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Chapter 8: Problem 38

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

In a study to establish the absolute threshold of hearing, 70 male college freshmen were asked to participate. Each subject was seated in a soundproof room and a \(150 \mathrm{H}\) tone was presented at a large number of stimulus levels in a randomized order. The subject was instructed to press a button if he detected the tone; the experimenter recorded the lowest stimulus level at which the tone was detected. The mean for the group was \(21.6 \mathrm{db}\) with \(s=2.1\). Estimate the mean absolute threshold for all college freshmen and calculate the margin of error.

Independent random samples were selected from populations 1 and 2 . The sample sizes, means, and variances are as follows: $$\begin{array}{lcc} & \multicolumn{2}{c} {\text { Population }} \\\\\cline { 2 - 3 } & 1 & 2 \\\\\hline \text { Sample Size } & 35 & 49 \\\\\text { Sample Mean } & 12.7 & 7.4 \\\\\text { Sample Variance } & 1.38 & 4.14\end{array}$$ a. Find a \(95 \%\) confidence interval for estimating the difference in the population means \(\left(\mu_{1}-\mu_{2}\right) .\) b. Based on the confidence interval in part a, can you conclude that there is a difference in the means for the two populations? Explain.

An experimenter fed different rations, \(A\) and \(B\), to two groups of 100 chicks each. Assume that all factors other than rations are the same for both groups. Of the chicks fed ration \(\mathrm{A}, 13\) died, and of the chicks fed ration \(\mathrm{B}, 6\) died. a. Construct a \(98 \%\) confidence interval for the true difference in mortality rates for the two rations. b. Can you conclude that there is a difference in the mortality rates for the two rations?

An increase in the rate of consumer savings is frequently tied to a lack of confidence in the economy and is said to be an indicator of a recessional tendency in the economy. A random sampling of \(n=200\) savings accounts in a local community showed a mean increase in savings account values of \(7.2 \%\) over the past 12 months, with a standard deviation of \(5.6 \% .\) Estimate the mean percent increase in savings account values over the past 12 months for depositors in the community. Find the margin of error for your estimate.

If 36 measurements of the specific gravity of aluminum had a mean of 2.705 and a standard deviation of .028 , construct a \(98 \%\) confidence interval for the actual specific gravity of aluminum.
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