Question

Exit polling is a popular technique used to determine the outcome of an election prior to results being tallied. Suppose a referendum to increase funding for education is on the ballot in a large town​ (voting population over​ 100,000). An exit poll of 200 voters finds that 94 voted for the referendum. How likely are the results of your sample if the population proportion of voters in the town in favor of the referendum is 0.52​? Based on your​ result, comment on the dangers of using exit polling to call elections.

Answer 1

P(X ≤ 94) = 0.09012

From what we observe; There is a probability of less than 94 people who voted for the referendum is 0.09012

Comment:

The result is unusual because the probability that p is equal to or more extreme than the sample proportion is greater than 5%. Thus, it is not unusual for a wrong call to be made in an election if the exit polling alone is considered.

Explanation:

From the information given :

An exit poll of 200 voters finds that 94 voted for the referendum.

How likely are the results of your sample if the population proportion of voters in the town in favor of the referendum is 0.52​? Based on your​ result, comment on the dangers of using exit polling to call elections.

This implies that ;

the Sample size n = 200

the probability p = 0.52

Let X be the random variable

So; the Binomial expression can be represented as:

X
`\sim` Binomial ( n = 200, p = 0.52)

Mean
`\mu` = np

Mean
`\mu` = 200 × 0.52

Mean
`\mu` = 104

The standard deviation
`\sigma` =
`√(np(1-p))`

The standard deviation
`\sigma` =
`√(200 * 0.52(1-0.52))`

The standard deviation
`\sigma` =
`√(200 * 0.52(0.48))`

The standard deviation
`\sigma` =
`√(49.92)`

The standard deviation
`\sigma` = 7.065

However;

P(X ≤ 94) because the discrete distribution by the continuous normal distribution values lies in the region of 93.5 and 94.5 .

The less than or equal to sign therefore relates to the continuous normal distribution of X < 94.5

Now;

x = 94.5

Therefore;

`z = (x- \mu)/(\sigma)`

`z = (94.5 - 104)/(7.065)`

`z = (-9.5)/(7.065)`

z = −1.345

P(X< 94.5) = P(Z < - 1.345)

From the z- table

P(X ≤ 94) = 0.09012

From what we observe; There is a probability of less than 94 people who voted for the referendum is 0.09012

Comment:

The result is unusual because the probability that p is equal to or more extreme than the sample proportion is greater than 5%. Thus, it is not unusual for a wrong call to be made in an election if the exit polling alone is considered.