I've been teaching myself some statistics and upon doing this problem I've stumbled at the very last part:
I've had no problem reaching $\mu\approx x=220.1$ and $\sigma\approx y= \frac{1}{2\sqrt{5}}$, which agrees with the solution document. I then reasoned that if I wanted an interval of 2 standard errors around the estimated mean, that would just be $x\pm2\cdot\frac{y}{\sqrt{5}}=220.1\pm\frac{1}{5}$, but this is apparently incorrect.
The correct calculation of 2 standard errors around the mean is shown as $220.1\pm2\cdot\frac{2}{\sqrt{5}}\cdot\frac{1}{2\sqrt{5}}=220.1\pm\frac{2}{5}$. Can you help me figure out why they have multiplied by a factor of $\frac{2}{\sqrt{5}}$ instead?
(A picture of the solutions I've been using are here and here)
$\hat \mu = \frac45T+\frac15(T_1+T_2+T_3+T_4)$ has expectation $\mu$ and variance $\frac45 \sigma^2$
$\hat \sigma^2 =(\frac1{\sqrt{5}}T-\frac1{\sqrt{5}}(T_1+T_2+T_3+T_4))^2$ has expectation $\sigma^2$, so $\frac45 \hat \sigma^2$ has expectation $\frac45 \sigma^2$
So it seems a reasonable confidence interval is $\hat \mu \pm k\sqrt{\frac45\hat \sigma^2}$ which here is $220.1 \pm k\sqrt{\frac45 \times (-\frac1{\sqrt{5}} 0.5)^2}= 220.1 \pm 0.2 k$, and for $k=2$ is $220.1 \pm 0.4$ i.e. $[219.7, 220.5]$