/*! 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 137 Election Poll In October before ... [FREE SOLUTION] | 91Ó°ÊÓ

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Election Poll In October before the 2008 US presidential election, \(A B C\) News and the Washington Post jointly conducted a poll of "a random national sample" and asked people who they intended to vote for in the 2008 presidential election. \(^{37}\) Of the 1057 sampled people who answered either Barack Obama or John McCain, \(55.2 \%\) indicated that they would vote for Obama while \(44.8 \%\) indicated that they would vote for MeCain. While we now know the outcome of the election, at the time of the poll many people were very curious as to whether this significantly predicts a winner for the election. (While a candidate needs a majority of the electoral college vote to win an election, we'll simplify things and simply test whether the percentage of the popular vote for Obama is greater than \(50 \% .\) ) (a) State the null and alternative hypotheses for testing whether more people would vote for Obama than MeCain. (Hint: This is a test for a single proportion since there is a single variable with two possible outcomes.) (b) Describe in detail how you could create a randomization distribution to test this (if you had many more hours to do this homework and no access to technology).

Short Answer

Expert verified
The null hypothesis is that the proportion of votes for Obama is equal to 50% and the alternative hypothesis is that the proportion is greater than 50%. A randomization distribution would be created by simulating multiple elections under the null hypothesis, where votes are assigned randomly to each candidate with equal probability, and the proportion of votes for Obama is calculated in each.

Step by step solution

01

State the null and alternative hypotheses

The null hypothesis, \(H_0\), is that the proportion of votes for Obama is equal to 50%, and the alternative hypothesis, \(H_1\), is that the proportion of votes for Obama is greater than 50%. Mathematically, these can be represented as: \n \[H_0: p = 0.50\]\n \[H_1: p > 0.50\] where \(p\) stands for the proportion of votes for Obama.
02

Describe the process of creating a randomization distribution

The process would involve simulating multiple elections under the assumption of the null hypothesis. Each simulated election would involve randomly assigning votes to Obama or McCain such that both candidates have an equal chance of getting a vote, i.e., the probability for each is 0.50. For each simulated election, the proportion of votes for Obama would be calculated. Repeating this process a large number of times would result in a randomization distribution that can be compared to the observed proportion of votes for Obama in the original sample.

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Key Concepts

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

Null and Alternative Hypotheses
In statistical analysis, the concept of null and alternative hypotheses is fundamental. Specifically, in the context of election polls, these hypotheses help in determining if a polling result genuinely reflects the voter preference, or if the observed outcome could simply be due to chance.

Put simply, the null hypothesis represents a default position that there is no effect or no difference. In the election poll scenario, we would express this by saying the proportion of votes for Obama (\(p\)) is equally likely to be 50% as it would be for any fair coin toss. Mathematically, we denote this as \(H_0: p = 0.50\).

Conversely, the alternative hypothesis competes with the null hypothesis and represents what the researcher really wants to prove. For our election poll example, the alternative hypothesis suggests that more than 50% of the vote would go to Obama, notated as \(H_1: p > 0.50\).

Assessing these hypotheses usually involves a statistical test to see whether the data significantly contradicts the null hypothesis, thus giving credence to the alternative hypothesis.
Single Proportion Test
A single proportion test is suitable for scenarios just like our election poll, where there are only two outcomes (Obama or McCain) and we need to assess the proportion for one of these outcomes. The goal here is to test whether the observed proportion of voters favoring one candidate is significantly different from a pre-specified null value—in this case, 50%.

To perform the test, you need the sample size and the number of successes—in our case, the number of people who said they would vote for Obama. Then, you use a test statistic (like the Z-score) to measure the difference between the observed sample proportion and the null hypothesis proportion, scaled by the standard error. The resulting p-value from the test informs us whether we should reject the null hypothesis in favor of the alternative. If the p-value is less than a chosen significance level (commonly 0.05), the null hypothesis is rejected, indicating the observed proportion is significantly different from 50%.
Randomization Distribution
Creating a randomization distribution is a way to model what the results of an election might look like under the null hypothesis. To build this distribution, you would simulate many artificial elections assuming the null hypothesis is true—that both candidates have an equal chance of receiving a vote. This process involves randomly allocating votes to each candidate with a 50% probability for each vote.

After many simulations, you end up with a distribution of vote proportions for Obama. This randomization distribution represents the sampling variability and shows how the proportion might differ purely by chance when there is no real difference in voter preference. By comparing the observed proportion of votes for Obama to this distribution, we can see if the observed result lies within the realm of typical fluctuations or if it stands out as being unusually high, which would suggest evidence against the null hypothesis.

