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Chi Square Practice Problems Name: _____________________________ Class: _____
Directions: Solve all problems using a chi square analysis. You must use statistics to support your answers.
- 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 Aggressive acts B, normal light 42 A, lower light 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 Aggressive acts
B, normal light 39
A, lower light 39
Total - 78
o e (o-e) (o-e)^2 (o-e)^2 /e Normal light 42 39 3 9 9/39 =. Lower light 36 39 -3 9 9/39 =.
X2 =.
Deg of freed – 1, critical value at .05 = 3.
.46 < 3.84, so accept the null – light intensity does not have an effect on
aggression.
- 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 Observed use Expected use #1 60 students 68.67 students #2 66 students 68.67 students #3 80 students 68.67 students
206 students 206/3 = 69 expected for each door
Door o e (o-e) (o-e)^2 (o-e)^2 /e #1 60 68.67 -8.67 75.1689 1. #2 66 68.67 -2.67 7.1289. #3 80 68.67 11.33 128.3689 1.
X2 = 3. Deg of freedom – 2, critical value at .05 = 5.
3.068 < 5.99, accept the null – no preferred door choice
- 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 Passed state test Failed state test New text 30x48/60 = 24 30x12/60 = 6 Old text 30x48/60 = 24 30x12/60 = 6
o e (o-e) (o-e)^2 (o-e)^2 /e
NT passed 26 24 2 4. NT failed 4 6 -2 4. OT passed 22 24 -2 4. OT failed 8 6 2 4.
X2 = 1.
Deg of freedom – 3, critical value at .05 = 7.
1.67< 7.82 accept the null – Success on the state test was not due to differences in the math textbook.
- 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 No relapse Relapse totals Placebo 370 698 1068 Antipsychotic 1488 639 2127 totals 1858 1337 3195
Expected results No relapse Relapse Placebo 621.01 447. Antipsychotic 1236.92 890.
o e (o-e) (o-e)^2 (o-e)^2 /e
Placebo no relapse
Placebo relapse 698 447 251 63,001 140.
Antipsych no relapse
Antipsych relapse 639 891 -252 63,504 71.
X2 = 364.
Deg of freedom – 3, critical value at .05 = 7.
364.59 > 7.82 Reject the null – The antipsychotic drug is significantly more effective in preventing relapse.
