Need help with this question Please show work Thanks It is 1

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It is 1581 Anno Domini. At the Undergraduate School of UMUC, besides Assistant Academic Director of Mathematics and Statistics, I am also the Undergraduate School-appointed CPA, Coffee Pot Attendant.

It is a very important office sponsored by the Holy See.

I have taken this job very seriously, because I believe that I am the key to increased productivity at the Undergraduate School. Why, by mid-morning, many of my colleagues act as if they were

It is imperative that I restore productivity via a secret naturally-occurring molecule, caffeine........

In order to see if my secret molecule works, I have observed the time, in hours, a random selection of ten of my colleagues

who could stay awake at the extremely long-winded Dean\'s meeting as soon as it started. Oh, yes, one fell asleep even before the meeting started!

1.9

0.8

1.1

0.1

-0.1

4.4

5.5

1.6

4.6

3.4

Now, I have to complete a report to the Provost\'s Office on the effectiveness of my secret molecule so that UMUC can file for a patent at the United Provinces Patent and Trademark Office as soon as possible. Oh, yes, I am waiting for a handsome reward from the Provost.....

But I need the following information:

What is a 95% confidence interval for the time my colleagues can stay awake on average for all of my colleagues?

Was my secret molecule effective in increasing their attention span, I mean, staying awake? And, please explain.....

Solution

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)=2.33
Standard deviation( sd )=2.0022
Sample Size(n)=10
Confidence Interval = [ 2.33 ± t a/2 ( 2.0022/ Sqrt ( 10) ) ]
= [ 2.33 - 2.262 * (0.633) , 2.33 + 2.262 * (0.633) ]
= [ 0.898,3.762 ]