# AP Statistics Curriculum 2007 Contingency Indep

(Difference between revisions)
 Revision as of 19:09, 14 June 2007 (view source)IvoDinov (Talk | contribs)← Older edit Revision as of 01:59, 4 March 2008 (view source)IvoDinov (Talk | contribs) Newer edit → Line 2: Line 2: === Contingency Tables: Independence and Homogeneity === === Contingency Tables: Independence and Homogeneity === - Example on how to attach images to Wiki documents in included below (this needs to be replaced by an appropriate figure for this section)! -
[[Image:AP_Statistics_Curriculum_2007_IntroVar_Dinov_061407_Fig1.png|500px]]
- ===Approach=== + The chi-square test may also be used to assess independence and association between variables. - Models & strategies for solving the problem, data understanding & inference. + - * TBD + ==Motivational example== + Suppose 200 randomly selected cancer patients were asked if their primary diagnosis was Brain cancer and if they owned a cell phone before their diagnosis.  The results are presented in the table below. - ===Model Validation=== + Suppose we want to analyze the association, if any, between ''brain cancer'' and ''cell phone use''. - Checking/affirming underlying assumptions. + The  2x2 table below lists two possible outcomes for each variable (each variable is dichotomous). We have the following population parameters: + : P(CP|BC) = true probability of owning a cell phone (CP) given that the patient had brain cancer (BC). This chance may be estimated by P(CP|BC) = 0.72. - * TBD + : P(CP|NBC) = true probability of owning a cell phone given that the patient had another cancer, which is estimated by P(CP|NBC) = 0.46 - ===Computational Resources: Internet-based SOCR Tools=== +
- * TBD + {| class="wikitable" style="text-align:center; width:25%" border="1" + |- + |  || || colspan=3| '''Brain cancer''' + |- + |  || || '''Yes''' || '''No''' || '''Total''' + |- + | rowspan=3| '''Cell Phone Use''' || '''Yes''' || 18 || 80 || 98 + |- + | '''No''' || 7 || 95 || 102 + |- + | '''Total''' || 25 || 175 || 200 + |} +
- ===Examples=== + Does it seem like there is an association between brain cancer and cell phone use? - Computer simulations and real observed data. + Of the brain cancer patients 18/25 = 0.72, owned a cell phone before their diagnosis. + ''P(CP|BC) = 0.72'',  estimated probability of owning a cell phone given that the patient has brain cancer. - * TBD + Of the other cancer patients, 80/175 = 0.46, owned a cell phone before their diagnosis. - + ''P(CP|NBC) = 0.46'', estimated probability of owning a cell phone given that the patient has another cancer. - ===Hands-on activities=== + - Step-by-step practice problems. + ==Calculations== + + Suppose there were ''N = 1064'' data measurements with ''Observed(Tall) = 787'' and ''Observed(Dwarf) = 277''. These are the O’s (observed values). To calculate the E’s (expected values), we will take the hypothesized proportions under $H_o$ and multiply them by the total sample size ''N''. Expected(Tall) = (0.75)(1064) = 798 and Expected(Dwarf) = (0.25)(1064) = 266. Quickly check to see if the expected total = N = 1064. + + * The hypotheses: + : $H_o$: there is no association between variable 1 and variable 2  (independence) + + : $H_a$: there is an association between variable 1 and variable 2 (dependence) + + * Test statistics: + The test statistic: + + :$\chi_o^2 = \sum_{all-categories}{(O-E)^2 \over E} \sim \chi_{(df=(\# rows – 1)(\# col – 1))}^2$ + + : Expected cell counts can be calculated by + :: $E = { (row-toral)(column-total)\over grand-total}$ + with ''df = (# rows – 1)(# col – 1)''. + + * P-values and critical values for the [http://socr.stat.ucla.edu/htmls/SOCR_Distributions.html Chi-Square distribution may be easily computed using SOCR Distributions]. + + * Results: + + + * [[SOCR_EduMaterials_AnalysisActivities_Chi_Goodness |SOCR Chi-square Calculations]]: + +
[[Image:SOCR_EBook_Dinov_ChiSquare_030108_Fig1.jpg|500px]]
+ + ==Examples== + + + ==Applications== - * TBD

- ===References=== + ==References== * TBD * TBD

## General Advance-Placement (AP) Statistics Curriculum - Contingency Tables: Independence and Homogeneity

### Contingency Tables: Independence and Homogeneity

The chi-square test may also be used to assess independence and association between variables.

## Motivational example

Suppose 200 randomly selected cancer patients were asked if their primary diagnosis was Brain cancer and if they owned a cell phone before their diagnosis. The results are presented in the table below.

Suppose we want to analyze the association, if any, between brain cancer and cell phone use. The 2x2 table below lists two possible outcomes for each variable (each variable is dichotomous). We have the following population parameters:

P(CP|BC) = true probability of owning a cell phone (CP) given that the patient had brain cancer (BC). This chance may be estimated by P(CP|BC) = 0.72.
P(CP|NBC) = true probability of owning a cell phone given that the patient had another cancer, which is estimated by P(CP|NBC) = 0.46
 Brain cancer Yes No Total Cell Phone Use Yes 18 80 98 No 7 95 102 Total 25 175 200

Does it seem like there is an association between brain cancer and cell phone use? Of the brain cancer patients 18/25 = 0.72, owned a cell phone before their diagnosis. P(CP|BC) = 0.72, estimated probability of owning a cell phone given that the patient has brain cancer.

Of the other cancer patients, 80/175 = 0.46, owned a cell phone before their diagnosis. P(CP|NBC) = 0.46, estimated probability of owning a cell phone given that the patient has another cancer.

## Calculations

Suppose there were N = 1064 data measurements with Observed(Tall) = 787 and Observed(Dwarf) = 277. These are the O’s (observed values). To calculate the E’s (expected values), we will take the hypothesized proportions under Ho and multiply them by the total sample size N. Expected(Tall) = (0.75)(1064) = 798 and Expected(Dwarf) = (0.25)(1064) = 266. Quickly check to see if the expected total = N = 1064.

• The hypotheses:
Ho: there is no association between variable 1 and variable 2 (independence)
Ha: there is an association between variable 1 and variable 2 (dependence)
• Test statistics:

The test statistic:

Failed to parse (lexing error): \chi_o^2 = \sum_{all-categories}{(O-E)^2 \over E} \sim \chi_{(df=(\# rows – 1)(\# col – 1))}^2

Expected cell counts can be calculated by
$E = { (row-toral)(column-total)\over grand-total}$

with df = (# rows – 1)(# col – 1).

• Results:

• TBD