Improve Your Understanding of College Level Statistics Problems

Find a 95 confidence interval for ${\displaystyle \theta }$ based on inverting the test statistic statistic ${\displaystyle \hat{\theta }}$.
For our data we have ${\displaystyle Y_i\sim N(\theta x_i,1)\text{for}i=1,\dots ,n.}$
Therefore it can be proven that the MLE for ${\displaystyle \theta }$ is given by
${\displaystyle \hat{\theta }=\frac{\sum x_i{Y}_i}{\sum x_i^2}}$
To find the confidence interval I should invert the test statistic ${\displaystyle \hat{\theta }}$.
The most powerful unbiased size ${\displaystyle \alpha =0,05}$ test for testing
${\displaystyle H_0:\mu ={\mu }_0\text{vs.}{H}_1:\mu \ne {\mu }_0}$
where ${\displaystyle X_1,\dots ,X_n\sim \text{ iid }n(\mu ,{\sigma }^2)}$ has acceptance region
$A\left({\mu }_0\right)=\left\{\mathbf{x}:|\overline{x}-{\mu }_0|\le 1,96\sigma /\sqrt{n}\right\}.$
Substituting my problem (I think) we get that the most powerful unbiased size ${\displaystyle \alpha =0,05}$ test for testing
${\displaystyle H_0:\theta =\hat{\theta }\text{vs.}{H}_1:\theta \ne \hat{\theta }}$
has acceptance region $\{A\left(\hat{\theta }\right)=\{\mathbf{y}:|\bar{y}-\hat{\theta }|\le 1,\frac{96}{\sqrt{n}}\}$
or equivalently, $A\left(\hat{\theta }\right)=\left\{\mathbf{y}:\frac{\sqrt{n}\overline{y}-1,96}{\sqrt{n}{x}_i}\le \hat{\theta }\le \frac{\sqrt{n}\overline{y}+1,96}{\sqrt{n}{x}_i}\right\}$
Substituting ${\displaystyle \hat{\theta }=\sum x_{i}Y\frac{i}{\sum }{x}_i^2}$ we obtain
$A\left(\hat{\theta }\right)=\left\{\mathbf{y}:\frac{\sqrt{n}\overline{y}-1,96}{\sqrt{n}{x}_i}\le \frac{\Sigma x_{i}Yi}{\Sigma x_i^2}\le \frac{\sqrt{n}\overline{y}+1,96}{\sqrt{n}{x}_i}\right\}$
This means that my ${\displaystyle 1-0,05=0,95\left(95\mathrm{\%}\right)}$ confindence interval is defined to be
${\displaystyle C\left(y\right)=\{\hat{\theta }:y\in A\left(\hat{\theta }\right)\}}$
But I can't seem to find anything concrete and I feel that I've made mistakes somewhere. What to do?

(a) Evaluate the standard deviation s for each set of data.

(b) Pool the data to obtain an absolute standard deviation

for the method.

the manager of an electrical supply store measured th diameters of the rolls of wire in the inventory. The diameter of the rolls are listed below 0.56,  0.622, 0.154, 0.412, 0.287, 0.118

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