SOCR EduMaterials Activities General CI Experiment - Revision history

Dinov: /* Confidence intervals using bootstrapping */

2023-06-20T15:14:37Z

‎Confidence intervals using bootstrapping

Dinov: /* Confidence intervals using bootstrapping */

2023-06-20T15:12:32Z

‎Confidence intervals using bootstrapping

Dinov: /* Confidence intervals for the population parameters of a distribution based on the asymptotic properties of maximum likelihood estimates */

2023-06-20T15:03:05Z

‎Confidence intervals for the population parameters of a distribution based on the asymptotic properties of maximum likelihood estimates

Dinov: /* See also */

2017-07-22T14:11:20Z

‎See also

IvoDinov: /* References */

2011-06-01T00:53:04Z

‎References

Jenny: /* Motivational example */

2010-06-29T18:05:12Z

‎Motivational example

IvoDinov: /* Confidence intervals for the population mean \mu with known population variance \sigma^2 */ typo

2010-06-02T15:39:41Z

‎Confidence intervals for the population mean \mu with known population variance \sigma^2: typo

IvoDinov: added see also section

2009-11-24T06:02:31Z

added see also section

IvoDinov: entered bug fixes from Dave Zes

2009-07-31T22:03:12Z

entered bug fixes from Dave Zes

IvoDinov: /* Confidence intervals (CI) for the population mean \mu of normal population with known population variance \sigma^2 */

2009-07-31T21:52:33Z

‎Confidence intervals (CI) for the population mean \mu of normal population with known population variance \sigma^2

@@ Line 592: / Line 592: @@
 <blockquote>
 rm(list=ls())  # clean the environment
 set.seed(123)
 data0 = rgamma(383,5,3)   # specify distribution, or reference a specific dataset, here we use simulated data
 mean(data0) # Sample statistics (e.g., mean)
@@ Line 601: / Line 604: @@
 meanB <- function(data, indices) {  # function to obtain the sample statistics (e.g., mean) for each sub-sample
    d <- data[indices] # allows boot to select sample
-    return(mean(d))
+}

@@ Line 607: / Line 607: @@
 results <- boot(data=data0, statistic=meanB, R=10000)  # bootstrapping with 10000 replications
-results; plot(results)  # explore the results of htge multiple sample statistics calculations
+results; plot(results)  # explore the results of the multiple sample statistics calculations
 boot.ci(results, type=c("norm", "basic", "perc", "bca")) # compute 95\% confidence interval

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	<center>[[Image:SOCR_Activities_General_CI_Activity_070709_Fig22.png\|300px]][[Image:SOCR_Activities_General_CI_Activity_070709_Fig23.png\|300px]]</center>		<center>[[Image:SOCR_Activities_General_CI_Activity_070709_Fig22.png\|300px]][[Image:SOCR_Activities_General_CI_Activity_070709_Fig23.png\|300px]]</center>
		+
		+	==Confidence intervals using bootstrapping==
		+
		+	Above we showed that asymptotically (for extremely large samples) the [[AP_Statistics_Curriculum_2007_Limits_CLT \| CLT asymptotic properties]] allow us to estimate the confidence intervals for various statistics, e.g., mean, median, range, etc. for most ''nice distributions'', e.g., Weibull, Gamma, Lognormal, etc. However, in practice, we generally have only small samples, and the data may not be “perfectly distributed” according to a fixed known distribution. In such situations, we commonly compute the '''bootstrap confidence interval (BCI)''' estimates using the following protocol:
		+
		+	* Resample from the observed dataset and generate new samples with replacement \(B\) times.
		+	* For each of these data sub-samples calculate the sample statistics (e.g., mean, median, etc.).
		+	* Calculate an appropriate bootstrap confidence interval for each sample statistics using one of many different types of [[SOCR_Resampling_HTML5_Project \| bootstrap confidence interval (BCI) recipes]].
		+
		+	In practice, we calculate several BCI and examine any discrepancies between them. Let's look at this an R implementation example:
		+
		+	<blockquote>
		+	rm(list=ls()) # clean the environment
		+	set.seed(123)
		+	data0 = rgamma(383,5,3) # specify distribution, or reference a specific dataset, here we use simulated data
		+	mean(data0) # Sample statistics (e.g., mean)
		+
		+	hist(data0) # Histogram of the data
		+
		+	library(boot)
		+
		+	meanB <- function(data, indices) { # function to obtain the sample statistics (e.g., mean) for each sub-sample
		+	d <- data[indices] # allows boot to select sample
		+	return(mean(d))
		+	}
		+
		+	results <- boot(data=data0, statistic=meanB, R=10000) # bootstrapping with 10000 replications
		+
		+	results; plot(results) # explore the results of htge multiple sample statistics calculations
		+
		+	boot.ci(results, type=c("norm", "basic", "perc", "bca")) # compute 95\% confidence interval
		+	</blockquote>
		+
		+	See the [https://socr.umich.edu/HTML5/Resampling_Webapp/ SOCR Resampling Webapp] demonstrating the bootstrap resampling applications.

