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The Sullivan Statistics Survey I asks, "Would you be willing to pay higher taxes if the tax revenue went directly toward deficit reduction?" Treat the survey respondents as a random sample of adult Americans. Go to www.pearsonhighered.com/sullivanstats to obtain the data file SullivanSurveyI using the file format of your choice for the version of the text you are using. The column "Deficit" has survey responses. Construct and interpret a \(90 \%\) confidence interval for the proportion of adult Americans who would be willing to pay higher taxes if the revenue went directly toward deficit reduction.

Short Answer

Expert verified
90% CI for the proportion is p̂ ± 1.645 × SE.

Step by step solution

01

- Obtain the Data

Go to the provided website www.pearsonhighered.com/sullivanstats to download the data file SullivanSurveyI in the file format that corresponds to your version of the text. Locate the column labeled 'Deficit' which contains the survey responses.
02

- Identify the Sample Proportion

Count the number of 'Yes' responses in the 'Deficit' column and denote this number as X. Let n be the total number of responses in the 'Deficit' column. The sample proportion (p̂) is then given by the ratio \(\frac{X}{n}\).
03

- Determine the Confidence Level and z-Value

For a 90% confidence level, the corresponding z-value (critical value) is 1.645. This value is obtained from the standard normal distribution table.
04

- Compute the Standard Error

Calculate the standard error (SE) of the sample proportion using the formula: \(\text{SE} = \sqrt{\frac{p̂(1 - p̂)}{n}}\).
05

- Construct the Confidence Interval

Construct the 90% confidence interval using the formula: \(\text{CI} = p̂ \pm (z \times \text{SE})\). This will give you the lower and upper bounds of the confidence interval.
06

- Interpret the Confidence Interval

Interpret the interval: With 90% confidence, it can be stated that the true proportion of adult Americans who would be willing to pay higher taxes if the revenue went directly toward deficit reduction lies within the calculated confidence interval.

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Key Concepts

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

Sample Proportion
The sample proportion is a fundamental concept in statistics. It represents the fraction of a sample that has a particular attribute. For this exercise, it’s the fraction of respondents who answered 'Yes' to the survey question about being willing to pay higher taxes for deficit reduction. First, you need the total number of 'Yes' responses, denoted as X. Then, divide X by the total number of responses, denoted as n. Mathematically, it’s represented as \(\hat{p} = \frac{X}{n} \). So, if 150 out of 500 respondents said 'Yes', the sample proportion \(\hat{p}\) would be \(\frac{150}{500} = 0.3\). This value gives you a first insight into what portion of the population might hold this view. Breaking it down into a step-by-step process makes it easier to understand:
  • Count the number of 'Yes' responses (X)
  • Count the total number of responses (n)
  • Divide X by n to get the sample proportion \(\hat{p}\)
Standard Error
Standard error measures the variability or the expected fluctuation in the sample proportion if we were to take multiple samples from the population. It provides insight into how much the sample proportion is likely to differ due to sampling variability. The formula for standard error (SE) is \( \text{SE} = \sqrt{ \frac{ \hat{p} ( 1 - \hat{p} )}{n} } \). For example, if the sample proportion \(\hat{p}\) is 0.3 and the sample size (n) is 500, the standard error would be \(\sqrt{ \frac{ 0.3 \times 0.7 }{ 500 } }\). Simplify the expression under the square root to get your final SE value. This step is essential because it helps in constructing confidence intervals:
  • Calculate \(\hat{p} \times ( 1 - \hat{p} ) \)
  • Divide by the total number of responses (n)
  • Take the square root of the result to get the standard error (SE)
z-value
The z-value is a critical value from the standard normal distribution corresponding to the desired confidence level. It tells how many standard deviations away from the mean our value sits. For a 90% confidence interval, the z-value is 1.645. This value comes from z-tables or statistical software. It is used in the final step of constructing the confidence interval, multiplying it by the standard error. Here’s a straightforward way to understand how it's applied:
  • Select the desired confidence level (e.g., 90%)
  • Refer to the z-table to find the corresponding z-value, here it’s 1.645
  • Use this z-value in calculating the confidence interval
In the formula for the confidence interval, \( \text{CI} = \hat{p} \pm ( z \times \text{SE})\), the z-value gives the range around the sample proportion within which we expect the true population proportion to lie with a specific level of confidence.

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Most popular questions from this chapter

A recent Gallup poll asked Americans to disclose the number of books they read during the previous year. Initial survey results indicate that \(s=16.6\) books. (a) How many subjects are needed to estimate the number of books Americans read the previous year within four books with \(95 \%\) confidence? (b) How many subjects are needed to estimate the number of books Americans read the previous year within two books with \(95 \%\) confidence? (c) What effect does doubling the required accuracy have on the sample size? (d) How many subjects are needed to estimate the number of books Americans read the previous year within four books with \(99 \%\) confidence? Compare this result to part (a). How does increasing the level of confidence in the estimate affect sample size? Why is this reasonable?

Dr. Paul Oswiecmiski wants to estimate the mean serum HDL cholesterol of all 20 - to 29 -year-old females. How many subjects are needed to estimate the mean serum HDL cholesterol of all 20 - to 29 -year-old females within 2 points with \(99 \%\) confidence assuming that \(s=13.4\) based on earlier studies? Suppose that Dr. Oswiecmiski would be content with \(95 \%\) confidence. How does the decrease in confidence affect the sample size required?

A school administrator is concerned about the amount of credit-card debt that college students have. She wishes to conduct a poll to estimate the percentage of full-time college students who have credit-card debt of \(\$ 2000\) or more. What size sample should be obtained if she wishes the estimate to be within 2.5 percentage points with \(94 \%\) confidence if (a) a pilot study indicates that the percentage is \(34 \% ?\) (b) no prior estimates are used?

A simple random sample of size \(n\) is drawn. The sample mean, \(\bar{x},\) is found to be \(35.1,\) and the sample standard deviation, \(s,\) is found to be \(8.7 .\) (a) Construct a \(90 \%\) confidence interval for \(\mu\) if the sample size, \(n,\) is \(40 .\) (b) Construct a \(90 \%\) confidence interval for \(\mu\) if the sample size, \(n\), is \(100 .\) How does increasing the sample size affect the margin of error, \(E ?\) (c) Construct a \(98 \%\) confidence interval for \(\mu\) if the sample size, \(n\), is \(40 .\) Compare the results to those obtained in part (a). How does increasing the level of confidence affect the margin of error, \(E ?\) (d) If the sample size is \(n=18\), what conditions must be satisfied to compute the confidence interval?

In a USA Today/Gallup poll, 768 of 1024 randomly selected adult Americans aged 18 or older stated that a candidate's positions on the issue of family values are extremely or very important in determining their vote for president. (a) Obtain a point estimate for the proportion of adult Americans aged 18 or older for which the issue of family values is extremely or very important in determining their vote for president. (b) Verify that the requirements for constructing a confidence interval for \(p\) are satisfied. (c) Construct a \(99 \%\) confidence interval for the proportion of adult Americans aged 18 or older for which the issue of family values is extremely or very important in determining their vote for president. (d) Is it possible that the proportion of adult Americans aged 18 or older for which the issue of family values is extremely or very important in determining their vote for president is below \(70 \%\) ? Is this likely? (e) Use the results of part (c) to construct a \(99 \%\) confidence interval for the proportion of adult Americans aged 18 or older for which the issue of family values is not extremely or very important in determining their vote for president.

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