Answers - Georgetown High School



More Chi Square ProblemsHere’s the chart you use: Degrees of FreedomProbability (p-value)0.900.500.250.100.050.0110.0160.461.322.713.846.642.0.211.392.774.615.999.2130.582.374.116.257.8211.3541.063.365.397.789.4913.2851.614.356.639.2411.0715.091. 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:?Spades404Hearts420Diamonds400Clubs376?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:?Spades404Hearts420Diamonds400Clubs356Jokers 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. 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??5. 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?6. A mutation causing short wings in fruit flies is thought to be autosomal recessive. A homozygous wild type female is crossed with a homozygous short winged male. All the offspring are long winged. The F2 generation is 102 longwinged females, 34 short winged females, 113 long winged males and 45 short winged males. Given that information, can we accept the null hypothesis?7. A mutation in rabbits causing crooked back legs is thought to be autosomal dominant. A homozygous wild type female is crossed with a homozygous crooked legged rabbit. All the offspring have crooked back legs. The F2 is 38 crooked legged females, 28 crooked legged males, 19 straight legged females and 14 straight legged males. Given that information can we accept the null hypothesis?8. A mutation in cats causing notched ears is believed to be autosomal recessive. An old lady has a male cat with notched ears and a female cat that is homozygous for normal ears. They have a 5 litters of kittens all with normal ears. If the next generation of cats numbers 1248 in the old lady’s house when the film crew from A&E’s Horders shows up, how many of them would you EXPECT to have notched ears?9. The film crew mentioned in question #8 captures all 1248 cats and 512 of them have notched ears. Can you accept the null hypothesis? Answers1. 1600 cards, 4 suits – ? chance of a suit being drawn. ? of 1600 =400.ObservedExpected(o-e)2/e404400.04420400140040003764001.44X2 = 2.48Degrees of freedom = 3Critical value = 7.82Because 2.48 is below 7.82, we can accept that the cards are being sorted randomly.2. 2 jokers for every 54 cards = .037 chance; .037 X 1622 = 61.5; so about 62 of the 1662 cards should be (are expected to be) jokers. The remaining 1600 should be evenly divided across the 4 suits = 400 expected of each suitObservedExpected(o-e)2/e404400.04420400140040003564004.8482626.45X2 = 12.33Degrees of freedom = 4Critical value = 9.49Because 12.33 is above 9.49, we must reject that the cards are being sorted randomly.3. The expected ratio is given in the problem (4 stripes, 3 spots, 9 mixed) To calculate expected take the total observed (176) and multiply it by the ratio expected.4/16 * 176 = expected # of stripes = 44 3/16 * 176 = expected # of spots = 33 9/16 * 176 = expected # stripes/spots = 99 ObservedExpected(o-e)2/e5044.8241331.9485991.98X2 = 4.74Degrees of freedom = 2Critical value = 5.99Because 4.74 is below 5.99, we can accept the geneticist’s hypothesis4. The null hypothesis is that we would expect no difference in number of acts of aggression when comparing low light to normal light. There were 78 total acts of aggression. So we would expect 39 acts to occur in each of the 2 rooms (78/2).ObservedExpected(o-e)2/e3639.234239.23X2 = .46Degrees of freedom = 1Critical value = 3.84Because .46 is below 3.84, we can accept the null hypothesis. (Which proves the zookeeper’s hypothesis as incorrect. 5. The null hypothesis is that we would expect no difference in which door is preferred by students. The doors were opened 206 times total. 206/3 = 68.67. ObservedExpected(o-e)2/e60691.1766690.1380691.75X2 = 3.05Degrees of freedom = 2Critical value = 5.99Because 3.05 is below 5.99, we can accept the null hypothesis. The students do not prefer any door. 6. There is an expected 3:1 ratio long-winged to short-winged. The total number of F2 flies is 294. ObservedExpected(o-e)2/e102+ 113 = 215294 X .75= 220.5.1434 + 45 = 79294 X .25= 73.5.41X2 = 0.55Degrees of freedom = 1Critical value = 3.84Because .55 is below 3.84, we can accept the null hypothesis 7. There is an expected 3:1 ratio crooked legs to straight legged. The total number of F2 rabbits is 99 ObservedExpected(o-e)2/e38 + 28=6699 X .75= 74.25.91619 + 14 = 3399 X .25 = 24.752.75X2 = 3.66Degrees of freedom = 1Critical value = 3.84Because 3.66 is below 3.84, we can accept the null hypothesis 8. There is an expected 3:1 ratio of normal to notched ears. The total number1248 cats would have an expected 312 with notched ears9. ObservedExpected(o-e)2/e73693642.7512312128.2X2 = 3.66Degrees of freedom = 1Critical value = 3.84Because 170.9 is above 3.84, we must reject the null hypothesis ................
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