Respuesta :
Answer:
385
Step-by-step explanation:
Lets say that p is the probability for the governor to be voted by a person.
Lets call X the random variable with value 1 if a person votes for the governor and 0 if not. X is a random variable with Bernoulli distribution of parameter p. X has mean p and variance p(1-p). Lets denote Y the mean of a sample of lenght n. Y has mean μ=p and variance σ² = p*(1-p)/n.
As a consecuence of the Central Limit Theorem, since Y is the mean of identically distributed independent random variables, we have that Y is approximately Normal, with parameters μ = p and σ = √(p(1-p)/n).
As a consecuence of the Law of Big Numbers, we have that
[tex]W = \frac{Y-p}{\sqrt(\frac{Y(1-Y)}{n})} \simeq \frac{Y-p}{\sqrt(\frac{p(1-p)}{n})} \simeq N(0.1)[/tex]
The values of Ф, the cummulative function of the normal density funtion, can be obtained from the attached file. We want a 95% confidence interval, so we want Z that satisfies
P(-Z < W < Z) = 0.95
Thus, P(W < Z) = 0.975
If we look the table, we will find that Z = 1.96, Hence
[tex]0.95 = P(-1.96 < W < 1.96) = P(-1.96 < \frac{Y-p}{\sqrt(\frac{Y(1-Y)}{n})} < 1.96 ) = \\P(-1.96 * \sqrt(\frac{Y(1-Y)}{n}) < Y-p < 1.96 * \sqrt(\frac{Y(1-Y)}{n})[/tex]
Now we move the Y and the - sign, that changes the inequalities, so we obtain
[tex]0.95 = P(Y-1.96 * \sqrt(\frac{Y(1-Y)}{n} < p < Y+ 1.96 * \sqrt(\frac{Y(1-Y)}{n})[/tex]
Therefore, the confidence interval is [tex] [Y-1.96 * \sqrt(\frac{Y(1-Y)}{n}, Y+1.96 * \sqrt(\frac{Y(1-Y)}{n})] [/tex], which has lenght [tex] 2*1.96*\sqrt(\frac{Y(1-Y)}{n}) [/tex] . In the worst case, Y(1-Y) takes the value 0.25 (when the sample proportion is 1/2). Thus, the lenght of the interval is bounded above by [tex] 2*1.96*\sqrt(\frac{0.25}{n}) = \frac{1.96}{\sqrt{n}} . [/tex]
In order for the lenght to be at most 0.1 (so, the margin of error is 0.05) we need n such that
[tex] \frac{1.96}{\sqrt{n}} \leq 0.1 [/tex]
Hence,
[tex] \sqrt{n} > 19.6 \rightarrow n > 19.6^2 = 384.1[/tex]
We take n = 385. None of the 4 given values would work according to the information we have. A sample size of lenght 200 would give us a confidence interval of lenght almost 0.2
