Microsoft Word - stu_handout_Chi square for dummies.doc
Chi-Squared For DummiesThe test: X2 "Chi-square test" is often shorthand for Pearson's chi-square test.The function: To decide whether the difference between observed and expected values is actually significant.The example: You’ve been told that in 1995, the average successful first kiss rate was 15%. With some digging, you found that last year, 27 of 476 young men successfully landed their first kiss in your hometown. Is this consistent with the average?To solve this problem:You’ll want to establish all of the categories. In this problem, we are comparing kissers to missers. This is a success-failure scenario, so those are the ONLY two categories.Establish your null and alternative hypotheses.Null Hypothesis = The average 15% success rate holds true because … (insert hypothesized reason here).Alternative Hypothesis = Due to (insert hypothesized reason here), the success rate has changed from the 15% average.REMEMBER: These are hypotheses. You don’t have to get them exactly right every time. That’s what experimentation is for. Essentially, the null is just supposed to say “there was no change” and the alternative hypothesis is supposed to say “there was a change.” Don’t forget what the function of a chi square test is.Since you now know what is to be compared, you’ll want to make a table of expected versus observed values. I’ve taken the liberty of completing the table for this problem.Successful Kisses Failed KissesObserved27449Expected67.35(or 15% of 476)449(observed - expected)- 40.350(o-e)21628.12250[(o-e2]/e24.1740REMEMBER: You MUST MUST MUST use actual data (numbers) rather than percentages or ratios. AND don’t use the data if any of the points are less than 5.Use the data you found by completing the table along with your knowledge of the X2 equation to calculate the X2 value. If this is the first time that you’ve encountered X2, I’ve included a copy of the equation below. The sigma (the funny looking E) means that you make the calculation for each category separately and then you ADD THEM ALL TOGETHER to get the final value.Use a probability chart like on the next page to find the P value. Anyway, in the context of this problem, you’ll find that X2 = 24.174 and P <0.0005. Using the standard P> 0.05, we can compare to find that our experimental P value of 0.0005. Since our value is waaaaaaaay smaller than 0.05, we can reject the null hypothesis and conclude that THE AVERAGE KISS SUCCESS RATE IN YOUR HOMETOWN IS LESS THAN IT WAS IN1995. Boom Shakka Lakka!ON-LINE CHI-SQUARED TABLEHow to Use the Chi-Squared DistributionThe first row of numbers indicates probability.For your degrees of freedom (df) read across that row until you find the next smallest number. Go to the top and find the probability.AN EXAMPLE:p value df 0.25 0.20 0.15 0.10 0.05 0.025 0.02 0.01 0.005 0.0025 0.001 0.0005 1 1.321.642.072.713.845.025.416.637.889.1410.8312.12 2 2.773.223.794.615.997.387.829.2110.6011.9813.8215.20 3 4.114.645.326.257.819.359.8411.3412.8414.3216.2717.73 4 5.395.596.747.789.4911.1411.6713.2314.8616.4218.4720.00 5 6.637.298.129.2411.0712.8313.3315.0916.7518.3920.5122.11 6 7.848.569.4510.6412.5314.4515.0316.8113.5520.2522.4624.10 7 9.045.8010.7512.0214.0716.0116.6218.4820.2822.0424.3226.02 8 10.2211.0312.0313.3615.5117.5318.1720.0921.9523.7726.1227.87 9 11.3912.2413.2914.6816.9219.0219.6321.6723.5925.4627.8329.67 10 12.5513.4414.5315.9918.3120.4821.1623.2125.1927.1129.5931.42 11 13.7014.6315.7717.2919.6821.9222.6224.7226.7628.7331.2633.14 12 14.8515.8116.9918.5521.0323.3424.0526.2228.3030.3232.9134.82 13 15.9315.5818.9019.8122.3624.7425.4727.6929.8231.8834.5336.48 14 17.1218.1519.421.0623.6826.1226.8729.1431.3233.4336.1238.11 15 18.2519.3120.6022.3125.0027.4928.2630.5832.8034.9537.7039.72 16 19.3720.4721.7923.5426.3028.8529.6332.0034.2736.4639.2541.31 17 20.4921.6122.9824.7727.5930.1931.0033.4135.7237.9540.7942.88 18 21.6022.7624.1625.9928.8731.5332.3534.8137.1639.4242.3144.43 19 22.7223.9025.3327.2030.1432.8533.6936.1938.5840.8843.8245.97 20 23.8325.0426.5028.4131.4134.1735.0237.5740.0042.3445.3147.50 21 24.9326.1727.6629.6239.6735.4836.3438.9341.4043.7846.8049.01 22 26.0427.3028.8230.8133.9236.7837.6640.2942.8045.2048.2750.51 23 27.1428.4329.9832.0135.1738.0838.9741.6444.1846.6249.7352.00 24 28.2429.5531.1333.2036.4239.3640.2742.9845.5648.0351.1853.48 25 29.3430.6832.2834.3837.6540.6541.5744.3146.9349.4452.6254.95 26 30.4331.7933.4335.5638.8941.9242.8645.6448.2950.8354.0556.41 27 31.5332.9134.5736.7440.1143.1944.1446.9649.6452.2255.4857.86 28 32.6234.0335.7137.9241.3444.4645.4248.2850.9953.5956.8959.30 29 33.7135.1436.8539.0942.5645.7246.6949.5952.3454.9758.3060.73 30 34.8036.2537.9940.2643.7746.9847.9650.8953.6756.3359.7062.16 40 45.6247.2749.2451.8155.7659.3460.4463.6966.7769.7073.4076.09 50 56.3353.1660.3563.1767.5071.4272.6176.1579.4982.6686.6689.56 60 66.9868.9771.3474.4079.0883.3084.5888.3891.9595.3499.61102.7 80 88.1390.4193.1196.58101.9106.6108.1112.3116.3120.1124.8128.3 100109.1111.7114.7118.5124.3129.6131.1135.8140.2144.3149.4153.2For example, if your df is 2 and your chi-square is 9.10, then your probability would be expressed as P<0.02; this would be significant. Therefore the number of “whatevers” you observed in each category is significantly different from what would be expected, based on the hypothesis you used to do the Chi- Squared test.Critical values ................
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