AP Statistics Curriculum 2007 Infer 2Means Dep
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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 uncountably many differnt types of dependencies prevents us from having a similar analysis protocol for all dependent sample cases. However, in one sepecific case  paired samples  we have the theory to generalize the significance testing analysis proitocol. 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 where 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 .
Paired Analysis Protocol
To study paired data we would like to examine the differences between each pair. Suppose {} and {} represent to paired samples. The we want to study the difference sample {}. Notice the effect of the pairings of each and .
Now we can clearly see that the group effect (group differences) are directly represented in the {d_{i}} sequence. The onesample T test is the proper strategy to analyze the difference sample {d_{i}}, if the and samples come from Normal distributions.
Because 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:
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_{o}:μ_{1} − μ_{2} < μ_{o}
 Double sided:
Test Statistics
 If the two populations that the {} and {} samples were drawn from are approximately Normal, then the Test statistics is:
 .
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 depict graphically the probability of interest, which represents the strenght 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 Smaple 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}): \overline{d} \pm t_{\alpha\over 2} SE(\overline {d}) = 0.045 \pm 1.833 \times 0.00833 = [0.0297 ; 0.0603].</math>
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 a effective in changing carotid artery plaque after treatment. The practical effect does apear 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 parameteric 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 hense 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:
Examples
Cavendish Mean Density of the Earth
A number of famous early experiments of measuring physical constants have later been shown to be biased. In the 1700's Henry Cavendish measured the Mean density of the Earth. Formulate and test null and research hypotheses about these data regarding the now know exact meandensity value = 5.517. These sample statistics may be helpful
 n = 23, sample mean = 5.483, sample SD = 0.1904
5.36  5.29  5.58  5.65  5.57  5.53  5.62  5.29  5.44  5.34  5.79  5.10  5.27  5.39  5.42  5.47  5.63  5.34  5.46  5.30  5.75  5.68  5.85 
References
 SOCR Home page: http://www.socr.ucla.edu
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