	==See also==		==See also==

@@ Line 582: / Line 582: @@
 ==See also==
 * The [[SOCR_EduMaterials_Activities_ConfidenceIntervals | Simple Confidence Interval Activity]].
 ==References==

@@ Line 592: / Line 592: @@
 * Ferguson, T., S.,  A Course in Large Sample Theory, Chapman & Hall (1996).
 * John Rice, Mathematical Statistics and Data Analysis, Third Edition, Duxbury Press (2006).
 {{translate|pageName=http://wiki.stat.ucla.edu/socr/index.php?title=SOCR_EduMaterials_Activities_General_CI_Experiment}}

@@ Line 15: / Line 15: @@
 ==Motivational example==
 [[Image:SOCR_Activities_General_CI_Activity_070709_Fig1.png|150px|thumbnail|right| [http://www.stat.ucla.edu/%7Edinov/courses_students.dir/04/Spring/Stat233.dir/HWs.dir/AD_NeuroPsychImagingData1.html Alzheimer's Disease Dataset]]]
-A 2005 study proposing a new computational brain atlas for Alzheimer’s disease (Mega et al., 2005) investigated the mean volumetric characteristics and the spectra of shapes and sizes of different cortical and subcortical brain regions for Alzheimer’s patients, individuals with minor cognitive impairment and asymptomatic subjects. This study estimated a number of centrality and variability parameters for these three populations. Based on these point- and interval-estimates, the study analyzed a number of digital scans to derive criteria for imaging-based classification of subjects based on the intensities of their 3D brain scans. Their results enabled a number of subsequent inference studies that quantified the effects of subject demographics (e.g., education level, familial history, APOE allele, etc.), stage of the disease and the efficacy of new drug treatments targeting Alzheimer’s disease. The Figure to the right illustrates the ''shape, center'' and ''distribution parameters'' for the 3D geometric structure of the right hippocampus in the Alzheimer’s disease brain atlas. New imaging data can then be coregistered and compared relative to the amount of anatomical variability encoded in this atlas. This enables automated, efficient and quantitative inference on large number of brain volumes. Examples of point and interval estimates computed in this atlas framework include the mean-intensity and mean shape location, and the standard deviation of intensities and the mean deviation of shape, respectively.
+A 2005 study proposing a new computational brain atlas for Alzheimer’s disease (Mega et al., 2005) investigated the mean volumetric characteristics and the spectra of shapes and sizes of different cortical and subcortical brain regions for Alzheimer’s patients, individuals with minor cognitive impairment and asymptomatic subjects. This study estimated a number of centrality and variability parameters for these three populations. Based on these point- and interval-estimates, the study analyzed a number of digital scans to derive criteria for imaging-based classification of subjects based on the intensities of their 3D brain scans. Their results enabled a number of subsequent inference studies that quantified the effects of subject demographics (e.g., education level, familial history, APOE allele, etc.), stage of the disease and the efficacy of new drug treatments targeting Alzheimer’s disease. The Figure to the right illustrates the ''shape, center'' and ''distribution parameters'' for the 3D geometric structure of the right hippocampus in the Alzheimer’s disease brain atlas. New imaging data can then be coregistered and compared relative to the amount of anatomical variability encoded in this atlas. This enables automated, efficient and quantitative inference on large number of brain volumes. Examples of point and interval estimates computed in this atlas framework include the mean-intensity and mean shape location, and the standard deviation of intensities and the mean deviation of shape.
 ==Activity==

