# AP Statistics Curriculum 2007 Infer 2Means Dep

(Difference between revisions)
 Revision as of 21:59, 9 February 2008 (view source)IvoDinov (Talk | contribs)← Older edit Revision as of 22:55, 9 February 2008 (view source)IvoDinov (Talk | contribs) Newer edit → Line 28: Line 28: 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 $d_i=X_i^1-X_i^2, \forall 1\leq i\leq n$. 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 $d_i=X_i^1-X_i^2, \forall 1\leq i\leq n$. - === Two-Sample Hypothesis Testing about a Mean=== + ===[[AP_Statistics_Curriculum_2007_Estim_S_Mean | 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 $(1-\alpha)100%$, where $\alpha$ is small (e.g., 0.1, 0.05, 0.025, 0.01, 0.001, etc.). Then a $(1-\alpha)100%$ confidence interval for $\mu_1 - \mu_2$ is defined in terms of the T-distribution: + : $CI(\alpha): \overline{x_1}-\overline{x_2} \pm t_{\alpha\over 2} SE(\overline {x_1}-\overline{x_2}) = \overline{d} \pm t_{\alpha\over 2} {1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(d_i-\overline{d})^2\over n-1}}$ + + * $t_{\alpha\over 2}$ is the [[AP_Statistics_Curriculum_2007_StudentsT | critical value for the T(df=sample-size -1) distribution at ${\alpha\over 2}$]]. + + ===[[EBook#Chapter_VIII:_Hypothesis_Testing | Hypothesis Testing]] about the difference of Means=== * Null Hypothesis: $H_o: \mu_1-\mu_2=\mu_o$ (e.g..., $\mu_1-\mu_2=0$) * Null Hypothesis: $H_o: \mu_1-\mu_2=\mu_o$ (e.g..., $\mu_1-\mu_2=0$) * Alternative Research Hypotheses: * Alternative Research Hypotheses: Line 39: Line 45: ===Example=== ===Example=== - Suppose we measure the thickness of plaque (mm) in the carotid artery of 10 randomly selected patients with [http://www.heartcheckamerica.com/cas_more.htm mild atherosclerotic disease].  Two measurements are taken, thickness before treatment with Vitamin E (baseline) and after two years of taking Vitamin E daily. + Suppose we measure the thickness of plaque (mm) in the carotid artery of 10 randomly selected patients with [http://www.heartcheckamerica.com/cas_more.htm 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 $\alpha=0.05$. *What makes this paired data rather than independent data? *What makes this paired data rather than independent data? Line 90: Line 96: We begin first by exploring the data visually using various [[AP_Statistics_Curriculum_2007_EDA_Plots | SOCR EDA Tools]]. We begin first by exploring the data visually using various [[AP_Statistics_Curriculum_2007_EDA_Plots | SOCR EDA Tools]]. - * [[SOCR_EduMaterials_Activities_IndexChart | Index plot]] of the differences + * [[SOCR_EduMaterials_Activities_LineChart | Line Chart]] of the two samples
[[Image:SOCR_EBook_Dinov_Infer_2Means_Dep_020908_Fig1.jpg|600px]]
[[Image:SOCR_EBook_Dinov_Infer_2Means_Dep_020908_Fig1.jpg|600px]]
- * [[SOCR_EduMaterials_Activities_IndexChart | Index plot]] of the differences + * [[SOCR_EduMaterials_Activities_BoxPlot| Box-And-Whisker Plot]] of the two samples
[[Image:SOCR_EBook_Dinov_Infer_2Means_Dep_020908_Fig2.jpg|600px]]
[[Image:SOCR_EBook_Dinov_Infer_2Means_Dep_020908_Fig2.jpg|600px]]
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[[Image:SOCR_EBook_Dinov_Infer_2Means_Dep_020908_Fig3.jpg|600px]]
[[Image:SOCR_EBook_Dinov_Infer_2Means_Dep_020908_Fig3.jpg|600px]]
- As the population variance is not given, we have to use the [[AP_Statistics_Curriculum_2007_StudentsT |T-statistics]]: $T_o = {\overline{x} - \mu_o \over SE(\overline{x})} \sim T(df=9)$ + ====Inference==== - : $T_o = {\overline{x} - \mu_o \over SE(\overline{x})} = {22.1 - 12 \over {{1\over \sqrt{10}} \sqrt{\sum_{i=1}^{10}{(x_i-22.1)^2\over 9}}})}=1.176$. + * Null Hypothesis: $H_o: \mu_{before}-\mu_{after}=0$ - : $p-value=P(T_{(df=9)}>T_o=1.176)=0.134882$ for this (one-sided) test. Therefore, we '''can not reject''' the null hypothesis at $\alpha=0.01$! 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 0.134882, which is larger than the initially set [[AP_Statistics_Curriculum_2007_Hypothesis_Basics | Type I]] error $\alpha = 0.01$ and we can not reject the null hypothesis. + * (One-sided) alternative Research Hypotheses: $H_1: \mu_{before} -\mu_{after}>0$. -
[[Image:SOCR_EBook_Dinov_Hypothesis_020508_Fig4.jpg|600px]]
+ + * Test statistics: We can use the sample summary statistics to compute the [[AP_Statistics_Curriculum_2007_Infer_2Means_Dep#Test_Statistics |T-statistic]]: $T_o = {\overline{d} - \mu_o \over SE(\overline{d})} \sim T(df=9)$ + : $T_o = {\overline{d} - \mu_o \over SE(\overline{d})} = {0.045 - 0 \over {{1\over \sqrt{10}} \sqrt{\sum_{i=1}^{10}{(d_i-0.045)^2\over 9}}})}= {0.045 \over 0.00833}=5.4022$. + + : $p-value=P(T_{(df=9)}>T_o=5.4022)=0.000216$ for this (one-sided) test. Therefore, we '''can reject''' the null hypothesis at $\alpha=0.05$! 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 [[AP_Statistics_Curriculum_2007_Hypothesis_Basics | Type I]] error $\alpha = 0.05$ and we reject the null hypothesis. +
[[Image:SOCR_EBook_Dinov_Infer_2Means_Dep_020908_Fig4.jpg|600px]]
+ + * You can also use the [http://socr.ucla.edu/htmls/SOCR_Analyses.html SOCR Analyses (One-Sample T-Test)] to carry out these calculations as shown in the figure below. +
[[Image:SOCR_EBook_Dinov_Infer_2Means_Dep_020908_Fig5.jpg|600px]]
+ + * This [[SOCR_EduMaterials_AnalysisActivities_OneT | SOCR One Smaple T-test Activity]] provides additional hands-on demonstrations of the one-sample hypothesis testing for the difference in paired experiments. + + * $95%=(1-0.05)100%$ ($\alpha=0.05$) Confidence interval (before-after): + : $CI(\mu_{before}-\mu_{after})$: \overline{d} \pm t_{\alpha\over 2} SE(\overline {d}) = 0.045 \pm 1.833 \times 0.00833 = [0.0297 ; 0.0603].[/itex] + + ====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 [[AP_Statistics_Curriculum_2007_NonParam_2MedianPair | 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 hense we may be justified in using the one-sample T-test in this case. - * You can see use the [http://socr.ucla.edu/htmls/SOCR_Analyses.html SOCR Analyses (One-Sample T-Test)] to carry out these calculations as shown in the figure below. + * [[AP_Statistics_Curriculum_2007_Normal_Prob#Assessing_Normality |Quantile-Quantile Data-Data plot]] of the two datasets: -
[[Image:SOCR_EBook_Dinov_Hypothesis_020508_Fig5.jpg|600px]]
+
[[Image:SOCR_EBook_Dinov_Infer_2Means_Dep_020908_Fig6.jpg|600px]]
- * This [[SOCR_EduMaterials_AnalysisActivities_OneT | SOCR One Smaple T-test Activity]] provides additional hands-on demonstrations of one-sample hypothesis testing. + * [[AP_Statistics_Curriculum_2007_Normal_Prob#Assessing_Normality | QQ-Normal plot]] of the before data: +
[[Image:SOCR_EBook_Dinov_Infer_2Means_Dep_020908_Fig7.jpg|600px]]
===Examples=== ===Examples===

