SMHS HypothesisTesting - Revision history

Glenbrau: /* Motivation */

2015-04-29T13:54:04Z

‎Motivation

Glenbrau: /* Overview */

2015-04-27T17:49:58Z

‎Overview

Glenbrau: /* Scientific Methods for Health Sciences - Hypothesis Testing */

2015-04-27T17:49:38Z

‎Scientific Methods for Health Sciences - Hypothesis Testing

Dinov: /* Testing a claim about a proportion */

2014-12-04T19:42:39Z

‎Testing a claim about a proportion

Dinov: /* Testing a claim about a proportion */

2014-12-04T19:42:16Z

‎Testing a claim about a proportion

Dinov: /* Testing a claim about variance (or standard deviation) */

2014-11-19T16:14:04Z

‎Testing a claim about variance (or standard deviation)

Dinov: /* Testing a claim about variance (or standard deviation) */

2014-11-19T16:13:36Z

‎Testing a claim about variance (or standard deviation)

Dinov: /* Theory */

2014-11-10T22:08:32Z

‎Theory

Dinov: /* Testing a claim about variance (or standard deviation) */

2014-10-26T16:12:53Z

‎Testing a claim about variance (or standard deviation)

Dinov: /* Testing a claim about a mean with large sample size */

2014-09-22T15:55:17Z

‎Testing a claim about a mean with large sample size

@@ Line 12: / Line 12: @@
 ===Motivation===
-In statistical data analysis, we are often encountered with the problem of making statistical decisions about populations or processes based on experimental data. And hypothesis testing will be the direct answer to questions like how well the findings fit he possibility that the chance along might be responsible for the observed discrepancy between theoretical model and empirical observations or what is the likelihood of the observed summary statistics if the data did come from the distribution specified by the null hypothesis? And what if it follows the distribution stated in the alternative hypothesis? In fact, one use of hypothesis testing is to decide whether experimental results contain enough information to cast doubt on conventional wisdom.
+In statistical data analysis, we often encounter the problem of making statistical decisions about populations or processes based on experimental data. Hypothesis testing will be the direct answer to questions like:
-Consider an example of testing whether the new production purse from a factory contains radioactive material. The null hypothesis is that there is no radioactive material in the purse and that all measured counts are due to ambient radioactivity typical of the surrounding air and harmless objects in the purse. We can then calculate how likely it is that the null hypothesis produce 10 count per minute. If it is likely, for example if the null hypothesis predicts on average 9 counts per minute, we say the purse is compatible with the null hypothesis, on the other hand, if the null hypothesis predicts for example 3 count per minute, then the purse is not compatible with the null hypothesis and there must be other factors responsible to produce the increased radioactive counts.
 ===Theory===

@@ Line 5: / Line 5: @@
 Important parts included in Hypothesis testing:
-* Decision (significance or no significance);
+* Decision (significance or no significance)
-* Parameter of interest;
+* Parameter of interest
-* Variable of interest;
+* Variable of interest
-* Population under study;
+* Population under study
-* p-value.
+* p-value
 ===Motivation===

@@ Line 5: / Line 5: @@
 Important parts included in Hypothesis testing:
-* decision (significance or no significance);
+* Decision (significance or no significance);
-* parameter of interest;
+* Parameter of interest;
-* variable of interest;
+* Variable of interest;
-* population under study;
+* Population under study;
 * p-value.

@@ Line 109: / Line 109: @@
 where $z_\frac{\alpha}{2}$ is the [[AP_Statistics_Curriculum_2007_Normal_Critical | critical value of a standard normal distribution]] at $\alpha/2$.
-: The standard error of <math>\hat{p}</math> (<math>\tilde{p}</math>) also needs a slight modification
+: The standard error of <math>\hat{p}</math> ( and <math>\tilde{p}</math>) also needs a slight modification
 $$SE_{\hat{p}} =  \sqrt{\hat{p}(1-\hat{p})\over n} \longrightarrow SE_{\tilde{p}} =  \sqrt{\tilde{p}(1-\tilde{p})\over n+z_{\alpha \over 2}^2}.$$

@@ Line 109: / Line 109: @@
 where $z_\frac{\alpha}{2}$ is the [[AP_Statistics_Curriculum_2007_Normal_Critical | critical value of a standard normal distribution]] at $\alpha/2$.
-: The standard error of <math>\hat{p}</math> also needs a slight modification
+: The standard error of <math>\hat{p}</math> (<math>\tilde{p}</math>) also needs a slight modification
 $$SE_{\hat{p}} =  \sqrt{\hat{p}(1-\hat{p})\over n} \longrightarrow SE_{\tilde{p}} =  \sqrt{\tilde{p}(1-\tilde{p})\over n+z_{\alpha \over 2}^2}.$$

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 *Test statistics: $\chi_0^2 =\frac{(n-1) s^2}{\sigma_0^2} \sim \chi_{df=n-1}^2$
 *Example: we have a random sample of 30 objects drawn form a normal distribution with sample variance \(s^2=5.\) Test at \(α=0.05\) level of significance if this is consistent with \(H_0: \sigma^2=2\).
-:: $\chi_0^2=\frac{(n-1) s^2}{\sigma_0^2} =(29*5)/2=72.5$, $\chi_L^2=16.047$ and $\chi_R^2=45.722$, since we have \(\chi_0^2 > \chi_R^2\), we reject the null hypothesis at 5% level of significance. The image below illustrates this calculation using the [http://socr.umich.edu/html/dist/SOCR_Distributions.html SOCR $\chi_{df=29}^2$ calculator]. Notice the Left and right limits for the central 95% confidence interval ($(\chi_L^2=16.047 : \chi_R^2=45.722)$).
+:: $\chi_0^2=\frac{(n-1) s^2}{\sigma_0^2} =(29*5)/2=72.5$, $\chi_L^2=16.047$ and $\chi_R^2=45.722$, since we have \(\chi_0^2 > \chi_R^2\), we reject the null hypothesis at 5% level of significance. The image below illustrates this calculation using the [http://socr.umich.edu/html/dist/SOCR_Distributions.html SOCR $\chi_{df=29}^2$ calculator]. Notice the Left and right limits for the central 95% confidence interval, $(\chi_L^2=16.047 : \chi_R^2=45.722)$.
 <center>

