Question 5 Believe it or not while writing this assignment I

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?

Solution

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

Assume that X_i ~ N(µ,²) i.i.d

Define Xbar = SUM(X_i)/n.

Define s² = (SUM(X_i - 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].