# Confidence Interval for $\mu$

Let $$X_i$$ be Uniformly distributed (i.i.d.) between [$$\mu-1,\mu+1$$]. For i element of the Natural numbers. A sample of $$X_1=0.2$$ and $$X_2=0.7$$ has been drawn.

$$\hat\mu=\frac{X_1+X_2}{2}$$ Calculate a 95% CI for $$\mu$$ in the form of $$[\mu-k,\mu+k]$$ for k>0.

Use the following $$X_1+X_2$$ follows a triangular distribution with mean $$2\mu$$ and a support of $$[2\mu-2,2\mu+2]$$

My ideas: can i use that $$\mu=0$$ So that F($$\hat\mu-\mu)=F(\hat\mu)$$ and also for the support of the triangular distribution. I have never heard the term support is it where the distribution is non zero?

I would have stated that (B being one of the values for the interval)

$$F(\hat\mu $$F(\frac{X_1+X_2}{2} can i do that?

i am so new to all of this and my maths background is not very profound, so my questions may be a little stupid.

Anyway many thanks

Thanks for including the 'Self Study' tag and showing your thoughts so far. Here is a method that organizes your idea towards a solution:

For the triangular distribution of $$T = X_1 + X_2,$$ which has support $$(2\mu - 2, 2\mu + 2),$$ find the number $$\delta$$ such that $$P(2\mu - 2 + \delta \le T \le 2\mu + 2 - \delta) = .95.$$ Draw a picture. Elementary geometry and algebra should suffice.

Then $$\bar X = T/2,$$ so that $$P(\mu - 1 + \delta/2 < \bar X \le \mu + 1 - \delta/2) = .95.$$

Finally, solve the inequality of the event so that you have $$P(\bar X -1 + \delta/2 \le \mu \le \bar X + 1 - \delta/2) = .95.$$ This is called 'pivoting'.

Finally, a 95% CI for $$\mu$$ is if the form $$(\bar X - 1 +\delta/2, \bar X + 1 - \delta/2).$$

Notes: (1) Just to check, I used my $$\delta,$$ del in the code, to simulate (in R) using $$\mu = 5$$ to see what proportion of a million such confidence intervals $$(L,U)$$ actually covers $$\mu.$$ Answer is consistent with 95%.

set.seed(324)
mu = 5;  del = sqrt(.2)       # How to evaluate 'del'?
x1 = runif(10^6, mu-1, mu+1)
x2 = runif(10^6, mu-1, mu+1)
a = (x1 + x2)/2  # mean
L = a - 1 + del/2
U = a + 1 - del/2
mean(L < mu & U > mu)
## 0.950169


(2) If $$T$$ is the sum of, say, $$n = 10$$ or more observations, then the distribution of $$T$$ is nearly normal (instead of triangular) with mean and SD that are fairly easy to find. Then you can use an appropriate CI based on a good normal approximation.

• Many thanks für your answer. Just to check if i got it all correct my Intervall would be (-0.7763,07763) – Ang Mar 25 at 5:04
• Only half awake by now, but I get, $\bar X = (.2+.7)/2 =.45, \delta/2 = \sqrt{.2}/2=0.2236068,$ and CI $(-0.3263932, 1.226393),$ which is centered at $\bar X.$ – BruceET Mar 25 at 6:20
• oh ok i am centering not on $\mu$ but instead around the result of the draws, this does make sense :) - many thanks for your help – Ang Mar 25 at 6:28
• Mean of draws is $\bar X = .45.$ Used $(\bar X - 1 +\delta/2, \bar X + 1 - \delta/2)$ as shown in Answ. Will look again in the morning, after coffee. – BruceET Mar 25 at 6:34
• This is so nice of you! But i think i get it now :) Have a nice morning coffee and day! – Ang Mar 25 at 6:46