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Question 2. Thank you. 2.A researcher takes a simple random sample of 900 studen

ID: 3231120 • Letter: Q

Question

Question 2. Thank you.

2.A researcher takes a simple random sample of 900 students from the 13,000 students at UMass. On the average, there are 0.52 cats per sample student, and the SD is 0.152. Say whether each of the following statements is true or false, and explain.

(a)The 0.52 is 0.152 or so off the average number of cats per student in the whole school.

(b)A 95%-confidence interval for the average number of cats per student in the sample is 0.216 to 0.824. (c) A 95%-confidence interval for the average number of cats per student in the whole school is 0.51 to 0.53.

(d)

(e)There is a 95% chance that the average number of cats per student at UMass falls within the 95% confidence interval.

(f)The 95%-confidence level is about right because the number of cats follows the normal curve.

(g)The 95%-confidence level is about right because, with 900 draws from the box, the probability histogram for the average number of cats in the sample follows the normal curve.

Explanation / Answer

Solution:-

(a)The 0.52 is 0.152 or so off the average number of cats per student in the whole school. False

(b)A 95%-confidence interval for the average number of cats per student in the sample is 0.216 to 0.824. True

(c) A 95%-confidence interval for the average number of cats per student in the whole school is 0.51 to 0.53. False.

(e)There is a 95% chance that the average number of cats per student at UMass falls within the 95% confidence interval. True

(f)The 95%-confidence level is about right because the number of cats follows the normal curve. False

(g)The 95%-confidence level is about right because, with 900 draws from the box, the probability histogram for the average number of cats in the sample follows the normal curve. False

95% confidenc einterval is for the average number of cats per student in the sample is C.I = (0.22, 0.812).

C.I = xbar + zalpha/2 × S.E

C.I = 0.52 + 1.96 × 0.152

C.I = 0.52 + 0.29792

C.I = (0.22, 0.812)

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