# AP Statistics Curriculum 2007 Infer 2Means Dep

(Difference between revisions)
 Revision as of 19:56, 17 June 2013 (view source)IvoDinov (Talk | contribs) (→Paired vs. Independent Testing)← Older edit Revision as of 19:58, 17 June 2013 (view source)IvoDinov (Talk | contribs) (→Paired vs. Independent Testing)Newer edit → Line 152: Line 152: ====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 T-test]]) rather than using this paired test (this would be an incorrect way of analyzing this ''before-after'' 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 T-test]]) rather than using this paired test (this would be an incorrect way of analyzing this ''before-after'' data). How would this change our results and findings? - : $T_o = {\overline{x}-\overline{y} - \mu_o \over SE(\overline{x}+\overline{y})} \sim T(df=17)$ + : $$T_o = {\overline{x}-\overline{y} - \mu_o \over SE(\overline{x}+\overline{y})} \sim T(df=17)$$ - : $T_o = {\overline{x}-\overline{y} - \mu_o \over SE(\overline{x}+\overline{y})} = {0.682 -0.637- 0 \over \sqrt{SE^2(\overline{x})+SE^2(\overline{y})}}= {0.682 -0.637\over \sqrt{{0.0742^2\over 10}+ {0.0709^2\over 10}}}={0.682 -0.637\over 0.0325}=1.38$ + : $$T_o = {\overline{x}-\overline{y} - \mu_o \over SE(\overline{x}+\overline{y})} = {0.682 -0.637- 0 \over \sqrt{SE^2(\overline{x})+SE^2(\overline{y})}}=$$ $${0.682 -0.637\over \sqrt{{0.0742^2\over 10}+ {0.0709^2\over 10}}}={0.682 -0.637\over 0.0325}=1.38$$ - : $p-value=P(T>1.38)= 0.100449$ and we would have failed to reject the null-hypothesis ('''incorrect!''') + : $$p-value=P(T>1.38)= 0.100449$$ and we would have failed to reject the null-hypothesis ('''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: - : $CI: {\overline{x}-\overline{y} - \mu_o \pm t_{(df=17, \alpha/2)} \times SE(\overline{x}+\overline{y})} = 0.045 \pm 1.740\times 0.0325 = [-0.0116 ; 0.1016]$ + : $$CI: {\overline{x}-\overline{y} - \mu_o \pm t_{(df=17, \alpha/2)} \times SE(\overline{x}+\overline{y})} =$$ $$0.045 \pm 1.740\times 0.0325 = [-0.0116 ; 0.1016].$$

## General Advance-Placement (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 {$X_1, X_2, X_3, \cdots , X_n$} of the process, the population mean may be estimated by the sample average, $\overline{X_n}={1\over n}\sum_{i=1}^n{X_i}$.
• The standard error of $\overline{x}$ is given by ${{1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}}$.

### Analysis Protocol for Paired Designs

To study paired data, we would like to examine the differences between each pair. Suppose {$X_1, X_2, X_3, \cdots , X_n$} and {$Y_1, Y_2, Y_3, \cdots , Y_n$} represent the 2 paired samples. Then we want to study the difference sample {$d_1=X_1-Y_1, d_2=X_2-Y_2, d_3=X_3-Y_3, \cdots , d_n=X_n-Y_n$}. Notice the effect of the pairings of each Xi and Yi.

Now we can clearly see that the group effect (group differences) is directly represented in the {di} sequence. The one-sample T test is the proper strategy to analyze the difference sample {di}, if the Xi and Yi 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 $\overline{d}$) of $\overline{d}={1\over n}\sum_{i=1}^n{d_i}$.

Thus, the standard error of $\overline{d}$ is given by ${{1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(d_i-\overline{d})^2\over n-1}}}$, where $d_i=X_i-Y_i, \forall 1\leq i\leq n$.

### 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 μ1 − μ2 is defined in terms of the T-distribution:

$CI(\alpha): \overline{x}-\overline{y} \pm t_{\alpha\over 2} SE(\overline {x}-\overline{y}) = \overline{d} \pm t_{\alpha\over 2} {1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(d_i-\overline{d})^2\over n-1}}$

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 not-parametric (distribution free test), even if the latter is less powerful.

### Hypothesis Testing about the Difference of Means

• Null Hypothesis: Ho1 − μ2 = μo (e.g., μ1 − μ2 = 0)
• Alternative Research Hypotheses:
• One sided (uni-directional): H11 − μ2 > μo, or H11 − μ2 < μo
• Double sided: $H_1: \mu_1 - \mu_2 \not= \mu_o$

#### Test Statistics

• If the two populations that the {Xi} and {Yi} samples were drawn from are approximately Normal, then the Test Statistics is:
$T_o = {\overline{d} - \mu_o \over SE(\overline{d})} = {\overline{d} - \mu_o \over {{1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(d_i-\overline{d})^2\over n-1}}}} \sim T_{(df=n-1)}$.

#### 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 t-test via s1 and s2. The original reason we paired was to try to control for some of this inter-subject variation, which is not of interest in the paired design. Notice that the inter-subject 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 T-test. 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.

#### Inference

• Null Hypothesis: Hobefore − μafter = 0
• (One-sided) Alternative Research Hypotheses: H1before − μafter > 0.
• Test statistics: We can use the sample summary statistics to compute the T-statistic: $T_o = {\overline{d} - \mu_o \over SE(\overline{d})} \sim T(df=9)$
$T_o = {\overline{d} - \mu_o \over SE(\overline{d})} = {0.045 - 0 \over {{1\over \sqrt{10}} \sqrt{\sum_{i=1}^{10}{(d_i-0.045)^2\over 9}}})}= {0.045 \over 0.00833}=5.4022$.
pvalue = P(T(df = 9) > To = 5.4022) = 0.000216 for this (one-sided) 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.

• This SOCR One Sample T-test Activity provides additional hands-on demonstrations of the one-sample hypothesis testing for the difference in paired experiments.
• 95% = (1 − 0.05)100% (α = 0.05) Confidence interval (before-after):
CIbefore − μafter): $\overline{d} \pm t_{\alpha\over 2} SE(\overline {d}) = 0.045 \pm 1.833 \times 0.00833 = [0.0297 ; 0.0603].$

#### 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 not-parametric (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 one-sample T-test in this case.

#### Paired vs. Independent Testing

Suppose we accidentally analyzed the groups independently (using the independent T-test) rather than using this paired test (this would be an incorrect way of analyzing this before-after data). How would this change our results and findings?

$$T_o = {\overline{x}-\overline{y} - \mu_o \over SE(\overline{x}+\overline{y})} \sim T(df=17)$$
$$T_o = {\overline{x}-\overline{y} - \mu_o \over SE(\overline{x}+\overline{y})} = {0.682 -0.637- 0 \over \sqrt{SE^2(\overline{x})+SE^2(\overline{y})}}=$$ $${0.682 -0.637\over \sqrt{{0.0742^2\over 10}+ {0.0709^2\over 10}}}={0.682 -0.637\over 0.0325}=1.38$$
$$p-value=P(T>1.38)= 0.100449$$ and we would have failed to reject the null-hypothesis (incorrect!)

Similarly, had we incorrectly used the independent design and constructed a corresponding Confidence interval, we would obtain an incorrect inference:

$$CI: {\overline{x}-\overline{y} - \mu_o \pm t_{(df=17, \alpha/2)} \times SE(\overline{x}+\overline{y})} =$$ $$0.045 \pm 1.740\times 0.0325 = [-0.0116 ; 0.1016].$$