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Suppose there are two cities. In the city A 52% of people are males and 48% are females. In the city B 47% of people are males and 53% are females. We conduct a survey: from each city a simple random sample of size 100 is taken. I need to find the probability that the survey will show greater percentage of males in city B than males in city A.

What kind of statistical test should I use? $\chi^2$ or $t$-test? And how can I find the asked probability?

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The asked probability can be estimated using the central limit theorem. $A_i$ and $B_i$ are each Bernoulli with expected value and variance

$$\begin{matrix}E(A_i)=.52&E(B_i)=.47\\ Var(A_i)=.52(.48)&Var(B_i)=.47(.53)\end{matrix}$$

Then by the central limit theorem, a sample of 100 from each group is approximately normally distributed

$$\bar A\sim \mathscr N\left(.52, \frac{.52(.48)}{100}\right)\\ \bar B\sim \mathscr N\left(.47, \frac{.47(.53)}{100}\right)$$

And the difference is also normal $\bar B-\bar A\sim \mathscr N\left(-.05, \frac{.52(.48)+.47(.53)}{100}\right)$

Then we can standardize to find the probability that the proportion of males in the sample from city B is greater than the proportion of males in sample A:

$$\begin{split}\Pr(\bar B>\bar A)&=\Pr\left(Z>\frac{.05}{\sqrt{\frac{.52(.48)+.47(.53)}{100}}}\right)\\ &=\Pr(Z>.708)\\ &=0.2394726\end{split}$$

This is only an approximation.

A somewhat better approximation can be obtained using the correction for continuity. (This is slightly more technical and might not be necessary for your purposes.)

$$A\sim \mathscr N\left(52, 52(.48)\right)\\ B\sim\mathscr N\left(47, 47(.53)\right)\\ B-A\sim \mathscr N\left(-5, 52(.48)+47(.53)\right)$$

The desired probability $$\begin{split}\Pr(B>A)&=\Pr(B-A\ge 1)\\ &=\Pr\left(Z\ge\frac{\frac 12+5}{\sqrt{52(.48)+47(.53)}}\right)\\ &=\Pr(Z>0.7788)\\ &=0.2180398\end{split}$$

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Let $X$ be the number of males in your sample from city A.

Then $X\sim B(100,0.52)$

Similarly let $Y$ be the number of males in your sample from city B.

Then $Y\sim B(100,0.47)$

You want to know the probability $P(Y>X)$

This can be calculated as $P(Y>X) =\Sigma^{x=100}_{x=0}P(Y>x)P(X=x)$

I did it in Excel and got 0.2179

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