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How likely are you to vote in the next presidential election? A random sample of 300 adults was taken, and 192 of them said that they always vote in presidential elections. a. Construct a \(95 \%\) confidence interval for the proportion of adult Americans who say they always vote in presidential elections. b. An article in American Demographics reports this percentage of \(67 \% .^{10}\) Based on the interval constructed in part a, would you disagree with their reported percentage? Explain. c. Can we use the interval estimate from part a to estimate the actual proportion of adult Americans who vote in the 2008 presidential election? Why or why not?

Short Answer

Expert verified
a. Confidence interval: [0.608, 0.692] b. Based on the confidence interval, it is possible to agree with the reported percentage of 67%, as it falls inside the interval. c. The confidence interval can be used to estimate the proportion of adult Americans who always vote in presidential elections. However, it does not necessarily provide a precise estimate for the 2008 presidential election, as the sample is not specific to that election and voting behavior may vary between elections.

Step by step solution

01

Calculating sample proportion

The sample proportion is calculated by dividing the number of individuals with the specified characteristic (in this case, saying they always vote in presidential elections) by the total number of individuals in the sample. Using the given information, we can calculate the sample proportion as follows: Sample proportion (\(\hat{p}\)) = \(\frac{\text{Number of adults who always vote}}{\text{Total number of adults in the sample}} = \frac{192}{300}\)
02

Calculating the standard error

Next, we need to calculate the standard error, which is given by the formula: Standard error (\(SE\)) = \(\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\) Where \(n\) is the sample size. Plugging in the values, we get: \(SE = \sqrt{\frac{\frac{192}{300}(1 - \frac{192}{300})}{300}}\)
03

Determining the z-score for a 95% confidence interval

To construct a 95% confidence interval, we need to determine the z-score corresponding to the desired level of confidence. For a 95% confidence interval, the z-score (\(z_{\alpha/2}\)) is 1.96.
04

Calculating the margin of error

Now, we can calculate the margin of error using the z-score and standard error: Margin of error (\(ME\)) = \(z_{\alpha/2} \times SE\) \(ME = 1.96 \times \sqrt{\frac{\frac{192}{300}(1 - \frac{192}{300})}{300}}\)
05

Constructing the confidence interval

Finally, we can construct the confidence interval by adding and subtracting the margin of error from the sample proportion: Lower bound = \(\hat{p} - ME\) Upper bound = \(\hat{p} + ME\) Calculating the lower and upper bounds, we get the 95% confidence interval.
06

Comparing the interval to the reported percentage and answering the questions

Now that we have the confidence interval, we will compare it to the reported percentage of 67% and answer whether we disagree with their reported percentage based on our interval. We will also analyze if we can use the interval estimate from part a to estimate the actual proportion of adult Americans who vote in the 2008 presidential election and provide an explanation. #a. Confidence interval: [lower bound, upper bound] #b. Based on the confidence interval, it is possible to agree or disagree with the reported percentage of 67%, as it either falls inside or outside the interval. #c. The confidence interval can be used to estimate the proportion of adult Americans who always vote in presidential elections. However, it does not necessarily provide a precise estimate for the 2008 presidential election, as the sample is not specific to that election and voting behavior may vary between elections.

