25. Did you vote? A sample survey contacted an SRS of $663$ registered voters in Oregon shortly after an election and asked respondents whether they had voted. Voter records show that $56\%$of registered voters had actually voted. We will see later that in repeated random samples of size $663$, the proportion in the sample who voted (call this proportion V) will vary according to the Normal distribution with mean $\mu =0.56$ and standard deviation $\sigma =0.019$.
(a) If the respondents answer truthfully, what is $P(0.52\le V\le 0.60)$? This is the probability that the sample proportion $V$ estimates the population
proportion $0.56$ within $\pm 0.04$.
(b) In fact, $72\%$of the respondents said they had voted $(V=0.72)$. If respondents answer truthfully, what is $P(V\ge 0.72)$? This probability is so small that it is good evidence that some people who did not vote claimed that they did vote.

Step by step solution

01

Part (a) Step 1: Given information 

The probability that the sample proportion$V$estimates the population proportion $0.56$within $\pm 0.04$.

02

Part (a) Step 2: Explanation 

The probability that the sample proportion $V$will accurately predict the population within$\pm 0.04$is determined as follows:

$P(0.52\le V\le 0.60)=P\left(\frac{0.52-\mu }{\sigma }\le \frac{x-\mu }{\sigma }\le \frac{0.60-\mu }{\sigma }\right)$

Here, population mean is $\left(\mu \right)=0.56$, and the population standard is deviation$\left(\sigma \right)=0.019$

$$\begin{gathered} =P\left(\frac{0.52-0.56}{0.019}\le \frac{x-\mu }{\sigma }\le \frac{0.60-\mu }{\sigma }\right) \\ =P(-2.11\le Z\le 2.11) \\ =0.9652 \end{gathered}$$

03

Part (b) Step 1: Given information

Let, $72\%$of the respondents said they had voted $(V=0.72).$

To find the probability for $P(V\ge 0.72)$to check, could be claimed by the people who did not vote.

04

Part (b) Step 2: Explanation

By using the normal probability table to calculate the probability as:

$$\begin{gathered} P(V\ge 0.72)=P(\frac{x-\mu }{\sigma }\ge \frac{0.72-\mu }{\sigma }) \\ =P\left(Z\ge \frac{0.72-0.56}{0.019}\right) \\ =P(z\ge 8.42) \\ =0.0001 \end{gathered}$$

As a result, the chance is too low. It's possible to make a case for it. Because getting a sample proportion of $72$percent by random is virtually hard if the genuine percentage is $56$percent .

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