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How to calculate confidence interval?
To calculate a confidence interval (two-sided), you need to follow these steps:
- Let's say the sample size is
100. - Find the mean value of your sample. Assume it's
3. - Determine the standard deviation of the sample. Let's say it's
0.5. - Choose the confidence level. The most common confidence level is
95%. - In the statistical table find the Z(0.95)-score, i.e., the 97.5th quantile of N(0,1) – in our case, it's
1.959. - Compute the standard error as
σ/√n = 0.5/√100 = 0.05. - Multiply this value by the z-score to obtain the margin of error:
0.05 × 1.959 = 0.098. - Add and subtract the margin of error from the mean value to obtain the confidence interval. In our case, the confidence interval is between 2.902 and 3.098.
What will decrease the width of a confidence interval?
The width of a confidence interval decreases when the margin of error decreases, which happens when the:
- Significance level decreases;
- Sample size increases; or
- Sample variance decreases.
The sample mean has no impact on the width of a confidence interval!
What will increase the width of a confidence interval?
The width of a confidence interval increases when the margin of error increases, which happens when the:
- Significance level increases;
- Sample size decreases; or
- Sample variance increases.
As $x \to \infty$, $e^x \to \infty$ so that $ 1 + e^x \to \infty$ and therefore $$\lim_{x \to \infty}\frac{1}{1+e^x} = 0$$ This is your lower bound since it occurs when the denominator is maximized. Now try to follow the same reasoning for $x \to -\infty$ to achieve a similar bound.
In case you were actually asking about $\frac{1}{1 + e^{-x}}$, you can follow a similar procedure. Take $x \to \infty$ and note that $e^{-x} = \frac{1}{e^x} \to 0$ so that $$\lim_{x \to \infty} \frac{1}{1 + e^{-x}} = 1$$
Let $x_1 , x_2 \in [-10,10] $ be arbitrary.
For $x_1<x_2$ ,
Observe that
$$e^{x_1}<e^{x_2}$$ since $e^x$ is a strictly increasing function.
$$1+e^{-x_1}>1+e^{-x_2}$$
$$1+e^{x_1}<1+e^{x_2}$$
$$\frac{1}{1+e^{x_1}}>\frac{1}{1+e^{x_2}}$$
Thus $$f(x)=\frac{1}{1+e^x}$$ is a strictly decreasing function.
So for $x \in [-10,10]$ ,
$$f(x)_{min}=f(x)=\frac{1}{1+e^{10}} >0 $$ and
$$f(x)_{max}=f(x)=\frac{e^{10}}{1+e^{10}} <1$$
You can show that $f_{min}$ and $f_{max}$ are its supremum and infimum . Hence upper bound and lower bounds.
In my class, most kids have a calculator that, when you go to binomCDF, gives you the option to input upper and lower bounds. Unfortunately, with my calculator, I can't do that. So, I was wondering how I am supposed to go about solving these problems with my calculator. Is there some other option I can do that would allow me to calculate it?
I need to be able to do this for my class, so I asked my teacher and we could not figure out how to do it. This was not the first time we had trouble with my calculator, as he doesn't know how to use it, and therefore doesn't really know how to teach me to use it. If anyone knows how to do this please let me know.
The process is that the function in the lower limit needs to be less than the function in the upper limit over the interval of integration. In your case, $y^2 \leq y+2$ for all $y$ in $[-1,2]$.
As for how to determine that inequality, you could do it by algebra, but the quickest way is probably with a graph. You know that $x=y^2$ is a parabola with its vertex at the origin, and which opens to the right. The line $y=x-2$ passes through $(0,-2)$ and $(2,0)$.
A few more comments:
You may think that you can just try it with the functions in one order, and if you get a negative number, just drop the minus sign and make it positive. This may work, but only if the graphs don't cross each other in the interval over which you're integrating.
In multivariable calculus, you really need graphing skills. Trying to do it all by just pattern matching and algebra doesn't help you gain any intuition.
You asked which “equation” should be in the lower and upper limits of integration, but you mean which function. For instance $y=x+2$ is an equation, but when you set up the integral it had to be manipulated into $x=y-2$, because you needed a function of $y$.
You should use $x=y^2$ for the lower bound and $x=y+2$ for the upper because integrals are direction dependent. Values of the function are added when moving in the direction x increased. This is for each fixed y just the fact that $\int_a^b f(x)dx = - \int_b^a f(x)dx$ Convention makes this positive if $a<b$ and $f(x)>0$.