Sean van der Merwe
But we are not alone, the whole world faces these problems, and we don’t need to each solve all the world’s problems. We can learn from each other, like we are doing today 😀
Because I’m no longer so worried about copying and trying to thwart dodgy students …
Hi Sean, I just wanted to thank you for the exciting things I learned in Bayesian Analysis during the first semester. I was particularly happy about the practical way in which you introduced us to Stan 🙂 and writing a test / presenting something every week also kept me on my toes. The course finally unlocked the fun ways of Bayesian analysis for me and I am happy that I can now apply this to my own research. Thanks again! Sincerely, Trudie
Before using these approaches all feedback received focussed on technical aspects of the courses, things like number of assessments, feedback approaches, workload, “number the pages in the notes” 😉. Little was related to what they actually learned.
After making the changes not only did the feedback change in the direction of course outcomes and skills, but there was an increase in delayed feedback - indicating a more lasting impact (and deep learning).
The module was interesting, and I believe that most of its basics will help me in my future research. It had too much of complicated R programming, however i believe i have fallen in love it, since i have noticed that it is currently being used in most contemporary fields that are taking over the world, such as data-analytics, big-data and data-Science. Thanks again for helping me in climbing the academic ladder … Stanley Lekata
Because we now have the power of R available we can do just about anything.
“Which one of the following four time series is most likely to be stationary?”
“Let \(X\) be a random variable on the domain 0 to 1, with density function \(f(x) = 1.5 (x^2 + 2/3 x)\). What is the \(P[X> 0.3]\)? [Accurate to 3 decimals]”
Solution:
\(F(x)=1.5\int_0^x u^2 + 2/3u = 1.5\left[\frac{1}{3} u^3 + \frac{2/3}{2} u^2\right]_0^x = 1.5\frac{2x^3 + 3*2/3*x^2}{6}\)
\(P[X>0.3] = F(1) - F(0.3) \approx 0.9415\)
“Let \(X\) be a random variable on the domain 0 to 1, with density function \(f(x) = 2 (x^2 + 1/3 x)\). What is the \(P[X> 0.5]\)? [Accurate to 3 decimals]”
Solution:
\(F(x)=2\int_0^x u^2 + 1/3u = 2\left[\frac{1}{3} u^3 + \frac{1/3}{2} u^2\right]_0^x = 2\frac{2x^3 + 3*1/3*x^2}{6}\)
\(P[X>0.5] = F(1) - F(0.5) \approx 0.8333333\)
Thank you for your time and attention.
I really hope I could inspire some people, or at least broaden minds as to what is possible.
Yes, new tools have a learning curve and require some creativity to implement at first, but the long run efficiency is worth it.
I created a new module evaluation system for my faculty by leveraging some new tools.
2022/09/12 - Randomised Assessment