95% Confidence Interval for Staying Awake at UMUC

[elove]

This is the word problem I am faced with... I have provided my answer. Please give input if I am right or wrong.

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. 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 zombies. 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

My work so far for the 1st question…

To answer this question I began by finding the mean:
1.9+.8+1.1+.1+-.1+4.4+5.5+1.6+4.6+3.4= 23.3/10= 2.33

I then calculated the standard deviation:
∑(𝑥−µ)2𝑛−1=4.009
4.009=2.0022

From there, I calculated the degree of freedom and used the Inverse t Distribution calculator to find the t for confidence interval:
𝐷𝐹=𝑛−1=9
T for confidence interval = 2.262

Finally, I calculated the lower limit:
2.33−(2.262)(2.002210)=.897

And the upper limit:
2.33+(2.262)(2.002210)=3.762

Therefore, the answer is:
.897< µ<3.762

I have no idea how to answer the 2nd question! Opinion based? or based off the 1st question's answer?

 

[Siron]

Your reasoning for question $(1)$ looks good. Generally, a $95\%$ confidence interval (thus significance level $\alpha = 5\%$) is given by (assuming normality of the data)
$$\left[\overline{x} - t_{\alpha/2}\frac{s}{\sqrt{n}}, \overline{x} + t_{\alpha/2}\frac{s}{\sqrt{n}}\right].$$
Your computation of the sample variance $s$ looks messy without LaTEX code. What do you obtain for $s/\sqrt{n}$?

 

[elove]

Your reasoning for question $(1)$ looks good. Generally, a $95\%$ confidence interval (thus significance level $\alpha = 5\%$) is given by (assuming normality of the data)
$$\left[\overline{x} - t_{\alpha/2}\frac{s}{\sqrt{n}}, \overline{x} + t_{\alpha/2}\frac{s}{\sqrt{n}}\right].$$
Your computation of the sample variance $s$ looks messy without LaTEX code. What do you obtain for $s/\sqrt{n}$?

I don't know how to properly use the coding on here.
I can see how you did it so let me see if I can write it better.

s = 2.0022

n = 9

$s/\sqrt{n}$

so 2.0022/ 3

= 0.6674

if I give you all my values can you properly code them on here? so I show my work better

[overline{x}] = 2.33

t = 2.262

a = .05