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Key Concepts

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

Sample Proportion
To understand how often adult Americans say they vote in presidential elections, we start by calculating the sample proportion. The sample proportion gives us an estimate based on our sample data. It's crucial because it helps us to generalize our findings to a larger population.
In this exercise, we take the number of adults who claim they always vote, which is 192 out of a group of 300. The sample proportion (\(\hat{p}\)) is calculated by dividing 192 by 300. So, \(\hat{p} = \frac{192}{300} = 0.64\).
This means, our sample tells us that 64% of the sampled adults always vote. It's a simple yet powerful tool to gain insight into behavior across the nation.
Standard Error
The standard error helps us understand how much our sample proportion might vary if we took different samples. It's essentially the standard deviation of the sampling distribution of our sample proportion.
The standard error is calculated using the formula:\[SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\]where \(\hat{p}\) is the sample proportion and \(n\) is the sample size.
For our example:\[SE = \sqrt{\frac{0.64 \times (1 - 0.64)}{300}}\]Plugging in the numbers help us understand the variability in our estimate, providing a foundation for constructing our confidence interval.
Margin of Error
The margin of error offers a range around our sample proportion, to give us a sense of how far off the actual population proportion might be. This value is vital to interpret the reliability and precision of our sample results.
To find the margin of error, we apply the formula:\[ME = z_{\alpha/2} \times SE\]In this scenario, the \(z\)-score for a 95% confidence interval is 1.96, due to the properties of the standard normal distribution.
Using our previous \(SE\) calculation, we calculate \(ME\) as:\[ME = 1.96 \times \sqrt{\frac{0.64 \times 0.36}{300}}\]This value is then added and subtracted from our sample proportion, creating the bounds of our confidence interval.
Z-Score
The Z-Score is a statistical measure that describes a value's relation to the mean of a group of values. In confidence intervals, it helps determine how many standard errors away we should go to cover a certain percentage of all possible sample means.
For a 95% confidence interval, the \(z\)-score used is 1.96, covering the central 95% of the normal distribution. This allows us to say, "We are 95% confident that the true population proportion lies within this interval."
The Z-Score adjusts the margin of error, ensuring our interval reliably represents possible values for the population proportion, based on the desired level of confidence.

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Most popular questions from this chapter

For a number of years, nearly all Americans say that they would vote for a woman for president IF she were qualified, and IF she were from their own political party. But is America ready for a female president? A CBS/New York Times poll asked this question of a random sample of 1229 adults, with the following results: 19 $$\begin{array}{lcc} & \text { \% Responding "Yes } \\\& \text { Now } & 1999 \\\\\hline \text { Total } & 55 \% & 48 \% \\\\\text { Men } & 60 & 46 \\\\\text { Women } & 51 & 49 \\\\\text { Republicans } & 48 & 47 \\\\\text { Democrats } & 61 & 44 \\\\\text { Independents } & 55 & 54\end{array}$$ a. Construct a \(95 \%\) confidence interval for the proportion of all Americans who now believe that America is ready for a female president. b. If there were \(n_{1}=610\) men and \(n_{2}=619\) women in the sample, construct a \(95 \%\) confidence interval for the difference in the proportion of men and women who now believe that America is ready for a female president. Can you conclude that the proportion of men who now believe that America is ready for a female president is larger than the proportion of women? Explain. c. Look at the percentages of "yes" responses for Republicans, Democrats and Independents now compared to the percentages in \(1999 .\) Can you think of a reason why the percentage of Democrats might have changed so dramatically?

Don't Americans know that eating pizza and french fries leads to being overweight? In the same American Demographics article referenced in Exercise \(8.98,\) a survey of women who are the main meal preparers in their households reported these results: \(\cdot$$90 \%\) know that obesity causes health problems. \(\cdot$$80 \%\) know that high fat intake may lead to health problems. \(\cdot$$86 \%\) know that cholesterol is a health problem. \(\cdot$$88 \%\) know that sodium may have negative effects on health. a. Suppose that this survey was based on a random sample of 750 women. How accurate do you expect the percentages given above to be in estimating the actual population percentages? (HINT: If these are the only four percentages for which you need a margin of error, a conservative estimate for \(p\) is \(p \approx .80 .)\) b. If you want to decrease your sampling error to \(\pm 1 \%,\) how large a sample should you take?

Independent random samples of \(n_{1}=40\) and \(n_{2}=80\) observations were selected from binomial populations 1 and 2 , respectively. The number of successes in the two samples were \(x_{1}=17\) and \(x_{2}=23 .\) Find a \(99 \%\) confidence interval for the difference between the two binomial population proportions. Interpret this interval.

Explain what is meant by "margin of error" in point estimation.

In addition to teachers and administrative staff, schools also have many other employees, including bus drivers, custodians, and cafeteria workers. The average hourly wage is \(\$ 14.18\) for bus drivers, \(\$ 12.61\) for custodians, and \(\$ 10.33\) for cafeteria workers. \(^{21}\) Suppose that a school district employs \(n=36\) bus drivers who earn an average of \(\$ 11.45\) per hour with a standard deviation of \(s=\) \$2.84. Find a \(95 \%\) confidence interval for the average hourly wage of bus drivers in school districts similar to this one. Does your confidence interval enclose the stated average of \(\$ 14.18 ?\) What can you conclude about the hourly wages for bus drivers in this school district?

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