/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 Predicting Election Results Thro... [FREE SOLUTION] | 91Ó°ÊÓ

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Predicting Election Results Throughout the US presidential election of \(2016,\) polls gave regular updates on the sample proportion supporting each candidate and the margin of error for the estimates. This attempt to predict the outcome of an election is a common use of polls. In each case below, the proportion of voters who intend to vote for each of two candidates is given as well as a margin of error for the estimates. Indicate whether we can be relatively confident that candidate A would win if the election were held at the time of the poll. (Assume the candidate who gets more than \(50 \%\) of the vote wins.) \(\begin{array}{lll}\text { (a) Candidate A: } 54 \% & \text { Candidate }\end{array}\) B: \(46 \%\) Margin of error: \(\pm 5 \%\) (b) Candidate A: \(52 \%\) Candidate B: \(48 \%\) Margin of error: \(\pm 1 \%\) \(\begin{array}{ll}\text { (c) Candidate A: } 53 \% & \text { Candidate }\end{array}\) B: \(47 \%\) Margin of error: \(\pm 2 \%\) \(\begin{array}{lll}\text { (d) Candidate A: } 58 \% & \text { Candidate }\end{array}\) B: \(42 \%\) Margin of error: \(\pm 10 \%\)

Short Answer

Expert verified
For options (b) and (c), we can be relatively confident that candidate A would win if the election were held at the time of the poll. For options (a) and (d), we cannot be confident about that.

Step by step solution

01

Predicting the outcome for option (a)

For option (a), the proportion for candidate A is 54% and the margin of error is ±5%. This means that candidate A's true proportion could be as low as 49% (54% - 5%) and as high as 59% (54% + 5%). Since the lowest limit for candidate A (49%) is above candidate B's highest limit (46% + 5% = 51%), we cannot be confident that candidate A would win.
02

Predicting the outcome for option (b)

For option (b), the proportion for candidate A is 52% and the margin of error is ±1%. This means that candidate A's true proportion could be as low as 51% (52% - 1%) and as high as 53% (52% + 1%). Since the lowest limit for candidate A (51%) is above candidate B's highest limit (48% + 1% = 49%), we can be relatively confident that candidate A would win.
03

Predicting the outcome for option (c)

For option (c), the proportion for candidate A is 53% and the margin of error is ±2%. This means that candidate A's true proportion could be as low as 51% (53% - 2%) and as high as 55% (53% + 2%). Since the lowest limit for candidate A (51%) is above candidate B's highest limit (47% + 2% = 49%), we can be relatively confident that candidate A would win.
04

Predicting the outcome for option (d)

For option (d), the proportion for candidate A is 58% and the margin of error is ±10%. This means that candidate A's true proportion could be as low as 48% (58% - 10%) and as high as 68% (58% + 10%). Since the lowest limit for candidate A (48%) is not above candidate B's highest limit (42% + 10% = 52%), we cannot be confident that candidate A would win.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Margin of Error
Polls often include a margin of error, a term that indicates the range within which the true value of a survey's result is expected to fall. When polling data is shared, it's crucial to communicate this variability to predict election outcomes more accurately. The margin of error represents a bit of wiggle room or flexibility. It's introduced because we survey a sample, not the entire population.
  • An important point to remember is that the larger the margin, the less precise the poll's estimate.
  • In election polls, the margin of error tells us how much a candidate's proportion could realistically go up or down.
For example, if a candidate has 54% support in a poll with a ±5% margin of error, the candidate's actual support could realistically be between 49% and 59%. Understanding the margin helps assess how strongly the poll predicts an election's result. A smaller margin confers higher certainty to the prediction.
Sample Proportion
When discussing polling predictions, the term 'sample proportion' frequently arises. It's vital to understand what it signifies in the context of election polling. The sample proportion is essentially the percentage of survey respondents indicating their support for a candidate.
  • This figure is based on the sample taken from a larger population.
  • Polls give us information on how many, out of this sample group, support a specific candidate.
For example, if in a poll of 1,000 voters, 540 indicated support for Candidate A, the sample proportion for Candidate A is 54%. Although this number is a useful snapshot, considering the margin of error is crucial, as the actual proportion in the whole population might slightly differ from what the sample shows.
Confidence Levels
Confidence levels are crucial when interpreting polling predictions because they indicate the likelihood that the poll's results reflect the true opinions of the total population. In simpler terms, a confidence level tells you how sure you can be about the poll's accuracy.
  • A standard confidence level is usually set at 95% in polling.
  • This means there's a 95% chance that the poll results are correct, or that they fall within the given margin of error.
For example, if a poll indicates that Candidate A has 52% support with a confidence level of 95%, it suggests a high probability that if all voters were surveyed, Candidate A's actual support would be very close to 52%. Confidence levels are essential for evaluating how reliable a poll is and, therefore, how much weight should be given to its predictions.

