Question 5 Believe it or not while writing this assignment I traveled forward in

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Question 5 Believe it or not while writing this assignment I traveled forward in time and collected a random sample of 20 submissions for this assignment The names on the assignments did not survive the return time-trip, but I could still calculate the average total score on the sampled assignments, which was 11, and their sample standard deviation, which was 2. On the basis of this information: a. Find a 95% confidence interval for the actual average score for the entire class. b. Construct a 95/95 tolerance interval for your score on this assignment Discuss any issues. c. While collecting the sample, I met my future self, who told me the actual mean and standard deviation of scores for all submitted assignments. Unfortunately, the time-travel sickness I experienced on my return trip to the present meant I was only able to remember the standard deviation, which was 2.25. Given this information, how would you change the interval you calculated in part (a), if at all?

Explanation / Answer

Let X1,X2,…..,Xn be a random sample of size n (=20) from the scores in assignment of the students of the class.

Assume that Xi ~ N(µ,2) i.i.d

Define Xbar = SUM(Xi)/n.

Define s2 = (SUM(Xi - Xbar)2)/(n-1)  

Given Xbar=11 and s=2.

(a)

Then Xbar ~ N(µ,2/n)

=> sqrt(n)*(Xbar- µ)/      ~N(0,1)

Then (n-1)*s2/ 2 ~ 2(n-1) independent of Xbar.

Then T= (sqrt(n)*(Xbar- µ)/ ) / sqrt(((n-1)*s2/ 2)/(n-1))

          = sqrt(n)*(Xbar - µ)/s    ~ t(n-1)

Then we know,

P[-t/2,n-1 < sqrt(n)*(Xbar - µ)/s < t/2,n-1] = 1-

=> P[Xbar - t/2,n-1 *s/sqrt(n) < µ < Xbar + t/2,n-1 *s/sqrt(n)] = 1-

Therefore a 100*(1-) % confidence interval for µ is given by

[Xbar - t/2,n-1 *s/sqrt(n), Xbar + t/2,n-1 *s/sqrt(n)].

Hence 95 % CI for average score of the class is obtained by putting =0.05 and the CI is   [10.065,11.935] (Using Statistical software).

(b)

The tolerance interval for your score in the assignment must be the same as that for the average score. Hence 95% tolerance interval for your score is given by [10.065,11.935].

(c)

Here is known to be 2.25 .

Then sqrt(n)(Xbar - µ)/ ~ N(0,1)

We know

P[-/2 < sqrt(n)(Xbar - µ)/ < /2 ] =1-

=> P[Xbar - /2 */sqrt(n) < µ < Xbar + /2 */sqrt(n)] = 1-

Therefore a 100*(1-) % confidence interval for µ is given by

[Xbar - /2 */sqrt(n), Xbar + /2 */sqrt(n)].

Using Statistical software we get the 95% CI to be [10.014,11.986].

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