Question . 3. One kind of plant has only blue flowers and white flowers. According to a genetic model, the offsprings of a certain cross have a 75% chance to be blue other. Two hundred seeds of such a cross are raised, and 142 turn out to be flowering, and a 25% chance to be white-flowering, independently of one an- blue-flowering. Are the data consistent 'with the model? Answer yes or no, and explain briefly. 4. One large course has 900 students, broken down into section meetings with 30 students each. The section meetings are led by teaching assistants. On the final, the class average is 63, and the SD is 20. However, in one section the average is only 55. The TA argues this way: If you took 30 students at random from the class, there is a pretty good chance they would average below 55 on the final. That's what happened to me-chance variation. Is this a good defense? Answer yes or no, and explain briefly. A newspaper article

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Transcribed Image Text: Transcribed image text: . 3. One kind of plant has only blue flowers and white flowers. According to a genetic model, the offsprings of a certain cross have a 75% chance to be blue other. Two hundred seeds of such a cross are raised, and 142 turn out to be flowering, and a 25% chance to be white-flowering, independently of one an- blue-flowering. Are the data consistent 'with the model? Answer yes or no, and explain briefly. 4. One large course has 900 students, broken down into section meetings with 30 students each. The section meetings are led by teaching assistants. On the final, the class average is 63, and the SD is 20. However, in one section the average is only 55. The TA argues this way: If you took 30 students at random from the class, there is a pretty good chance they would average below 55 on the final. That's what happened to me-chance variation. Is this a good defense? Answer yes or no, and explain briefly. A newspaper article

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Solution:-P is the probability af having blue folower 75%=0.75 Sample size =n=200number at blue flower from the sampling Plant =x=142Noll and alfernatéve hippothesis:{:[H_(0):P=0.75],[H_(a):P!=0.75],[" significance lavel "=alpha=0.05]:}This is a two failed teetteet eratiatics:z=(( hat(p))-p)/(sqrt(p(1-p)[n))=(0.71-0.75)/(sqrt(0.75(1-0.75)//200))=-1.31Pvalue: Varue at z to the left of -1.31 from the z table =0.08951since the pualue 0.095i>0.05, so fail to reject the noll hypotuelis at 0.05 isnificance level:yes, The siven sample dat a is consistent witu tue Model.conclucior. Since the P value 0.09.51 > 0.05, So There is n o evidence to concinde tuat tue siven sumple provide difberent resut than having a. 0.75 propertion ab blue flowering.(4)aू: P:1Givan'.{:[" A verase sumple "=55],[" SD box "=20],[" Number of draws "=30]:}rarr we will make 30 d raws from a box containing a ficicet per studunt, while The ticket-contains tue final score ab the standut.rarr The noll hypotheris trates That The averase al the box is equal the Claimed a verage at 63 , while tue alter native hypotueris states that the average of the bor is lers ... See the full answer