@@ Line 107: / Line 107: @@
 == Confidence intervals for the population mean <math>\mu</math> with known population variance <math>\sigma^2</math>==
-From the [[AP_Statistics_Curriculum_2007_Limits_CLT |central limit theorem]] we know that when the sample size is large (usually <math>n \ge 30</math>) the distribution of the sample mean <math>\bar X</math> approximately follows <math>\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}</math>. Therefore, the confidence interval for the population mean <math>\mu</math> is approximately given by the expression we previously discussed:
+From the [[AP_Statistics_Curriculum_2007_Limits_CLT |central limit theorem]] we know that when the sample size is large (usually <math>n \ge 30</math>) the distribution of the sample mean <math>\bar X</math> approximately follows <math>\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})</math>. Therefore, the confidence interval for the population mean <math>\mu</math> is approximately given by the expression we previously discussed:
 : <math>P\left(\bar x -z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} \le \mu \le
 \bar x + z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} \right) \approx 1-\alpha.</math>

← Older revision		Revision as of 06:02, 24 November 2009
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	<center>[[Image:SOCR_Activities_General_CI_Activity_070709_Fig22.png\|300px]][[Image:SOCR_Activities_General_CI_Activity_070709_Fig23.png\|300px]]</center>		<center>[[Image:SOCR_Activities_General_CI_Activity_070709_Fig22.png\|300px]][[Image:SOCR_Activities_General_CI_Activity_070709_Fig23.png\|300px]]</center>
		+
		+	==See also==
		+	* The [[SOCR_EduMaterials_Activities_ConfidenceIntervals \| Simple Confidence Interval Activity]].

