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Influencing Voters: Is a Phone Call More Effective? Suppose, as in Exercise \(4.38,\) that we wish to compare methods of influencing voters to support a particular candidate, but in this case we are specifically interested in testing whether a phone call is more effective than a flyer. Suppose also that our random sample consists of only 200 voters, with 100 chosen at random to get the flyer and the rest getting a phone call. (a) State the null and alternative hypotheses in this situation. (b) Display in a two-way table possible sample results that would offer clear evidence that the phone call is more effective. (c) Display in a two-way table possible sample results that offer no evidence at all that the phone call is more effective. (d) Display in a two-way table possible sample results for which the outcome is not clear: there is some evidence in the sample that the phone call is more effective but it is possibly only due to random chance and likely not strong enough to generalize to the population.

Short Answer

Expert verified
The null hypothesis is that there is no difference in effectiveness between a flyer and a phone call, while the alternative hypothesis is that the phone call is more effective. Hypothetical data tables were provided demonstrating clear evidence supporting the alternative hypothesis, no evidence to support it, and uncertain results.

Step by step solution

01

Formulate Hypotheses

The null hypothesis (\( H_0 \)) would be that there is no difference in the effectiveness of influencing voters between the phone call and the flyer. The alternative hypothesis (\( H_a \)) would be that a phone call is more effective than a flyer in influencing voters to support a candidate.
02

Results that show phone call is more effective

Imagine a two-way table where columns represent the method used - flyer vs phone call, and rows represent the outcomes - voting for or against the candidate. An example of a result that clearly supports the phone call being more effective could be: 30 voters influenced by the flyer and 70 by phone call voted for the candidate, versus 70 voters influenced by the flyer and 30 by phone call voted against the candidate.
03

Results that show no difference in effectiveness

An example of a result offering no evidence that the phone call is more effective might be: 50 voters influenced by each method voted for the candidate and 50 voters influenced by each method voted against the candidate.
04

Results that show uncertain effectiveness

An example of an unclear result might be: 45 voters influenced by the flyer and 55 by phone call voted for the candidate, versus 55 voters influenced by the flyer and 45 by phone call voted against the candidate. Here there is slight evidence that phone calls might be more effective, but it could just be due to random chance as well, and not strong enough to conclude consistently that phone calls are more effective.

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Key Concepts

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

Null and Alternative Hypotheses
Understanding the null and alternative hypotheses is essential when designing an experiment or analyzing data in statistics. When we pose these hypotheses, we're setting up a framework for testing whether there's evidence to support a specific claim.

Take our voter influence example. The null hypothesis (\( H_0 \) ) serves as the baseline assumption that there is no difference in effectiveness between a phone call and a flyer in swaying voters. It's like a presumption of innocence in the court of law; unless proven otherwise, we assume no effect. In statistical analysis, disproving the null is often the goal to substantiate an alternative claim.

The alternative hypothesis (\( H_a \) ) represents what we're aiming to provide evidence for, which in this case is that phone calls are more effective than flyers. Crafting these hypotheses with precision is pivotal as they guide our entire analysis process and influence how we collect and interpret data.
Two-Way Table
A two-way table is a powerful tool in statistics for organizing and displaying categorical data from two different variables. It's akin to a spreadsheet with rows and columns that summarize information, allowing us to visualize relationships and make comparisons easily.

In our voter case study, the two variables could be the method of influence (phone call or flyer) and the voter's response (voted for or against). By filling in the two-way table with the number of voters influenced by each method and their corresponding response, we create a clear picture of our sample results. This visual aid is especially helpful in communicating complex data succinctly and facilitating further statistical analysis, such as chi-square tests.
Sample Results Interpretation
The interpretation of sample results often involves looking beyond the numbers to understand what they might imply about the population as a whole. It's crucial to consider the context of the data and the statistical methods used.

In interpreting our voter influence sample, we visually represented potential outcomes in two-way tables. We looked at scenarios from clear evidence - a substantial difference in the number of voters swayed by phone calls, to no evidence - an equal number influenced by both methods, and finally to uncertain evidence, where a slight difference might not be enough to overcome the variability inherent in random sampling. Making interpretations involves a keen eye for detail and a deep understanding of statistical concepts to draw meaningful conclusions.
Statistical Evidence
Statistical evidence is the backbone of making data-driven decisions or claims in research. It involves collecting data, analyzing it with proper statistical techniques, and then determining whether that data is statistically significant — typically meaning it is unlikely to have occurred by random chance.