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 *Hypothesis testing about a mean: large samples
-: $ H_{0}: \mu=\mu_{0}$ (e.g.,$\mu_{0}=0)$; one sided $H_{1}:\mu>\mu_{0}$  or $ \mu<\mu_{0} $ ;two sided $H_{1}:\mu≠\mu_{0} $.
+: $ H_{0}: \mu=\mu_{0}$ (e.g., $\mu_{0}=0)$; one sided $H_{1}:\mu>\mu_{0}$  or $ \mu<\mu_{0} $; two sided $H_{1}:\mu≠\mu_{0} $.
 :Test statistics:
 ::(1) with known variance: $Z_0=\frac{\bar x -\mu_{0}} {\frac{\sigma}{\sqrt n}}$ $ \thicksim N (0,1)$
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 *Example: consider we are testing if the population mean equal to 20 at $\alpha$=0.05 using a double sided alternative test. $H_{0}$:$\mu=20$ vs.$H_{1}:\mu≠20$. The sample data is given: 16, 9, 14, 11, 17, 12, 99, 18, 13, 12, 5, 9, 17, 6, 11, 17, 18, 20, 6, 14, 7, 11, 12, 5, 18, 6, 4, 13, 11, and 12. Population variance not given.
-: $T_{0}=\frac{\bar x -\mu_{0}} {SE(\bar x)}=\frac{\bar x -\mu_{0}} {{\frac{1} {\sqrt {n}}  \sqrt {\displaystyle \sum_{i=1}^{n} \frac{(x_{i}-\bar x)^{2}}{n-1}}}}$ $\frac{22.1-12}{{\frac{1} {\sqrt {10}} \sqrt {\displaystyle \sum_{i=1}^{10} \frac{(x_{i}-22.1)^{2}}{10-1}}}}$ $= 1.176$
+ a <- c( 16, 9, 14, 11, 17, 12, 99, 18, 13, 12, 5, 9, 17, 6, 11, 17, 18, 20, 6, 14, 7, 11, 12, 5, 18, 6, 4, 13, 11, 12)
 :: From the sample we have $\bar x=14.77, s=16.54$.

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 ====Testing a claim about variance (or standard deviation)====
 *Recall that the sample variance $s^2$ is an unbiased point estimate for the population variance $σ^2$, similarly for the standard deviation. The sample variance is roughly chi-square distributed: $\chi_0^2=\frac{(n-1) s^2}{\sigma_0^2} \sim \chi_{df=n-1}^2.$
-**Hypothesis testing about variance
+*Hypothesis testing about variance
 :: $H_0: σ^2=σ_0^2$ vs. $H_1:σ^2≠σ_0^2$. Given that the chi-square distribution is not symmetric, there are two critical values \(χ_L^2\)  and \(χ_R^2.\)
 *Test statistics: $\chi_0^2 =\frac{(n-1) s^2}{\sigma_0^2} \sim \chi_{df=n-1}^2$
-**Example: we have a random sample of 30 objects drawn form a normal distribution with sample variance \(s^2=5.\) Test at \(α=0.05\) level of significance if this is consistent with \(H_0: \sigma^2=2\).
+*Example: we have a random sample of 30 objects drawn form a normal distribution with sample variance \(s^2=5.\) Test at \(α=0.05\) level of significance if this is consistent with \(H_0: \sigma^2=2\).
 :: $\chi_0^2=\frac{(n-1) s^2}{\sigma_0^2} =(29*5)/2=72.5$, $\chi_L^2=16.047$ and $\chi_R^2=45.722$, since we have \(\chi_0^2 > \chi_R^2\), we reject the null hypothesis at 5% level of significance.
 ===Applications===

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 ::(1) with known variance: $Z_0=\frac{\bar x -\mu_{0}} {\frac{\sigma}{\sqrt n}}$ $ \thicksim N (0,1)$
-::(2) with unknown variance $T_{0}=\frac{\bar x -\mu_{0}} {SE(\bar x)}=\frac{\bar x -\mu_{0}} {{\frac{1} {\sqrt n}  \sqrt {\displaystyle \sum_{i=1}^{n} \frac{(x_{i}-\bar x)^{2}}{n-1}}}}$ $ \thicksim T_{df}=n-1$
+::(2) with unknown variance $T_{0}=\frac{\bar x -\mu_{0}} {SE(\bar x)}=\frac{\bar x -\mu_{0}} {{\frac{1} {\sqrt n}  \sqrt {\displaystyle \sum_{i=1}^{n} \frac{(x_{i}-\bar x)^{2}}{n-1}}}}$ $ \thicksim T_{df=n-1}$
 *Example: consider we are testing if the population mean equal to 20 at $\alpha$=0.05 using a double sided alternative test. $H_{0}$:$\mu=20$ vs.$H_{1}:\mu≠20$. The sample data is given: 16, 9, 14, 11, 17, 12, 99, 18, 13, 12, 5, 9, 17, 6, 11, 17, 18, 20, 6, 14, 7, 11, 12, 5, 18, 6, 4, 13, 11, and 12. Population variance not given.