<aside> ℹ️ For each $p$ with $0<p<1$, the critical value $z_p$ of the standard normal distribution is such that $P(Z \geq z_p)=p$, where $z\backsim N(0,1)$.
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The most frequently used critical values are:
As a result, for example, the area to the right of the point $1.645$ is equal to $0.05$:
The critical points are very comparable to the quantiles of the standard normal distribution, the difference being that quantiles use ≤ (left side area) and critical values use ≥ (right side area).
In good newspaper articles about election polls, they will not only tell you how many votes each party is projected to get but also what the error is.
For example, for the New York Times:
The margin of sampling error is plus or minus 6 point for both Democratic and Republican primary voters.
But, what does this ‘plus or minus’ mean?
In essence, the polling company has asked a lot of people about their preference, and used their answers to estimate the number of seats the party would get. Since the measurements contain randomness, these results are not exact. Therefore, the polling company can only give an estimate on the number of seats a party will get, and not the true value. The plus-minus part is used to give an idea of the size of the error. They are basically saying that, with a high probability, the number of seats the party will get will rely on the interval $[x-6,x+6]$. where $x$ is the estimated number of setas. This interval is called a confidence interval.
Every time you measure something, you get a slightly different result, based on thousands of little effect: try standing twice in a row on the same scale to measure your weight and you almost get the minimum result. Some people take the minimum of these two results as their true weight, but you can make better inferences.
Suppose the scale is such that the result it gives is normally distributed with mean your actual weight. The standard deviation $\sigma$ equals half a kg. So, our measurement $X \backsim N(\mu,\frac{1}{2}^2)$.
If you measure your weight twice in a row, it gives the results $x_1=60.1$ and $x_2=60.9$kg. The average of these two numbers is $\overline{x}=60.5\approx\mu$, which will be an estimate of our true weight. But, what is the error?
Since you don’t know your true weight, you cannot calculate the error exactly. But, as we will calculate in a moment, a $95\%$ confidence interval of your weight is the interval from $59.81$ to $61.19$.
That is, if you were to repeat the experiment of weighing yourself twice and then calculating such an interval, many times over, in $95\%$ of the cases, the interval you calculate will contain your true weight.