4. A teacher did a poll of the students attending Happy Valley School as to whet

Question

4. A teacher did a poll of the students attending Happy Valley School as to whether the student believed in Santa Claus or not. The results of the poll are summarized in the table below Believe in Santa Claus YES 27 27 19 GRADE in School Kindergarten 1st Grade nd Grade 3rd Grade 4th Grade 5th Grade TOTAL NO TOTAL 15 31 30 78 186 Fill in the missing values in the table. (3 points) A student is randomly chosen from the sample; what is the probability that the student does not believe in Santa Claus? Is this a joint, marginal, or conditional probability? (3 points) If a second grader is chosen, what is the probability that he/she believes in Santa? Is this a joint, marginal, or conditional probability? (3 points) What is the probability that a randomly chosen student is in 3rd grade and does not believe in Santa? Is this a marginal, joint or conditional probability? write out the hypotheses for the Chi-Square (2) test of this data a) b) c) d) e) f Finish the table of expected counts. Expected Cell Counts Believe in Santa Claus GRADE in School Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade YES 16.8387 19.1613 NO 12.1613 11.7419 20.3226 14.6774 17.4194 g) Finish the table of partial 2 values. Partial Chi-Square Believe in Santa Claus GRADE in School Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade YES NO 6.1318 4.4401 0.6403 0.3077 0.4624 0.2222 2.6385 7.486010.3653 h) What is the value of the 2 statistic? what are its degrees of freedom? i) using = 0.05, state the conclusion of the Chi-Square (2) test in the context of this problem. State your

Explanation / Answer

(a)

(b) required probability=number(No)/Total=78/186=0.4194

this is marginal probability

(c) required probability=19/186=0.6786

this is conditional probability

(d)required probability=15/31=0.4839

this is conditional probability

(e) grade and believe are indpendent

(f)

(g)

(h) chi-square=48.0446 and df=(r-1)*(c-1)=(6-1)*(2-1)=5

(i) critical chi-square(0.05,5)=11.0705 is less than calculated chi-square=48.0446, so we fail to accept null hypothesis H0 and conclude that belive and grade are not independent.

Belive in santa claus grade in school Yes No total Kindergarten 27 2 29 1st 27 6 33 2nd 19 9 28 3rd 16 15 31 4th 13 22 35 5th 6 24 30 Total 108 78 186

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