	==References==		==References==

@@ Line 99: / Line 99: @@
 * The observations selected from the distribution for the construction of each of the 50 intervals shown in blue on the top-left graph panel.
 * The confidence intervals shown as red line segments on the bottom-left panel.
-* The green dots represent instances of confidence intervals that do not include the estimated parameter (in this case population mean of 5).
+* The green dots represent instances of confidence intervals that do not include the parameter (in this case population mean of 5).
 * All the parameters and simulation results are summarized on the right panel of the applet.
@@ Line 357: / Line 357: @@
 <center>[[Image:SOCR_Activities_General_CI_Activity_070709_Fig2_t.png|300px]]</center>
-As before, the area <math>1-\alpha</math> is called ''confidence level''.  The values of <math>t_{\frac{\alpha}{2};n-1}</math> can be found from:
+As before, the area <math>1-\alpha</math> is called the ''confidence level''.  The values of <math>t_{\frac{\alpha}{2};n-1}</math> can be found from:
 * [http://socr.ucla.edu/htmls/dist/StudentT_Distribution.html SOCR Studnet's T-distribution], interactive applet, or
 * [http://socr.ucla.edu/Applets.dir/T-table.html SOCR T-table], below are some examples:
@@ Line 369: / Line 369: @@
 | 0.95 || 21 || 2.086
 |-
-| 0.98 || 31 || 2.457e
+| 0.98 || 31 || 2.457
 |-
 | 0.99 || 61 || 2.660
@@ Line 401: / Line 401: @@
 ==Confidence interval for the population proportion ''p''==
-Let <math>X_1, X_2, \cdots, X_n</math> be a random sample from the Bernoulli distribution with probability of success ''p''.  To construct a confidence interval for ''p'' the following result is used based on normal approximation:  <math>\frac{X-np}{\sqrt{np(1-p)}} \sim N(0,1).</math> Therefore,
+Let <math>Y_1, Y_2, \cdots, Y_n</math> be a random sample from the Bernoulli distribution with probability of success ''p''.  To construct a confidence interval for ''p'' the following result is used based on the normal approximation:  <math>\frac{X-np}{\sqrt{np(1-p)}} \sim N(0,1),</math> where <math>X=\sum_{i=1}^n{Y_i}</math> is the total number of successes in the ''n'' experiments.
 : <math>P\left(-z_{\frac{\alpha}{2}}  \le \frac{X-np}{\sqrt{np(1-p)}}  \le z_{\frac{\alpha}{2}} \right)=1-\alpha,</math>
 : where <math>-z_{\frac{\alpha}{2}}</math> and <math>z_{\frac{\alpha}{2}}</math> defined as above.
@@ Line 420: / Line 421: @@
 The basic problem we will address now is how to determine the sample size needed so that the resulting confidence interval will have a fixed margin of error ''E'' with confidence level <math>1-\alpha</math>.
-In the expression <math>\hat p \pm z_{\frac{\alpha}{2}} \sqrt{\frac{\hat p(1-\hat p)}{n}}</math>, the width of the confidence interval is given by the margin of error <math>\frac{\hat p(1-\hat p)}{n}</math>.  We can simply solve for ''n'':
+In the expression <math>\hat p \pm z_{\frac{\alpha}{2}} \sqrt{\frac{\hat p(1-\hat p)}{n}}</math>, the width of the confidence interval is given by the ''margin of error'' <math>z_{\frac{\alpha}{2}} \sqrt{\frac{\hat p(1-\hat p)}{n}}</math>.  We can simply solve for ''n'':
 : <math>E=z_{\frac{\alpha}{2}} \sqrt{\frac{\hat p(1-\hat p)}{n}} \Rightarrow n = \frac {z_{\frac{\alpha}{2}}^2\hat  p (1-\hat p)}{E^2}.</math>
@@ Line 455: / Line 456: @@
 : where, ''x'' is the number of successes among ''n'' trials, and <math>F_{a,b,c}</math> is the ''a'' quantile of the ''F'' distribution with numerator degrees of freedom ''b'' and denominator degrees of freedom ''c''.
-==Confidence interval for the population variance <math>\sigma^2</math> of normal distribution==
+==Confidence interval for the population variance <math>\sigma^2</math> of the normal distribution==
 Again, let <math>X_1, X_2, \cdots, X_n</math> random sample from <math>N(\mu, \sigma^2)</math>.  It is known that <math>\frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1}</math>.  Therefore,
@@ Line 507: / Line 508: @@
 We observe that regardless of the sample size the 95% CI(<math>\sigma^2</math>) coverage is poor. In these situations, (sampling from non-normal populations) an asymptotically distribution-free confidence interval for the variance can be obtained using the following large sample theory result:
