MHB  95% Confidence Interval for Staying Awake at UMUC

AI Thread Summary
The discussion revolves around calculating a 95% confidence interval for the average time colleagues can stay awake during a meeting at UMUC, using a sample of ten observations. The calculated mean is 2.33 hours, with a standard deviation of approximately 2.0022, leading to a confidence interval of 0.897 to 3.762 hours. Participants discuss the methodology for calculating the confidence interval and the significance of the results in determining the effectiveness of caffeine. There is also confusion regarding how to interpret the results for the second question about the effectiveness of caffeine. The conversation highlights the need for clarity in statistical calculations and their implications for productivity at the university.
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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?
 
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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 $s/\sqrt{n}$?
 
Siron said:
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
 
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