## 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 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 { $X_1, X_2, X_3, \cdots , X_n$} 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 ${{1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}}$.

### Paired Analysis Protocol

To study paired data we would like to examine the differences between each pair. Suppose { $X_1^1, X_2^1, X_3^1, \cdots , X_n^1$} and { $X_1^2, X_2^2, X_3^2, \cdots , X_n^2$} represent to paired samples. The we want to study the difference sample { $d_1=X_1^1-X_1^2, d_2=X_2^1-X_2^2, d_3=X_3^1-X_3^2, \cdots , d_n=X_n^1-X_n^2$}. Notice the effect of the pairings of each $X_i^1$ and $X_i^2$.

Now we can clearly see that the group effect (group differences) are directly represented in the {di} sequence. The one-sample T test is the proper strategy to analyze the difference sample {di}, if the $X_i^1$ and $X_i^2$ 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 $\overline{d}$) 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 $d_i=X_i^1-X_i^2, \forall 1\leq i\leq n$.

### 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: $CI(\alpha): \overline{x_1}-\overline{x_2} \pm t_{\alpha\over 2} SE(\overline {x_1}-\overline{x_2}) = \overline{d} \pm t_{\alpha\over 2} {1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(d_i-\overline{d})^2\over n-1}}$

### 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 Ho1 − μ2 < μo
• Double sided: $H_1: \mu_1 - \mu_2 \not= \mu_o$

#### Test Statistics

• If the two populations that the { $X_i^1$} and { $X_i^2$} samples were drawn from are approximately Normal, then the Test statistics is: $T_o = {\overline{d} - \mu_o \over SE(\overline{d})} = {\overline{x} - \mu_o \over {{1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(d_i-\overline{d})^2\over n-1}}})} \sim N(0,1)$.

### 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 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.  • This SOCR One Smaple 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].[/itex]

#### 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 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 hense we may be justified in using the one-sample T-test in this case.  ### 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 mean-density 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 5.1 5.27 5.39 5.42 5.47 5.63 5.34 5.46 5.3 5.75 5.68 5.85