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Most popular questions from this chapter

A Sampling Distribution for Performers in the Rock and Roll Hall of Fame Exercise 3.38 tells us that 206 of the 303 inductees to the Rock and Roll Hall of Fame have been performers. The data are given in RockandRoll. Using all inductees as your population: (a) Use StatKey or other technology to take many random samples of size \(n=10\) and compute the sample proportion that are performers. What is the standard error of the sample proportions? What is the value of the sample proportion farthest from the population proportion of \(p=0.68 ?\) How far away is it? (b) Repeat part (a) using samples of size \(n=20\). (c) Repeat part (a) using samples of size \(n=50\). (d) Use your answers to parts (a), (b), and (c) to comment on the effect of increasing the sample size on the accuracy of using a sample proportion to estimate the population proportion.

How Many Apps for Your Smartphone? Exercise 3.20 describes a study about smartphone users in the US downloading apps for their smartphone. Of the \(n=355\) smartphone users who had downloaded an app, the average number of apps downloaded was 19.7 (a) Give notation for the parameter of interest, and define the parameter in this context. (b) Give notation for the quantity that gives the best estimate and give its value. (c) What would we have to do to calculate the parameter exactly?

Is a Car a Necessity? A random sample of \(n=1483\) adults in the US were asked whether they consider a car a necessity or a luxury, \({ }^{31}\) and we find that a \(95 \%\) confidence interval for the proportion saying that it is a necessity is 0.83 to \(0.89 .\) Explain the meaning of this confidence interval in the appropriate context.

Exercises 3.71 to 3.73 consider the question (using fish) of whether uncommitted members of a group make it more democratic. It has been argued that individuals with weak preferences are particularly vulnerable to a vocal opinionated minority. However, recent studies, including computer simulations, observational studies with humans, and experiments with fish, all suggest that adding uncommitted members to a group might make for more democratic decisions by taking control away from an opinionated minority. \({ }^{36}\) In the experiment with fish, golden shiners (small freshwater fish who have a very strong tendency to stick together in schools) were trained to swim toward either yellow or blue marks to receive a treat. Those swimming toward the yellow mark were trained more to develop stronger preferences and became the fish version of individuals with strong opinions. When a minority of five opinionated fish (wanting to aim for the yellow mark) were mixed with a majority of six less opinionated fish (wanting to aim for the blue mark), the group swam toward the minority yellow mark almost all the time. When some untrained fish with no prior preferences were added, however, the majority opinion prevailed most of the time. \({ }^{37}\) Exercises 3.71 to 3.73 elaborate on this study. How Often Does the Fish Majority Win? In a school of fish with a minority of strongly opinionated fish wanting to aim for the yellow mark and a majority of less passionate fish wanting to aim for the blue mark, as described under Fish Democracies above, a \(95 \%\) confidence interval for the proportion of times the majority wins (they go to the blue mark) is 0.09 to \(0.26 .\) Interpret this confidence interval. Is it plausible that fish in this situation are equally likely to go for either of the two options?

Topical Painkiller Ointment The use of topical painkiller ointment or gel rather than pills for pain relief was approved just within the last few years in the US for prescription use only. \({ }^{13}\) Insurance records show that the average copayment for a month's supply of topical painkiller ointment for regular users is \$30. A sample of \(n=75\) regular users found a sample mean copayment of \(\$ 27.90\). (a) Identify each of 30 and 27.90 as a parameter or a statistic and give the appropriate notation for each. (b) If we take 1000 samples of size \(n=75\) from the population of all copayments for a month's supply of topical painkiller ointment for regular users and plot the sample means on a dotplot, describe the shape you would expect to see in the plot and where it would be centered. (c) How many dots will be on the dotplot you described in part (b)? What will each dot represent?

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