Exit polling is a popular technique used to determine the outcome of an election

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 106 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 0.49? Based on your result, comment on the dangers of using exit polling to call elections. How likely are the results of your sample if the population proportion of voters in the town in favor of the referendum is 0.49? The probability that more than 106 people voted for the referendum is Comment on the dangers of using exit polling to call elections. Choose the correct answer below. The result is not 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 exit polling alone is considered. The result is not unusual because the probability that p is equal to or more extreme than the sample proportion is less than 5%. Thus, it is unusual for a wrong call to be made in an election if exit polling alone is considered. The result is unusual because the probability that p is equal to or more extreme than the sample proportion is the made in an election if exit polling alone is considered

Explanation / Answer

We first get the z score for the critical value:          
          
x = critical value =    106.5      
u = mean = np =    98      
          
s = standard deviation = sqrt(np(1-p)) =    7.069653457      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    1.202322016      
          
Thus, the left tailed area is          
          
P(z <   1.202322016   ) =    0.114619395 [ANSWER]

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As P > 0.05, then it is not unusual. Hence,

OPTION A. [ANSWER]

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Hi! I used continuity correction here. If you don't use that in class, this is the alternative solution:

We first get the z score for the critical value:          
          
x = critical value =    106      
u = mean = np =    98      
          
s = standard deviation = sqrt(np(1-p)) =    7.069653457      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    1.131597192      
          
Thus, the left tailed area is          
          
P(z <   1.131597192   ) =    0.128901911 [ALTERNATE ANSWER]

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