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ELEN [110]
3 years ago
15

A recent survey based on a random sample of n=470 voters, predicted that independent candidate for the mayoral election will get

24% of the vote, but he actually got 27%. discuss whether the claim that suggest the survey was done incorrectly! explain your rational.
Mathematics
1 answer:
julsineya [31]3 years ago
7 0

I have an expression

\sigma = \sqrt{p(1-p)/n}

floating around in my head; let's see if it makes sense.

The variance of binary valued random variable b that comes up 1 with probability p (so has mean p) is

E( (b-p)^2 ) =  (-p)^2(1-p) + (1-p)^2p = p(1-p)

That's for an individual sample. For the observed average we divide by n, and for the standard deviation we take the square root:

\sigma = \sqrt{p(1-p)/n}

Plugging in the numbers,

\sigma = \sqrt{.24(1-.24)/490} = 0.019 = 1.9\%

One standard deviation of the average is almost 2% so a 27% outcome was 3/1.9 = 1.6 standard deviations from the mean, corresponding to a two sided probability of a bit bigger than 10% of happening by chance.

So this is borderline suspect; most surveys will include a two sigma margin of error, say plus or minus 4 percent here, and the results were within those bounds.

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Check the picture below.

since the vertical distance, namely the y-coordinate, is twice as much as the horizontal, then if the horizontal is "x", the vertical one must be 2x.

let's find the hypotenuse first.

\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=\stackrel{adjacent}{x}\\ b=\stackrel{opposite}{2x}\\ \end{cases} \\\\\\ c=\sqrt{x^2+(2x)^2}\implies c=\sqrt{x^2+4x^2}\implies c=\sqrt{5x^2}\implies c=x\sqrt{5} \\\\[-0.35em] ~\dotfill

\bf sin(\theta )=\cfrac{\stackrel{opposite}{2~~\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }}{\stackrel{hypotenuse}{~~\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ \sqrt{5}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{\cfrac{2}{\sqrt{5}}\cdot \cfrac{\sqrt{5}}{\sqrt{5}}\implies \cfrac{2\sqrt{5}}{(\sqrt{5})^2}\implies \cfrac{2\sqrt{5}}{5}}

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