An object moves in simple harmonic motion with

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?

Rotation matrix to construct canonical form of a conic
${\displaystyle C:9x^2+4xy+6y^2-10=0.}$
I've found C is a non-degenerate ellipses (computing the cubic and the quadratic invariant), and then I've studied the characteristic polynomial 
$p\left(t\right)=\det \left(\begin{array}{cc}9-t& 2\\ 2& 6-t\\ & \end{array}\right)$
The eigenvalue are ${\displaystyle t_1=5,t_2=10}$, with associated eigenvectors ${\displaystyle (-1,2),(2,1)}$. Thus I construct the rotation matrix R by putting in columns the normalized eigenvectors (taking care that ${\displaystyle det\left(R\right)=1}$):
$R=\frac{1}{\sqrt{5}}\left(\begin{array}{cc}1& 2\\ -2& 1\\ & \end{array}\right)$
Then ${\displaystyle {(x,y)}^t=R{(x^{\prime },y^{\prime })}^t}$, and after some computations I find the canonical form
${\displaystyle \frac{1}{2}{x}^{\prime 2}+\frac{4}{5}{y}^{\prime 2}=1.}$

Use the method of your choice to factor the trinomial, or state that the trinomial is prime. Check each factorization using FOIL multiplication. 

6y^2 − 23y + 15

Assume that a procedure yields a binomial distribution with a trial repeated n=13${\displaystyle n=13}$ times. Use either the binomial probability formula (or technology) to find the probability of k=8${\displaystyle k=8}$ successes given the probability p=0.73${\displaystyle p=0.73}$ of success on a single trial.

(Report answer accurate to 4 decimal places.)

P(X=k)=

Prove that ${\displaystyle n\mid \varphi (a^n-1)}$. Any natural solution that uses $Aut\left(G\right)$

Brämhults Juice, based in Sweden, produces chilled fruit and vegetable juices. A manager at Brämhults is considering purchasing a new machine to bottle 16-fluid-ounce (473-milliliter) bottles of 100% pure carrot juice and wants an estimate of the differ- ence in the mean filling weights between the new machine and the old machine. Random samples of bottles of carrot juice that had been filled by both machines were obtained. Estimate the difference in the mean filling weights between the new and the old machines? Discuss the assumptions. Use a = 0.10.

Decreasing the sample size, while holding the confidence level the same, will do what to the length of the confidence interval?