Believe it or not, I traveled forward in time and collected a random sample of 2

Question

Believe it or not, I traveled forward in time and collected a random sample of 25 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 10, and their sample standard deviation, which was 1.8. On the basis of this information: Find a 95% confidence interval for the actual average score mu for the entire class. Construct a 99/95 tolerance interval for your score on this assignment (i.e., an interval that includes 99% of possible observations with 95% confidence). Discuss any issues. 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 2.0. Given this information, how would you change the interval you calculated in part if at all?

Explanation / Answer

a) The 95% c.i=xbar+-tcritical (s/sqrt N), where, s/sqrt N refer to standard error of mean, xbar is sample mean, s is sample standard deviation, N is sample size and tcritical corresponds to critical t value at df=24 (N-1).

=10+-2.064(1.8/sqrt 25)

=9.26 to 10.74

b) Compute (1-p)/2=(1-0.95)/2=0.025

Z0.025=1.96

Find lower critical value of Chi-square, for df=(25-1=24), v=24 and alpha=0.01. X^224,0.01=42.980.

Calculate, K2=sqrt [{v(1+1/N)z^2(1-p)/2}/X^224,0.01]

=sqrt [{24(1+1/25)*1.96^2}/42.980]

=2.23

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