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A poll in California (done by the Public Policy Institute) asked whether the government should regulate greenhouse gases, and 751 out of 1138 likely voters said yes. However, when a different polling agency asked whether stricter environmental controls are worth the cost, 523 of 1138 likely voters said yes; the alternative was that the laws hurt the economy and cost too many jobs. (Source: Ventura County Star, March 21,2013 ) a. Find both sample proportions and compare them. Comment on the similarities of the two questions. b. Determine whether the two sample proportions are significantly different at a \(0.05\) significance level. Comment on the effect of changing the wording of this question. c. Using methods learned in Chapter 7 , find a \(95 \%\) confidence interval for the difference between the two percentages, and interpret it. Does it capture \(0 ?\) What does that mean?

Short Answer

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The first sample proportion is approximately 0.66, while the second is approximately 0.46. After conducting a hypothesis test, if the p-value is less than \(0.05\), then the sample proportions are significantly different, suggesting that question phrasing can significantly affect responses. The 95% confidence interval for the difference between the proportions would be calculated using the given formula. If it includes zero, it suggests that the two sample proportions may not be significantly different.

Step by step solution

01

Find the sample proportions

The sample proportion for a set of data is calculated as the number of success outcomes divided by the total number of outcomes. In this case, the success outcomes are the number of likely voters who responded 'yes'. Thus, the sample proportion for the first question is calculated as \(751/1138 \approx 0.66\) and for the second question as \(523/1138 \approx 0.46\).
02

Test for significant difference

This step involves conducting a hypothesis test for the difference of two proportions. The null hypothesis for this test is usually that the two populations are the same, implying the difference of the proportions is zero. It's important to denote the significance level, which is stated as \(0.05\). This implies we would reject the null hypothesis if the calculated probability (or p-value) is less than \(.05\). Here we use the formula for testing the difference between two proportions: \(Z = \frac{(p1 – p2) - 0}{\sqrt{p * ( 1 - p ) * [ (1/n1) + (1/n2) ]} }\). Calculating this value, if Z score leads to a p-value less than 0.05, null hypothesis will be rejected indicating significant difference. It means the way of asking questions can have an impact on the responses of the voters.
03

Calculate the confidence interval

Now, find the 95% confidence interval for the difference by using the following formula: \((p1 - p2) ± Z_{0.025} * \sqrt{\frac{p1*(1 - p1)}{n1} + \frac{p2*(1 - p2)}{n2}} \). If this computed confidence interval includes zero, it indicates that there's a possibility for the true difference between the populations to be zero, i.e., the two samples might not be significantly different.

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Key Concepts

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

Hypothesis Testing
Understanding hypothesis testing is crucial when analyzing sample proportions from polls or surveys. At its core, hypothesis testing assesses whether there’s enough evidence to support a particular belief about a population parameter. For instance, if a poll suggests a certain percentage of voters favor a policy, hypothesis testing can determine whether this percentage significantly differs from another poll’s results.

In the exercise, students are to test the hypothesis that there's no significant difference between the proportion of California voters who support regulating greenhouse gases and those who think stricter environmental controls are not worth the economic costs. The null hypothesis (\(H_0\)) states that the true difference in population proportions (\(p1 - p2\)) is zero. To test this, a standard score (Z score) is calculated and compared against a critical value related to the chosen significance level (\(0.05\) in the exercise). This level defines the threshold of the rarity of the Z score that would lead us to reject the null hypothesis.

When conducting such tests, the type of question and its phrasing is extremely important, as they can influence the responses and, consequently, the outcome of the hypothesis test. This impact is highlighted in the exercise when comparing two differently-worded questions regarding environmental policy.
Confidence Interval
A confidence interval provides a range of values within which we can expect the true population parameter to fall with a certain level of confidence. It’s a way of expressing uncertainty about estimated effects or differences. The width of the interval reflects the precision of the estimate; narrower intervals represent more precise estimates.

In the classroom exercise, students learned how to calculate a 95% confidence interval for the difference between two sample proportions. This interval gives a range where the true difference in the population is likely to be found if we repeated the study many times. By including the value of zero within this interval, we acknowledge that the observed difference might be due to random chance, and there's a possibility that no true difference exists in the broader population. Conversely, if zero is not within the interval, it suggests the difference observed is unlikely to be due to random chance alone.

By calculating the confidence interval, students are not just determining whether the poll results are different but also gauging the reliability and precision of this difference. This is a core concept in interpretations that relies on sample statistics to make inferences about the whole population.
Polling and Surveys
Polling and surveys are foundational tools for gauging public opinion and making inferences about the attitudes and behaviors of a larger population. When designing and interpreting polls, essential considerations include question wording, sample size, and sampling method. These factors can dramatically influence the outcomes and their accuracy.

As evidenced by the California poll in the exercise, changing the phrasing of a question can yield different responses. When the question emphasizes benefits (regulating greenhouse gases), support appears higher than when potential downsides (economic costs) are highlighted. This hints at the importance of neutral wording to avoid leading responses and capture true public sentiment.

To ensure the representativeness of the survey results, the sample should be randomly chosen and ideally large enough to estimate the population parameter with reasonable precision. Sampling errors, biases, and the phrasing of questions are all critical in evaluating the quality and trustworthiness of poll results. The poll disparities addressed in the exercise provide an excellent opportunity to discuss these intricacies with students and to stress the comprehensive nature of interpreting survey data.

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Most popular questions from this chapter

For each of the following, state whether a one-proportion z-test or a two- proportion z-test would be appropriate, and name the populations. a. A marketing manager asks a random sample of cricketers and a random sample of soccer players whether they support television commercials during games. The manager wants to determine whether the proportion of cricketers who support commercials is less than the proportion of soccer players who support these. b. A survey takes a random sample to determine the proportion of students in India who support the Women's Reservation Bill.

A fair coin is flipped 80 times, and it turns up heads 30 times. You want to test the hypothesis that the coin does not turn up heads one-half of the time. Pick the correct null hypothesis. i. \(\mathrm{H}_{0}: p=3 / 8\) ii. \(\mathrm{H}_{0}: p=1 / 2\) iii. \(\mathrm{H}_{0}: \hat{p}=3 / 8\) iv. \(\mathrm{H}_{0}=\hat{p}=1 / 2\)

Many polls have asked people whether they are trying to lose weight. A Gallup Poll in November of 2008 showed that \(22 \%\) of men said they were seriously trying to lose weight. In \(2006,24 \%\) of men (with the same average weight of 194 pounds as the men polled in 2008 ) said they were seriously trying to lose weight. Assume that both samples contained 500 men. a. Determine how many men in the sample from 2008 and how many in the sample from 2006 said they were seriously trying to lose weight. b. Determine whether the difference in proportions is significant at the \(0.05\) level. c. Repeat the problem with the same proportions but assuming both sample sizes are now 5000 . d. Comment on the different p-values and conclusions with different sample sizes.

An education board declared that \(64 \%\) of all stu- dents who appeared for examination had passed in 2015, whereas \(58 \%\) passed in 2016 . Why can you not use this report for a hypothesis test?

Suppose you are testing someone to see if he or she can tell Coke from Pepsi, and you are using 20 trials, half with Coke and half with Pepsi. The null hypothesis is that the person is guessing. The alternative is onesided: \(\mathrm{H}_{\mathrm{a}}: p_{0}>0.5 .\) The person gets 13 right out of 20 . The p-value comes out to be \(0.090 .\) Explain the meaning of the p-value.

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