Chi Square Practice Problems Name: _____________________________ Class: _____ Directions: Solve all problems using a chi square analysis. You must use statistics to support your answers. 1. A zookeeper hypothesizes that changing the intensity of the light in the primate exhibits will reduce the amount of aggression between the baboons. In exhibit A, with a lower light intensity, he observes 36 incidences of aggression over a one month period. In exhibit B, with normal lights, he observes 42 incidences of aggression. Should he support or reject his hypothesis? Null Hypoth: There is no statistically significant difference between the observed results and the expected results. (Changing light intensity does not reduce baboon aggression.) Light intensity B, normal light A, lower light Aggressive acts 42 36 Total - 78 You would expect the values to be equal if there was no light difference, so just divide by 2 to get expected results… Expected values B, normal light A, lower light Normal light Lower light Aggressive acts 39 39 Total - 78 o 42 36 e 39 39 (o-e) 3 -3 (o-e) 9 9 2 2 (o-e) /e 9/39 = .23 9/39 = .23 X2 = .46 Deg of freed – 1, critical value at .05 = 3.84 .46 < 3.84, so accept the null – light intensity does not have an effect on aggression. 2. At a high school, students can choose to enter one of three doors. Custodians noticed that door #3 was always getting broken and suggested that more students use that door because it has a hands-free opener. Science minded students counted the number of students entering each door to see if the custodians were right. Door #1 had 60 students enter | Door #2 had 66 students enter | Door #3 had 80 students enter. Were the custodians right? Null Hypoth: There is no statistically significant difference between the observed results and the expected results. (No door is chosen in preference to the others.) Door #1 #2 #3 Observed use 60 students 66 students 80 students 206 students Door #1 #2 #3 o 60 66 80 Expected use 68.67 students 68.67 students 68.67 students 206/3 = 69 expected for each door e 68.67 68.67 68.67 (o-e) -8.67 -2.67 11.33 2 (o-e) 75.1689 7.1289 128.3689 2 (o-e) /e 1.0946 .1038 1.8694 X2 = 3.068 Deg of freedom – 2, critical value at .05 = 5.99 3.068 < 5.99, accept the null – no preferred door choice 3. A scientist predicts that the kittens born with a congenital birth defect will be 25% based on the hypothesis that it is caused be a recessive gene in that breed of cat. After surveying several litters, he found that 44 out of 125 kittens had the defect. Is his hypothesis correct? Null Hypoth: There is no statistically significant difference between the observed results and the expected results. (The birth defect seen in these kittens is not due to anything other than Mendelian Genetics.) Type of kitten born Normal Congenital defect Observed 81 44 Expected 94 31 125 total kittens observed Normal kittens Congenital defect o 81 44 e 94 31 (o-e) -13 13 (o-e) 169 169 2 2 (o-e) /e 1.798 5.45 X2 = 7.25 Deg of freedom – 1, critical value at .05 = 3.84 7.25 > 3.84, reject the null – something other than Mendelian Genetics with recessive genes is affecting the birth of kittens with a congenital defect. 4. Suppose you take a random sample of 30 students who are using a new math text and a second sample of 30 students who are using a more traditional text. You compare student achievement on the state test given to all students at the end of the course. Based on state test performance, would you recommend the new math book? Observed results Passed State Test Failed State Test New Textbook 26 4 Old Textbook 22 8 Null Hypoth: There is no statistically significant difference between the observed results and the expected results. (Success on the state test was not due to differences in the math textbook.) Now you need to determine the expected values to get a comparison…this is done with a simple addition of what you have from the observed data…here’s a formula for how to do that! Expected results New text Old text NT passed NT failed OT passed OT failed Passed state test 30x48/60 = 24 30x48/60 = 24 o 26 4 22 8 Failed state test 30x12/60 = 6 30x12/60 = 6 e 24 6 24 6 (o-e) 2 -2 -2 2 (o-e) 4 4 4 4 2 2 (o-e) /e .167 .667 .167 .667 X2 = 1.67 Deg of freedom – 3, critical value at .05 = 7.82 1.67< 7.82 accept the null – Success on the state test was not due to differences in the math textbook. 5. In a study of the effectiveness of an antipsychotic drug, patients treated with the drug were compared to patients receiving a placebo. In terms of the number relapsing, 698 of 1,068 patients relapsed after taking the placebo while 639 out of 2,127 patients relapsed after taking the antipsychotic drug. Test the prediction that the antipsychotic is significantly more effective in preventing relapse than the placebo. Null Hypoth: There is no statistically significant difference between the observed results and the expected results. (The antipsychotic drug is no more effective than the placebo.) (use the tables for formulas given in #4)… Observed results Placebo Antipsychotic totals No relapse 370 1488 1858 Relapse 698 639 1337 Expected results Placebo Antipsychotic No relapse 621.01 1236.92 Relapse 447.26 890.74 Placebo no relapse Placebo relapse Antipsych no relapse Antipsych relapse totals 1068 2127 3195 2 2 o 370 e 621 (o-e) -251 (o-e) 63,001 (o-e) /e 101.45 698 1488 447 1237 251 251 63,001 63,001 140.94 50.93 639 891 -252 63,504 71.27 X2 = 364.59 Deg of freedom – 3, critical value at .05 = 7.82 364.59 > 7.82 Reject the null – The antipsychotic drug is significantly more effective in preventing relapse. 6. A student makes a monohybrid cross with Drosophila (fruit flies). She crosses two heterozygotes for the white eye. Ww x Ww. She expects to see a 3:1 phenotypic ratio of Red eyes (WW and Ww) to white eyes (ww) – this is her null hypothesis. She rears the next generation through to adult flies and counts the following numbers: White eyes 210 Wild type (red eyes) 680 Perform a chi square analysis on these results and find out if it is close enough to 3:1 to fail to reject her null hypothesis. Make sure to show all work and explain your conclusions. Null Hypoth: There is no statistically significant difference between the observed results and the expected results. (Eye color in fruit flies follows Mendelian Genetics and nothing else is affecting it.) Eye color White Red White Red Observed 210 680 o 210 680 Expected 223 667 e 223 667 (o-e) -13 13 (o-e) 169 169 2 2 (o-e) /e .76 .25 X2 = 1.01 Deg of freedom – 1, critical value at .05 = 3.84 1.01 < 3.84, Accept the null – eye color in fruit flies follows Mendelian Genetics.
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