Using this method, no high-tech software is necessary, though without it, creating a thorough randomization distribution could be exceedingly time-consuming. It's a detailed method to illustrate the principles of statistical testing and reinforces the concept of chance in our interpretation of data.

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Most popular questions from this chapter

Flipping Coins We flip a coin 150 times and get 90 heads, so the sample proportion of heads is \(\hat{p}=90 / 150=0.6 .\) To test whether this provides evidence that the coin is biased, we create a randomization distribution. Where will the distribution be centered? Why?

In Exercises 4.150 to \(4.152,\) 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 99\% confidence interval for \(\mu: 134\) to 161 (a) \(H_{0}: \mu=100\) vs \(H_{a}: \mu \neq 100\) (b) \(H_{0}: \mu=150\) vs \(H_{a}: \mu \neq 150\) (c) \(H_{0}: \mu=200\) vs \(H_{a}: \mu \neq 200\)

Radiation from Cell Phones and Brain Activity Does heavy cell phone use affect brain activity? There is some concern about possible negative effects of radiofrequency signals delivered to the brain. In a randomized matched-pairs study, \(^{24}\) 47 healthy participants had cell phones placed on the left and right ears. Brain glucose metabolism (a measure of brain activity) was measured for all participants under two conditions: with one cell phone turned on for 50 minutes (the "on" condition) and with both cell phones off (the "off" condition). The amplitude of radiofrequency waves emitted by the cell phones during the "on" condition was also measured. (a) Is this an experiment or an observational study? Explain what it means to say that this was a "matched-pairs" study. (b) How was randomization likely used in the study? Why did participants have cell phones on their ears during the "off" condition? (c) The investigators were interested in seeing whether average brain glucose metabolism was different based on whether the cell phones were turned on or off. State the null and alternative hypotheses for this test. (d) The p-value for the test in part (c) is 0.004 . State the conclusion of this test in context. (e) The investigators were also interested in seeing if brain glucose metabolism was significantly correlated with the amplitude of the radiofrequency waves. What graph might we use to visualize this relationship? (f) State the null and alternative hypotheses for the test in part (e). (g) The article states that the p-value for the test in part (e) satisfies \(p<0.001\). State the conclusion of this test in context.

Exercise Hours Introductory statistics students fill out a survey on the first day of class. One of the questions asked is "How many hours of exercise do you typically get each week?" Responses for a sample of 50 students are introduced in Example 3.25 on page 207 and stored in the file ExerciseHours. The summary statistics are shown in the computer output. The mean hours of exercise for the combined sample of 50 students is 10.6 hours per week and the standard deviation is 8.04 . We are interested in whether these sample data provide evidence that the mean number of hours of exercise per week is different between male and female statistics students. Variable Gender N Mean StDev Minimum Maximum \(\begin{array}{lllllll}\text { Exercise } & \text { F } 30 & 9.40 & 7.41 & 0.00 & 34.00\end{array}\) \(\begin{array}{llll}20 & 12.40 & 8.80 & 2,00\end{array}\) Discuss whether or not the methods described below would be appropriate ways to generate randomization samples that are consistent with \(H_{0}: \mu_{F}=\mu_{M}\) vs \(H_{a}: \mu_{F} \neq \mu_{M} .\) Explain your reasoning in each case. (a) Randomly label 30 of the actual exercise values with " \(\mathrm{F}^{\prime \prime}\) for the female group and the remaining 20 exercise values with " \(\mathrm{M} "\) for the males. Compute the difference in the sample means, \(\bar{x}_{F}-\bar{x}_{M}\) (b) Add 1.2 to every female exercise value to give a new mean of 10.6 and subtract 1.8 from each male exercise value to move their mean to 10.6 (and match the females). Sample 30 values (with replacement) from the shifted female values and 20 values (with replacement) from the shifted male values. Compute the difference in the sample means, \(\bar{x}_{F}-\bar{x}_{M}\) - (c) Combine all 50 sample values into one set of data having a mean amount of 10.6 hours. Select 30 values (with replacement) to represent a sample of female exercise hours and 20 values (also with replacement) for a sample of male exercise values. Compute the difference in the sample means, \(\bar{x}_{F}-\bar{x}_{M}\)

Exercises 4.117 to 4.122 give null and alternative hypotheses for a population proportion, as well as sample results. Use StatKey or other technology to generate a randomization distribution and calculate a p-value. StatKey tip: Use "Test for a Single Proportion" and then "Edit Data" to enter the sample information. Hypotheses: \(H_{0}: p=0.6\) vs \(H_{a}: p>0.6\) Sample data: \(\hat{p}=52 / 80=0.65\) with \(n=80\)

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