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A sample of 20 managers was taken, and they were asked whether or not they usually take work home. The responses of these managers are given below, where yes indicates they usually take work home and no means they do not. \(\begin{array}{llllllll}\text { Yes } & \text { Yes } & \text { No } & \text { No } & \text { No } & \text { Yes } & \text { No } & \text { No } & \text { No } & \text { No } \\ \text { Yes } & \text { Yes } & \text { No } & \text { Yes } & \text { Yes } & \text { No } & \text { No } & \text { No } & \text { No } & \text { Yes }\end{array}\) Make a \(99 \%\) confidence interval for the percentage of all managers who take work home.

Short Answer

Expert verified
The 99% confidence interval for the proportion of managers who take work home is calculated based on the given sample. The exact intervals depend on the results from steps 1-3.

Step by step solution

01

Calculation of the sample proportion

First, tally the 'Yes' and 'No' responses. Count the number of 'Yes' responses and divide it by the total number of responses (20) to get the sample proportion \( p \).
02

Calculation of the standard error

Standard error (SE) can be calculated using the formula: SE = \( \sqrt{\frac{p (1 - p)}{n}} \) where \( n \) is the total number of responses.
03

Calculate the confidence interval

For a 99% confidence interval, the critical value from the Z-table is 2.58. The confidence interval is then calculated as \( p \pm (2.58 \times SE) \). The result will be your lower and upper bound for the confidence interval.

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Key Concepts

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

Sample Proportion
To understand a sample proportion, let's first break down the survey. We start by looking at the responses to see how many managers said 'yes,' meaning they usually take work home. Add up all the 'Yes' answers from the given responses to find the total number. In this example, there are 8 managers who answered 'Yes'. This number is then divided by the total number of managers surveyed, which is 20 in this exercise. This gives us the sample proportion, represented by the symbol \( p \). So, here, \( p = \frac{8}{20} = 0.4 \), or 40%. This means that in the sample, 40% of the managers usually take work home.
Understanding the sample proportion is crucial because it helps estimate the proportion of a larger population, based on a small sample. In statistics, the sample proportion is often used to make predictions about an entire population's behavior or characteristics.
Standard Error
The standard error is a measure that tells us how much the sample proportion might vary from the actual population proportion. It gives an idea of the uncertainty or the potential error in our estimate.
To calculate the standard error for a proportion, use the formula: \( SE = \sqrt{\frac{p(1 - p)}{n}} \). This formula requires the sample proportion \( p \) and the total number of responses \( n \).
In our example, with \( p = 0.4 \) and \( n = 20 \), the calculation would be:
  • \( SE = \sqrt{\frac{0.4 \times (1 - 0.4)}{20}} \)
This standard error helps us understand the variability of the sample, offering insight into how well the sample proportion represents the larger population. The smaller the standard error, the closer the sample proportion is likely to be to the population proportion.
Z-table Critical Value
The Z-table critical value is a number from the Z-table that helps set the bounds for a confidence interval. It represents the number of standard deviations a data point is from the mean in a normal distribution. Choosing the right Z-value depends on the level of confidence we want to have in our interval estimate.
For a 99% confidence interval, as noted in the exercise, the critical value is 2.58. This high Z-value reflects the high level of certainty we want when estimating the population proportion.
To use the critical value, multiply it by the standard error, then add and subtract the result from the sample proportion \( p \) to find the confidence interval. This gives you the range within which the true population proportion is likely to fall with 99% confidence. The calculation looks like this:
  • Confidence interval = \( p \pm (Z \times SE) \)
The larger the Z-value, the wider the confidence interval, reflecting increased certainty regarding the estimate, even as the specific population numbers remain uncertain.

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Most popular questions from this chapter

A computer company that recently developed a new software product wanted to estimate the mean time taken to learn how to use this software by people who are somewhat familiar with computers. A random sample of 12 such persons was selected. The following data give the times taken (in hours) by these persons to learn how to use this software. $$ \begin{array}{llllll} 1.75 & 2.25 & 2.40 & 1.90 & 1.50 & 2.75 \\ 2.15 & 2.25 & 1.80 & 2.20 & 3.25 & 2.60 \end{array} $$ Construct a \(95 \%\) confidence interval for the population mean. Assume that the times taken by all persons who are somewhat familiar with computers to learn how to use this software are approximately normally distributed.

Suppose, for a sample selected from a population, \(\bar{x}=25.5\) and \(s=4.9\). a. Construct a \(95 \%\) confidence interval for \(\mu\) assuming \(n=47\). b. Construct a \(99 \%\) confidence interval for \(\mu\) assuming \(n=47\). Is the width of the \(99 \%\) confidence interval larger than the width of the \(95 \%\) confidence interval calculated in part a? If yes, explain why. c. Find a \(95 \%\) confidence interval for \(\mu\) assuming \(n=32\). Is the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=32\) larger than the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=47\) calculated in part a? If so, why? Explain.

In June 2008 , SBRI Public Affairs conducted a telephone poll of 1004 adult Americans aged 18 and older. One of the questions asked was, "In the past year, was there ever a time when you ...?" Respondents could choose more than one of the answers mentioned. Of the respondents, \(64 \%\) said "cut back on vacations or entertainment because of their cost," \(37 \%\) said "failed to pay a bill on time," and \(25 \%\) said "have not gone to a doctor because of the cost." (Source: http://www.srbi.com/AmericansConcernEconomic.html.) Using these results, find a \(95 \%\) confidence interval for the corresponding population percentage for each answer. Write a one-page report to present these results to a group of college students who have not taken statistics. Your report should answer questions such as: (1) What is a confidence interval? (2) Why is a range of values more informative than a single percentage? (3) What does \(95 \%\) confidence mean in this context? (4) What assumptions, if any, are you making when you construct each confidence interval?

A random sample of 16 airline passengers at the Bay City airport showed that the mean time spent waiting in line to check in at the ticket counters was 31 minutes with a standard deviation of 7 minutes. Construct a \(99 \%\) confidence interval for the mean time spent waiting in line by all passengers at this airport. Assume that such waiting times for all passengers are normally distributed.

A sample of 1500 homes sold recently in a state gave the mean price of homes equal to \(\$ 299,720\). The population standard deviation of the prices of homes in this state is \(\$ 68,650\). Construct a \(99 \%\) confidence interval for the mean price of all homes in this state.

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