Suppose that a group of 200 colleagues is having a contest to see who can take t

Question

Suppose that a group of 200 colleagues is having a contest to see who can take the most steps over the course of a week. You are interested in the average number of steps that are taken by the members of the group. You randomly sample the number of steps taken over the course of the week from 10 members of the group. You see that the average number of steps taken by your sample of 10 people (in thousands) is 92.45 thousand steps with a variance of 32.36 (thousand steps)^2. Assume that the number of steps taken by a person in the contest is normally distributed. Construct a 99% confidence interval for the mean number of steps taken by the members of the group. Be sure to give an interpretation of your interval in the context of the problem.

Explanation / Answer

Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=92.45
Standard deviation( sd )=5.689
Sample Size(n)=10
Confidence Interval = [ 92.45 ± t a/2 ( 5.689/ Sqrt ( 10) ) ]
= [ 92.45 - 3.25 * (1.799) , 92.45 + 3.25 * (1.799) ]
= [ 86.603,98.297 ]

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