Chi-square Practice Problems

CHI-SQUARE PRACTICE PROBLEMS (Answers are also on this page at the bottom) 1. A poker-dealing machine is supposed to deal cards at random, as if from an infinite deck. In a test, you counted 1600 cards, and observed the following: Spades Hearts Diamonds Clubs 404 420 400 376 Could it be that the suits are equally likely? Or are these discrepancies too much to be random? 2. Same as before, but this time jokers are included, and you counted 1662 cards, with these results: Spades Hearts Diamonds Clubs Jokers 404 420 400 356 82 a. How many jokers would you expect out of 1662 random cards? How many of each suit? b. Is it possible that the cards are really random? Or are the discrepancies too large? 3.A genetics engineer was attempting to cross a tiger and a cheetah. She predicted a phenotypic outcome of the traits she was observing to be in the following ratio 4 stripes only: 3 spots only: 9 both stripes and spots. When the cross was performed and she counted the individuals she found 50 with stripes only, 41 with spots only and 85 with both. According to the Chi-square test, did she get the predicted outcome? 4. In the garden pea, yellow cotyledon color is dominant to green, and inflated pod shape is dominant to the constricted form. Considering both of these traits jointly in self-fertilized dihybrids, the progeny appeared in the following numbers: 193 green, inflated 184 yellow constricted 556 yellow, inflated 61 green, constricted Do these genes assort independently? Support your answer using Chi-square analysis. ANSWERS 1. expected expected observed (percent) (counts) 404 0.25 400 420 0.25 400 400 0.25 400 376 0.25 400 z 0.200 1.000 0.000 -1.200 chi-square-> 2.480 critical value-> 7.815 Compute each z from its own row as (observed-expected)/sqrt(expected). Be sure to use the counts in this formula, not the percentages. The chi-square statistic is the sum of the squares of the z-values. The number of degrees of freedom is 3 (number of categories minus 1). The critical value is from a table you’ll have on the exam (using a = 0.05). 2. expected expected observed (percent) (counts) 404 0.2407 400.1 420 0.2407 400.1 400 0.2407 400.1 356 0.2407 400.1 82 0.0370 61.6 1662 1662 z 0.194 0.994 -0.006 -2.205 2.606 chi-square-> 12.680 critical value-> 9.488 This time, the chi-square statistic (12.68) is above the a=0.05 critical value, so you could reject the null hypothesis and declare that the cards are not random. The problem is clearly that there are too many jokers at the expense of clubs – you can see that from the z statistics. 3. A genetics engineer was attempting to cross a tiger and a cheetah. She predicted a phenotypic outcome of the traits she was observing to be in the following ratio 4 stripes only: 3 spots only: 9 both stripes and spots. When the cross was performed and she counted the individuals she found 50 with stripes only, 41 with spots only and 85 with both. According to the Chi-square test, did she get the predicted outcome? Chi-square = (O-E)2/E D.F. Value 1 3.841 2 5.991 3 7.815 Set up a table to keep track of the calculations: Expected ratio Observed # Expected # O-E (O-E)2 4 stripes 3 spots 9 stripes/spots 16 total 50 41 85 176 total 44 33 99 176 total 6 8 -14 0 total 36 64 196 4/16 * 176 = expected # of stripes = 44 3/16 * 176 = expected # of spots = 33 9/16 * 176 = expected # stripes/spots = 99 Degrees of Freedom = 3 - 1 = 2 (3 different characteristics - stripes, spots, or both) Since 4.74 is less than 5.991, I can accept the null hypothesis put forward by the engineer. (O-E)2/E 0.82 1.94 1.98 Sum = 4.74