  0.637  0.045 
SD  0.0742  0.0709  0.0264 
Exploratory Data Analysis
We begin first by exploring the data visually using various SOCR EDA Tools.
 Line Chart of the two samples
 BoxAndWhisker Plot of the two samples
 Index plot of the differences
Inference
 Null Hypothesis: H_{o}:μ_{before} − μ_{after} = 0
 (Onesided) Alternative Research Hypotheses: H_{1}:μ_{before} − μ_{after} > 0.
 Test statistics: We can use the sample summary statistics to compute the Tstatistic:
 .
 p − value = P(T_{(df = 9)} > T_{o} = 5.4022) = 0.000216 for this (onesided) test.
Therefore, we can reject the null hypothesis at α = 0.05! The left white area at the tails of the T(df=9) distribution depicts graphically the probability of interest, which represents the strength of the evidence (in the data) against the Null hypothesis. In this case, this area is 0.000216, which is much smaller than the initially set Type I error α = 0.05 and we reject the null hypothesis.
 You can also use the SOCR Analyses (OneSample TTest) to carry out these calculations as shown in the figure below.
 This SOCR One Sample Ttest Activity provides additional handson demonstrations of the onesample hypothesis testing for the difference in paired experiments.
 95% = (1 − 0.05)100% (α = 0.05) Confidence interval (beforeafter):
 CI(μ_{before} − μ_{after}):
Conclusion
These data show that the true mean thickness of plaque after two years of treatment with Vitamin E is statistically significantly different than before the treatment (p =0.000216). In other words, vitamin E appears to be an effective in changing carotid artery plaque after treatment. The practical effect does appear to be < 60 microns; however, this may be clinically sufficient and justify patient treatment.
Paired Test Validity
Both the confidence intervals and the hypothesis testing methods in the paired design require Normality of both samples. If these parametric assumptions are invalid, we must use a notparametric (distribution free test), even if the latter is less powerful.
The plots below indicate that Normal assumptions are not unreasonable for these data, and hence we may be justified in using the onesample Ttest in this case.
 QuantileQuantile DataData plot of the two datasets:
 QQNormal plot of the before data:
Paired vs. Independent Testing
Suppose we accidentally analyzed the groups independently (using the independent Ttest) rather than using this paired test (this would be an incorrect way of analyzing this beforeafter data). How would this change our results and findings?
 \(T_o = {\overline{x}\overline{y}  \mu_o \over SE(\overline{x}+\overline{y})} \sim T(df=17)\)
 \(T_o = {\overline{x}\overline{y}  \mu_o \over SE(\overline{x}+\overline{y})} = {0.682 0.637 0 \over \sqrt{SE^2(\overline{x})+SE^2(\overline{y})}}= \) \({0.682 0.637\over \sqrt{{0.0742^2\over 10}+ {0.0709^2\over 10}}}={0.682 0.637\over 0.0325}=1.38\)
 \(pvalue=P(T>1.38)= 0.100449\) and we would have failed to reject the nullhypothesis (incorrect!)
Similarly, had we incorrectly used the independent design and constructed a corresponding Confidence interval, we would obtain an incorrect inference:
 \(CI: {\overline{x}\overline{y}  \mu_o \pm t_{(df=17, \alpha/2)} \times SE(\overline{x}+\overline{y})} = \) \(0.045 \pm 1.740\times 0.0325 = [0.0116 ; 0.1016].\)
Problems
 SOCR Home page: http://www.socr.ucla.edu
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