r - Matlab test of independence -

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For 1,000,000 comments, I had seen a discrete event, X, 3 times for the control group and 10 times for the test group.

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I have to prepare a Chihu Square Test of Independence. How would you do this in R: 

  m < - rbind (c (3, 1000000-3), c (10, 1000000-10)) # [, 1] [, 2] # [1,] 99 99 7 # [2,] 10 99 99 9 chisq .test (m)

R function returns CHI-squared = 2.7692, df = 1, p-value = 0.0961.

Should I use the Matlab function to do this or should I?

This is my own implementation that I use: 

  Function [hNull pValue X2] = Chasquestest (O, Alpha)% # Christmas Pearson's Chi-Square Examination Freedom #% # @ Ultimate O Two Similar Events (Attributes)% # # Importance Level% for the Ultimate Alpha Testing # @ Circulation Hunnel H = 1: Null hypothesis approved (independent)% # hNull = 0: null hypothesis rejected (dependent)% # @ Return Pivotal Examination P-value (% #) Problems of survival) frequencies observed under HN ull)% # @ Returns X2 Square Value for% #% #: observed frequency% #E: expected frequency% #DOOF: degree of independence [RC] = Size (O); Dof = (r-1) * (C-1); % # E = (calculation (a = ai) * calculation (b = bi)) / n = yoga (o, 2) * sum (o, 1) / zodiac (o (:)); % # [Sum_r [sum_c ((o_ij-e_ij) ^ 2 / e_ij)]] x2 = sum (zodiac (o-e). ^ 2 ./i)); % # P-value is required to reject hnull at the importance level with dof pValue = 1 - chi2cdf (x2, dof); HNull = (PLALA> alpha); The% # x2 value is required to reject hnull at the importance level with dof% # x2table = chi2inv (1-alpha, doof); % # HNull = (X2table & gt; X2);

For example and example: 

  t = [3 99 99 7 7; 10 99 99 9] [HNL PVL X2] = Chasquestest (T, 0.05) hNull = 1 pVal = 0.052203 X2 = 3.7693

Note that the results are different from you because Chisq.test ? According to Chisq.test

one corrects an improvement: A logical indicates that continuous improvement is applied when calculating test stats for 2x2 Or not tables: one half is subtracted from all. O - E | Difference. Alternatively, if you have actual comments of two incidents in question, you can use that function which calculates the contingency table and chi 2 and p. -Remains the following measures: 

  X = Randie ([1 2], [1000 1]); Y = Randie ([1 2], [1000 1]); [Tx2 pVal] = crosstab (x, y) t = 22 9 247 257 267 x 2 = 0.087581 pyil = 0.76728

R will be equal to: 

 < Note: Statistics toolbox is required for both the above (MATLAB) approach


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