For instance, in our exercise dealing with voter influence, statistical evidence would be necessary to confidently state that phone calls are a more effective method of influencing voters than flyers. This would require a significant difference in the number of voters who were persuaded by each method, beyond what could be expected by random chance. The presence or absence of statistical evidence shapes the confidence with which we can support our claims and often dictates our course of action in response to the data.

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Most popular questions from this chapter

A confidence interval for a sample is given, followed by several hypotheses to test using that sample. In each case, use the confidence interval to give a conclusion of the test (if possible) and also state the significance level you are using. A \(90 \%\) confidence interval for \(p_{1}-p_{2}: 0.07\) to 0.18 (a) \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1} \neq p_{2}\) (b) \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1}>p_{2}\) (c) \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1}

Indicate whether it is best assessed by using a confidence interval or a hypothesis test or whether statistical inference is not relevant to answer it. (a) What percent of US voters support instituting a national kindergarten through \(12^{\text {th }}\) grade math curriculum? (b) Do basketball players hit a higher proportion of free throws when they are playing at home than when they are playing away? (c) Do a majority of adults riding a bicycle wear a helmet? (d) On average, were the 23 players on the 2014 Canadian Olympic hockey team older than the 23 players on the 2014 US Olympic hockey team?

Exercise 4.19 on page 269 describes a study investigating the effects of exercise on cognitive function. \({ }^{31}\) Separate groups of mice were exposed to running wheels for \(0,2,4,7,\) or 10 days. Cognitive function was measured by \(Y\) maze performance. The study was testing whether exercise improves brain function, whether exercise reduces levels of BMP (a protein which makes the brain slower and less nimble), and whether exercise increases the levels of noggin (which improves the brain's ability). For each of the results quoted in parts (a), (b), and (c), interpret the information about the p-value in terms of evidence for the effect. (a) "Exercise improved Y-maze performance in most mice by the 7 th day of exposure, with further increases after 10 days for all mice tested \((p<.01)\) (b) "After only two days of running, BMP ... was reduced \(\ldots\) and it remained decreased for all subsequent time-points \((p<.01)\)." (c) "Levels of noggin ... did not change until 4 days, but had increased 1.5 -fold by \(7-10\) days of exercise \((p<.001)\)." (d) Which of the tests appears to show the strongest statistical effect? (e) What (if anything) can we conclude about the effects of exercise on mice?

Approval Rating for Congress In a Gallup poll \(^{51}\) conducted in December 2015 , a random sample of \(n=824\) American adults were asked "Do you approve or disapprove of the way Congress is handling its job?" The proportion who said they approve is \(\hat{p}=0.13,\) and a \(95 \%\) confidence interval for Congressional job approval is 0.107 to 0.153 . If we use a 5\% significance level, what is the conclusion if we are: (a) Testing to see if there is evidence that the job approval rating is different than \(14 \%\). (This happens to be the average sample approval rating from the six months prior to this poll.) (b) Testing to see if there is evidence that the job approval rating is different than \(9 \%\). (This happens to be the lowest sample Congressional approval rating Gallup ever recorded through the time of the poll.)

The same sample statistic is used to test a hypothesis, using different sample sizes. In each case, use StatKey or other technology to find the p-value and indicate whether the results are significant at a \(5 \%\) level. Which sample size provides the strongest evidence for the alternative hypothesis? Testing \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1}>p_{2}\) using \(\hat{p}_{1}-\hat{p}_{2}=0.45-0.30=0.15\) with each of the following sample sizes: (a) \(\hat{p}_{1}=9 / 20=0.45\) and \(\hat{p}_{2}=6 / 20=0.30\) (b) \(\hat{p}_{1}=90 / 200=0.45\) and \(\hat{p}_{2}=60 / 200=0.30\) (c) \(\hat{p}_{1}=900 / 2000=0.45\) and \(\hat{p}_{2}=600 / 2000=0.30\)

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