# Intervals to estimate the mean of normal distribution

$$X_1, \dots, X_n$$ are independent random variables distribute $$N(c,1)$$ with $$c$$ unknown.

How to construct the following intervals to estimate $$c$$,

The first interval is $$[i_1(X_1), j_1(X_1)]$$ s.t $$p(c \in [i_1(X_1), j_1(X_1)]) = 95\%$$

The second interval is $$[i_n(X_1, \dots, X_n), j_n(X_1, \dots, X_n)]$$ s.t $$p(c \in [i_n(X_1, \dots, X_n), j_n(X_1, \dots, X_n)]) = 95\%$$

I don't know from where to start! How I can use only the percentage $$95\%$$ to construct both intervals?

The idea with confidence intervals is to consider testing a hypothesis and find the value of the parameters where the conclusion of the test transitions from retaining the hypothesis to rejecting the hypothesis.

For estimating the mean of the normal distribution $$N(c,1)$$, we consider the hypothesis $$c=c_0$$. For general estimation we test it against the alternative hypothesis $$c \neq c_0$$ (what is called a two-tailed test).

Now the idea is to say: if $$c$$ were equal to $$c_0$$, what's the probability that a sample mean deviates from $$c_0$$ by more than some $$x>0$$? The answer is to use the fact that the sample mean has $$N(c_0,1/\sqrt{n})$$ distribution (second argument being standard deviation here), so

$$P(|\overline{X}-c_0|>x)=P(\overline{X}-c_0>x)+P(\overline{X}-c_0<-x)=2(1-\Phi(\sqrt{n}x))$$

using the symmetry of the normal distribution. If you now set that equal to $$0.05$$ then you can (using a table or a computer) solve for $$x$$. You will find that $$x \approx 1.96/\sqrt{n}$$.

Now we say that if our sample deviates by more than that $$x$$ from the proposed $$c_0$$, then we reject the hypothesis $$c=c_0$$ in favor of $$c \neq c_0$$, "with 95% confidence". Mathematically this means that if $$c=c_0$$ then only 5% of samples would have their sample mean deviating from $$c_0$$ by as much as we observed or more.

The 95% confidence interval for a particular sample consists of all the values of $$c_0$$ such that we would retain the hypothesis $$c=c_0$$. Thus if the sample mean is $$\overline{X}$$ then the confidence interval is $$(\overline{X}-1.96/\sqrt{n},\overline{X}+1.96/\sqrt{n})$$. If you think about it, $$c$$ will fall in this interval with probability 0.95, because $$c$$ falls in this interval if and only if $$\overline{X}$$ is within $$1.96/\sqrt{n}$$ of $$c$$.

• First of all, thanks a lot for the answer. I think the sample mean here $\bar{X} \sim N(c, 1 /n)$ not $1/ \sqrt{n}$. Additionally, I don't understand two things in the answer: $1$ The question asks to construct two intervals but while the answer gives only one. $2$ How you see the interval $(\bar{X}-1.96 \sqrt{n}, \bar{X}+1.96 \sqrt{n} )$ is similar to $[i_1(X_1), j_1(X_1)]$ and $[i_n(X_1, \dots, X_n), j_n(X_1, \dots, X_n)]$ Feb 27, 2020 at 18:13
• @user8003788 Depending on the way you paramrtrize either of us is right, I was parametrizing by the standard deviation. The way you were parametrizing isn't apparent from the question since $1^2=1$. As for the two intervals, you just have two samples of different sizes so you use the same method either way.
– Ian
Feb 27, 2020 at 19:09
• So you mean, generally, the interval $(i_z(X_z), j_z(X_z)) = (\sum_{z=1}^n \frac{X_z}{z} +1.96, \sum_{z=1}^n \frac{X_z}{z} +1.96 )$? Feb 27, 2020 at 19:28
• @user8003788 Your notation doesn't really make sense, I'm saying that $i_1(X_1)=X_1-1.96,i_n(I_1,\dots,X_n)=\frac{1}{n} \sum_{i=1}^n X_i - \frac{1.96}{\sqrt{n}}$ and similar for the $j$'s.
– Ian
Feb 28, 2020 at 2:29