# constructing a 95% confidence interval - manipulating inequalities

given the asymptotic distribution of $\hat{\theta_1}$ construct a 95% confidence interval for $\theta$ for large samples:

$\hat{\theta_1} = \frac{\hat{\theta_1}-\theta}{\frac{\theta}{\sqrt{n}}}$

I know that the confidence interval will be :

$-1.96 \leq \frac{\hat{\theta_1}-\theta}{\frac{\theta}{\sqrt{n}}} \leq 1.96$

and hence:

$\frac{\hat{\theta_1}}{1+\frac{1.96}{\sqrt{n}}} \leq \theta \leq \frac{\hat{\theta_1}}{1-\frac{1.96}{\sqrt{n}}}$

Question:

How to get from $-1.96 \leq \frac{\hat{\theta_1}-\theta}{\frac{\theta}{\sqrt{n}}} \leq 1.96$ to $\frac{\hat{\theta_1}}{1+\frac{1.96}{\sqrt{n}}} \leq \theta \leq \frac{\hat{\theta_1}}{1-\frac{1.96}{\sqrt{n}}}$ ? I am aware it is just inequality manipulation, however i am not able to solve it. Please provide detailed steps as my mathematical background is rather weak.

• You can get displayed equations by enclosing them in double dollar signs instead of single dollars signs. Especially with nested fractions, that makes them a lot easier to read. – joriki Jul 28 '18 at 9:57

For ease, let $\hat{\theta_1}=x, \theta=y$.

Divide by $\sqrt{n}$: $$-\frac{1.96}{\sqrt{n}}\le \frac{x}{y}-1 \le \frac{1.96}{\sqrt{n}}.$$ Add $1$: $$1-\frac{1.96}{\sqrt{n}}\le \frac{x}{y} \le 1+\frac{1.96}{\sqrt{n}}.$$ Raise to power $-1$: $$\frac{1}{1-\frac{1.96}{\sqrt{n}}}\ge \frac{y}{x} \ge \frac{1}{1+\frac{1.96}{\sqrt{n}}}.$$ Note: $2<3 \iff \frac12>\frac13$.

Now multiply by $x$ to get the final result.

Treat each of the inequalities like you'd treat an equation: Multiply both sides by the denominator; bring mulitples of $\theta$ to one side and everything else to the other; divide by the coefficient of $\theta$. The only thing specific to inequalities is to make sure you reverse the inequality when you multiply or divide by a negative quantity. In the present case, assuming that $\theta\gt0$ and $n\ge4$, no reversal occurs. In the end, you can put the two inequalities back together.

• thanks for the answer, but i still cannot solve this. I am getting stuck on the fact that there is 2 inequality signs and i do multiply by negative -1.96, therefor reversal occurs. I must be doing something wring. I would very much appreciate an little more elaborate answer that shows the aforementioned operations – user1607 Jul 28 '18 at 11:19
• @user1607: There shouldn't be two inequality signs if you follow the answer. You should be treating $A\le B\le C$ as two separate inequalities $A\le B$ and $B\le C$ and performing all operations exclusively on single inequalities up to the very end where you put them together again. You shouldn't be multiplying by negative $-1.96$, either -- in which of the steps in the answer does that happen? – joriki Jul 28 '18 at 11:22
• $$-1.96 \leq \frac{\hat{\theta_1}-\theta}{\frac{\theta}{\sqrt{n}}}$$ then $$-1.96 \frac{\theta}{\sqrt{n}} \leq \hat{\theta_1} - \theta$$ and here i devide by -1.96 so $$\frac{\theta}{\sqrt{n}} + \theta >= \frac{\hat{\theta_1}}{-1.96}$$ – user1607 Jul 28 '18 at 11:32
• @user1607: First, irrespective of what I suggested in the answer, that's not a valid transformation -- you'd have to divide the entire inequality by $-1.96$, including $\theta$. But it's also not what I suggested to do -- the next step at this point was to divide by the coefficient of $\theta$. What's the coefficient of $\theta$ in $-1.96\frac\theta{\sqrt n}+\theta$? – joriki Jul 28 '18 at 11:52
• @user1607: Please note the slight correction I made to the answer -- the statement that no reversal occurs was only correct for $n\ge4$, since otherwise $1-1.96/\sqrt n$ will be negative. – joriki Jul 28 '18 at 11:59