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?

individual plays a game of tossing a coin where he wins Rs 2 if head turns up and nothing if tail turns up.On the basis of the given information, find (i) The expected value of the game. (4) (ii) The risk premium this person will be willing to pay to avoid the risk associated with the game.

Using the integral test, find the positive values of p for which the series ∑∞ k=2 1/ (k(ln(k)))^p converges. Show your work and explain your reasoning.

A study compared women who viewed high levels of television violence as children with those who did not in order to study the differences with regard to physical abuse of their partners as adults. Use the table shown below to calculate the observed value of the chi-square statistic. The table shows both actual counts and expected counts (in parentheses).
$\begin{array}{|ccc|}\hline & \text{ High TV Violence }& \text{ Low TV Violence }\\ \text{ Yes, Physical Abuse }& 14(8.42)& 26(31.58)\\ \text{ No Physical Abuse }& 22(27.58)& 109(103.42)\\ \hline\end{array}$
The chi-square statistic is ${\chi }^2=\square$

Consider an electric refrigerator located in a room. Determine the direction of the work and heat interactions (in  or out) when the following are taken as the system: 

(a) the contents of the refrigerator,  

(b) all parts of the refrigerator including the contents, and  

(c) everything contained within the room during a winter day.