-: <math>\sqrt{n}(s^2-\sigma^2)  \rightarrow N\left(0, \sqrt{\mu_4-\sigma^4}\right),</math>
+: <math>\sqrt{n}(s^2-\sigma^2)  \rightarrow N\left(0, \mu_4-\sigma^4\right),</math>
 : That is, <math>\frac{\sqrt{n}(s^2-\sigma^2)}{ \sqrt{\mu_4-\sigma^4}}  \rightarrow N(0,1),</math>
 : where, <math>\mu_4=E(X-\mu)^4</math> is the [[AP_Statistics_Curriculum_2007_Estim_MOM_MLE | fourth moment of the distribution]].  Of course, <math>\mu_4</math> is unknown and will be estimated by the fourth sample moment <math>\mu_4=\frac{1}{n}\sum_{i=1}^n(X_i-\bar X)^4</math>.  The confidence interval for the population variance is computed as follows:
@@ Line 513: / Line 514: @@
 s^2 + z_{\frac{\alpha}{2}} \frac{\sqrt{m_4-s^4}}{\sqrt{n}}.</math>
-Using the [http://socr.ucla.edu/htmls/exp/Confidence_Interval_Experiment_General.html SOCR CI Applet], with exponential distribution (<math>\lambda=2</math>), sample size 300, number of intervals 50, confidence level 0.95) we see that the coverage of ''this interval'' is approximately 95%.
+Using the [http://socr.ucla.edu/htmls/exp/Confidence_Interval_Experiment_General.html SOCR CI Applet], with exponential distribution (<math>\lambda=2</math>), sample size 300, number of intervals 50, and confidence level 0.95, we see that the coverage of ''this interval'' is approximately 95%.
 <center>[[Image:SOCR_Activities_General_CI_Activity_070709_Fig18.png|300px]][[Image:SOCR_Activities_General_CI_Activity_070709_Fig19.png|300px]]</center>
@@ Line 521: / Line 522: @@
 ==Confidence intervals for the population parameters of a distribution based on the asymptotic properties of maximum likelihood estimates==
-To construct confidence intervals for  parameter of a distribution the following method can be used based on the large sample theory of maximum likelihood estimates. As the sample size ''n'' increases it can be shown that the maximum likelihood estimate <math>\hat \theta</math> of a parameter <math>\theta</math> follows approximately normal distribution with mean <math>\theta</math> and variance equal to the lower bound of the [http://en.wikipedia.org/wiki/Cramér–Rao_bound Cramer-Rao inequality].
+To construct confidence intervals for a parameter of some distribution, the following method can be used based on the large sample theory of maximum likelihood estimates. As the sample size ''n'' increases it can be shown that the maximum likelihood estimate <math>\hat \theta</math> of a parameter <math>\theta</math> follows approximately the normal distribution with mean <math>\theta</math> and variance equal to the lower bound of the [http://en.wikipedia.org/wiki/Cramér–Rao_bound Cramer-Rao inequality].
-: <math>\hat \theta \sim N\left(\theta, \sqrt{\frac{1}{nI(\theta)}}\right)</math>
+: <math>\hat \theta \sim N\left(\theta, \frac{1}{nI(\theta)}\right)</math>
 : where <math>\sqrt{\frac{1}{nI(\theta)}}</math> is the lower bound of the Cramer-Rao inequality.
@@ Line 548: / Line 549: @@
 : <math>\frac{\partial^2{lnf(x)}}{\partial \lambda^2}=-\frac{x}{\lambda^2}.</math>
-Therefore, <math>\frac{1}{-nE\left(\frac{\partial^2 lnf(x)}{\partial \lambda^2}\right)}=\frac{1}{-nE(-\frac{X}{\lambda^2})}= \frac{\lambda^2}{n}=\frac{\lambda}{n}</math>. When ''n'' is large, <math>\hat \lambda</math> follows approximately <math>\hat \lambda \sim N\left(\lambda, \sqrt{\frac{\lambda}{n}}\right)</math>. Because <math>\lambda</math> is unknown, we replace it with its MLE estimate <math>\hat \lambda</math>:
+Therefore, <math>\frac{1}{-nE\left(\frac{\partial^2 lnf(x)}{\partial \lambda^2}\right)}=\frac{1}{-nE(-\frac{X}{\lambda^2})}= \frac{\lambda^2}{\lambda n}=\frac{\lambda}{n}</math>. When ''n'' is large, <math>\hat \lambda</math> follows approximately <math>\hat \lambda \sim N\left(\lambda, \sqrt{\frac{\lambda}{n}}\right)</math>. Because <math>\lambda</math> is unknown, we replace it with its MLE estimate <math>\hat \lambda</math>:
 : <math>\hat \lambda \sim N\left(\bar X, \sqrt{\frac{\bar X}{n}}\right).</math>

@@ Line 38: / Line 38: @@
 | 0.98 || 2.325
 |-
-| 0.99 || &2.575
+| 0.99 || 2.575
 |}
 </center>