Flat Tire and missed class: A classic tale involves four car-pooling students wh
ID: 2951456 • Letter: F
Question
Flat Tire and missed class: A classic tale involves four car-pooling students who missed a testand gave as an excuse a flat tire. On the makeup test, theinstructor asked the student to identify the particular tire thatwent flat.If they really didn't have a flat tire, would they beable to identify the same tire?The author asked 41 other studentsto identify the tire they would select. The results are listed inthe following table ( except for one student who selected thespare).Use a 0.05 significance level to test the author's claimthat the results fit a uniform distribution. What does theresult suggest about the ability of the four studentsto select the same tire when they really didn't have a flat? Tire : Leftfront Rightfront Leftrear Right rear Number : 11 15 8 6 selected Help please with this question. Show step-step solutionplease.... Flat Tire and missed class: A classic tale involves four car-pooling students who missed a testand gave as an excuse a flat tire. On the makeup test, theinstructor asked the student to identify the particular tire thatwent flat.If they really didn't have a flat tire, would they beable to identify the same tire?The author asked 41 other studentsto identify the tire they would select. The results are listed inthe following table ( except for one student who selected thespare).Use a 0.05 significance level to test the author's claimthat the results fit a uniform distribution. What does theresult suggest about the ability of the four studentsto select the same tire when they really didn't have a flat? Tire : Leftfront Rightfront Leftrear Right rear Number : 11 15 8 6 selected Help please with this question. Show step-step solutionplease....Explanation / Answer
Null Hypothesis; The results fit a uniform distribution Alternative Hypothesis: The results does not fit a uniformdistribution We have to apply chi-square analysis for this data Consider the given data in the below format Observedvalues(Oi) Expectedvalues(Ei) (Oi-Ei)^2 (Oi-Ei)^2/Ei Leftfront 11 10 1 0.1 Rightfront 15 10 25 2.5 Leftrear 8 10 4 0.4 Rightrear 6 10 16 1.6 Total 4.6 Degrees of freedom = 4-1 = 3 The p-value of chi-square at =0.05 level of significanceis 0.216 Since the p-value is greater than , therefore we accept thenull hypothesis and conclude that the results fit a uniformdistribution
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