Question:
The average time per worker to perform a task has been 100 minutes with a standard deviation of 15 minutes. A modification was introduced to reduce this time, and after a few months, a sample of 16 workers was selected, measuring the execution time of each. A sample average time of 90 minutes and a standard deviation of 16 minutes were obtained. Assuming normality.
a) Estimate the new average execution time for a credibility interval of 95% probability.
b) Estimate the new population variance through a credibility interval with a 98% probability.
So, I was doing this exercise but I got really confused. It looks pretty simple from the classical perspective, but I don't really know how to do in a Bayesian way. What's the likelihood and the prior here? Could I consider $\bar{X}\ -> N(u, \frac{15^2}{16})$ and a Jeffreys Prior for A and $\bar{X}\ -> N(100, \frac{\sigma^2}{16})$ with another Jeffreys Prior for B? Don't know how to use the sample information that was given.
The prior is $\mu \sim N(\mu_0, \sigma_0^2)$ where $\mu_0=100$ and $\sigma_0^2=15^2$. Given $\mu$, you observe $$(x_i \mid \mu) \overset{\text{i.i.d.}}{\sim} N(\mu, \sigma^2)$$ for $i=1,\ldots, n$ for some fixed $\sigma^2$.
The posterior distribution of $\mu$ is $$N\left(\left(\frac{n}{\sigma^2} + \frac{1}{\sigma_0^2}\right)^{-1}\left(\frac{\mu_0}{\sigma_0^2} + \frac{n\bar{x}}{\sigma^2}\right), \left(\frac{n}{\sigma^2} + \frac{1}{\sigma_0^2}\right)^{-1}\right)$$ where $\bar{x} := \frac{1}{n} \sum_{i=1}^{n} x_i$. [Wikipedia]
You can use this posterior to construct a credible interval for $\mu$. I am not sure how to construct a credible interval for $\sigma^2$ though.