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In a poll to estimate presidential popularity, each person in a random sample of 1,000 voters was asked to agree with one of the following statements: 1\. The President is doing a good job. 2\. The President is doing a poor job. 3\. I have no opinion. A total of 560 respondents selected the first statement, indicating they thought the President was doing a good job. a. Construct a 95 percent confidence interval for the proportion of respondents who feel the President is doing a good job. b. Based on your interval in part (a), is it reasonable to conclude that a majority (more than half) of the population believes the President is doing a good job?

Short Answer

Expert verified
Yes, the interval (0.5302, 0.5898) indicates a majority believes the President is doing a good job.

Step by step solution

01

Identify the Sample Proportion

First, determine the sample proportion of respondents who think the President is doing a good job. This is given by dividing the number of respondents who think the President is doing a good job by the total number of respondents. Thus, \( \hat{p} = \frac{560}{1000} = 0.56 \).
02

Calculate the Standard Error

The standard error of the sample proportion is calculated using the formula: \( SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \), where \( \hat{p} = 0.56 \) and \( n = 1000 \). Therefore, \( SE = \sqrt{\frac{0.56 \times 0.44}{1000}} = 0.0157 \).
03

Determine the Z-Score for 95% Confidence

For a 95% confidence interval, the Z-score is 1.96. This value corresponds to the critical value for a two-tailed test covering 95% of the normal distribution.
04

Calculate the Confidence Interval

The confidence interval is calculated as \( \hat{p} \pm Z \times SE \). Substituting the values, we have \( 0.56 \pm 1.96 \times 0.0157 \). This results in the interval \( (0.5302, 0.5898) \).
05

Interpret the Confidence Interval

The 95% confidence interval for the proportion of respondents who feel the President is doing a good job is between 53.02% and 58.98%. Since the entire interval is above 50%, it suggests that it is reasonable to conclude that more than half of the population believes the President is doing a good job.

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Key Concepts

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

Sample Proportion
When we're estimating public opinion or any proportion of a population, the sample proportion (\( \hat{p} \) ) is our starting point. It's essentially a snapshot of our random sample that we hope closely represents the actual population.
For instance, in the original exercise, 560 out of 1,000 poll respondents said the President is doing well. To find \( \hat{p} \), we divide 560 by 1,000, giving us 0.56. This means 56% of respondents in our sample believe the President is doing a good job.
The sample proportion helps us estimate the true proportion of the entire population. But remember, it comes from a sample, so there’s naturally some uncertainty attached. That's where confidence intervals come into play, allowing us to account for possible sampling variations.
Standard Error
The standard error (SE) is crucial for measuring how much our sample proportion might vary from the actual population proportion. It's like a built-in error checker for our estimates!
In the context of our exercise, the formula for the standard error is: \[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]
Here, \( \hat{p} \) is 0.56 and \( n \) is 1,000. Plugging these into the formula gives us a standard error of approximately 0.0157.
What does this number tell us? It provides insight into how much we'd expect our sample proportion (56%) to vary if we took another sample from the same population. Smaller SE indicates more precision, while a larger SE suggests more variability.
Z-Score
The Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. When constructing a confidence interval, the Z-score helps determine how far away from the sample mean (or proportion) our interval should extend, considering the variability of the data.
In our exercise, we're using a 95% confidence level which implies a Z-score of 1.96. This value is widely used because it corresponds to covering the central 95% of a standard normal distribution, leaving 2.5% in each tail.
By using this Z-score, along with our standard error, we can compute a confidence interval that provides a range around our sample proportion, offering a likely span within which the true population proportion may lie.
Normal Distribution
Normal distribution is a foundational concept in statistics, often referred to as a bell curve due to its shape. It's used to model a variety of phenomena and plays a vital role in confidence interval calculations.
This distribution is symmetrical, centered around the mean, and characterized by its standard deviation which determines the width of the curve. Large samples (like our 1,000 respondents) most often show properties of normal distribution according to the Central Limit Theorem.
In the exercise, we assume the sample proportion distribution is approximately normal, which lets us use the Z-score for creating the confidence interval. This assumption is valid due to both the sample's size and its binary nature (yes/no).
The normal distribution helps ensure that the intervals calculated are reliable and enable us to make informed conclusions about the underlying population.

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Most popular questions from this chapter

The attendance at the Savannah Colts minor league baseball game last night was \(400 .\) A random sample of 50 of those in attendance revealed that the mean number of soft drinks consumed per person was 1.86 with a standard deviation of \(0.50 .\) Develop a 99 percent confidence interval for the mean number of soft drinks consumed per person.

The First National Bank of Wilson has 650 checking account customers. A recent sample of 50 of these customers showed 26 to have a Visa card with the bank. Construct the 99 percent confidence interval for the proportion of checking account customers who have a Visa card with the bank.

The Huntington National Bank, like most other large banks, found that using automatic teller machines (ATMs) reduces the cost of routine bank transactions. Huntington installed an ATM in the corporate offices of the Fun Toy Company. The ATM is for the exclusive use of Fun's 605 employees. After several months of operation, a sample of 100 employees revealed the following use of the ATM machine by Fun employees in a month. $$\begin{array}{|cc|}\hline \text { Number of Times } & \\\\\text { ATM Used } & \text { Frequency } \\\\\hline 0 & 25 \\\1 & 30 \\\2 & 20 \\\3 & 10 \\\4 & 10 \\\5 & 5 \\\\\hline\end{array}$$ a. What is the estimate of the proportion of employees who do not use the ATM in a month? b. Develop a 95 percent confidence interval for this estimate. Can Huntington be sure that at least 40 percent of the employees of Fun Toy Company will use the ATM? c. How many transactions does the average Fun employee make per month? d. Develop a 95 percent confidence interval for the mean number of transactions per month. e. Is it possible that the population mean is 0? Explain.

As a condition of employment, Fashion Industries applicants must pass a drug test. Of the last 220 applicants 14 failed the test. Develop a 99 percent confidence interval for the proportion of applicants that fail the test. Would it be reasonable to conclude that more than 10 percent of the applicants are now failing the test? In addition to the testing of applicants, Fashion Industries randomly tests its employees throughout the year. Last year in the 400 random tests conducted, 14 employees failed the test. Would it be reasonable to conclude that less than 5 percent of the employees are not able to pass the random drug test?

The estimate of the population proportion is to be within plus or minus \(.10,\) with a 99 percent level of confidence. The best estimate of the population proportion is .45. How large a sample is required?

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