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Construct the appropriate confidence interval. A simple random sample of size \(n=300\) individuals who are currently employed is asked if they work at home at least once per week. Of the 300 employed individuals surveyed, 35 responded that they did work at home at least once per week. Construct a \(99 \%\) confidence interval for the population proportion of employed individuals who work at home at least once per week.

Short Answer

Expert verified
The 99% confidence interval is (0.0679, 0.1654).

Step by step solution

01

- Identify the sample proportion

The sample proportion (\(\hat{p}\) is the number of individuals who work at home at least once per week divided by the total number of individuals surveyed. Here, \(n = 300\) and the number of individuals who work at home is 35. So, \(\hat{p} = \frac{35}{300} = 0.1167\).
02

- Determine the standard error

The standard error (SE) of the sample proportion is calculated using the formula \(\text{SE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\). Plugging in the values, \(\text{SE} = \sqrt{\frac{0.1167 \(1 - 0.1167\)}{300}} \approx 0.0189\).
03

- Find the Z-value for the confidence level

For a 99% confidence interval, the Z-value is approximately 2.576.
04

- Calculate the margin of error

The margin of error (ME) is found using the formula \(\text{ME} = Z \times \text{SE}\). So, \(\text{ME} = 2.576 \times 0.0189 \approx 0.0487\).
05

- Construct the confidence interval

The confidence interval is given by \(\text{CI} = \hat{p} \pm \text{ME}\). Substituting the values, \(0.1167 \pm 0.0487\). Thus the confidence interval is (0.0679, 0.1654).

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Key Concepts

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

sample proportion
In the context of survey statistics, the sample proportion is a measure used to estimate the proportion of a certain characteristic within a population. To calculate the sample proportion, you would divide the number of individuals exhibiting the characteristic of interest by the total number of individuals surveyed. For example, in our exercise, a total of 300 employed individuals were surveyed, and 35 of them reported working from home at least once per week. Therefore, the sample proportion, denoted as \( \hat{p} \), can be computed as follows: \ \[ \hat{p} = \frac{35}{300} = 0.1167 \ \]. This means that approximately 11.67% of the sampled employees work from home at least once per week. Calculating the sample proportion is the first essential step in constructing a confidence interval.
standard error
The standard error of a sample proportion measures the variability or precision of the sample proportion as an estimate of the population proportion. It tells us how much the sample proportion might differ from the true population proportion. The smaller the standard error, the more precise our sample proportion estimate is.
To calculate the standard error, we use the formula: \[ \text{SE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \ \].
Here, \ \( \hat{p} \) is our sample proportion (0.1167) and \ \( \text{n} \) is the sample size (300). Plugging in these values, we get: \[ \text{SE} = \sqrt{\frac{0.1167 \(1 - 0.1167 \)}{300}} \ \approx 0.0189 \].
This result indicates that the standard error for our sample proportion is 0.0189, and it will be used to calculate the margin of error next.
margin of error
The margin of error helps to determine the range within which the true population proportion is likely to fall. It provides us with a measure of uncertainty around our sample proportion estimate. To calculate the margin of error, we use the formula: \ \[ \text{ME} = Z \times \ \text{SE} \ \].
Here, \ \( Z \) is the Z-value corresponding to our desired confidence level, and \ \( \text{SE} \) is the standard error calculated in the previous step. For a 99% confidence interval, the Z-value is approximately 2.576.
Plugging in the values, we get: \[ \text{ME} = 2.576 \times 0.0189 \]
\[ \text{ME} \ \approx 0.0487 \]
This margin of error means that there is a predicted range of error of about 4.87% in our sample proportion estimate.
Z-value
The Z-value, also known as the Z-score, is a critical value derived from the standard normal distribution. It represents the number of standard deviations a data point is from the mean. In the context of confidence intervals, the Z-value corresponds to the desired level of confidence that the true population proportion lies within the calculated interval.
For a 99% confidence interval, the Z-value is approximately 2.576. This value can be found in Z-tables, which provide the critical values for different confidence levels. Here's why it is important:
  • The Z-value ensures that our interval has the correct level of confidence.
  • It reflects how rare or common the sample outcome is in the context of the normal distribution.
  • It is crucial to use the Z-value accurately so that we reliably capture the desired confidence level in our interval.
    Using this Z-value along with the standard error, we can determine the margin of error, which then allows us to construct our confidence interval.

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Most popular questions from this chapter

Suppose the following data represent the heights (in inches) of a random sample of males: 68,72,73,70,75,71 . Which of the following could be a possible bootstrap sample? (a) 68,72,72,68,70,71 (b) 75,72,73,73,68 (c) 70,71,68,73,73,71,68 (d) 72,73,75,71,73,63 (e) 68,72,73,71,72,73

Explain why the \(t\) -distribution has less spread as the number of degrees of freedom increases.

Certain statistics are difficult to bootstrap. One such statistic is the median. Consider the following to see why. (a) Simulate obtaining a random sample of 12 IQ scores. Recall IQ scores are approximately normally distributed with mean 100 and standard deviation \(15 .\) (b) Given that IQ scores are normally distributed, what is the median IQ score? (c) Obtain 1000 bootstrap samples from the data in part (a). Find the median of each bootstrap sample. (d) Draw a histogram of the bootstrap medians from part (c). What do you notice about the distribution? Find the \(95 \%\) confidence interval based on the 1000 bootstrap medians using the percentile method. (e) Repeat parts (a) through (d) using a random sample of 13 IQ scores. (f) Conclude that finding confidence intervals for medians is best if done where the sample size is even.

A Rasmussen Reports national survey of 1000 adult Americans found that \(18 \%\) dreaded Valentine's Day. The margin of error for the survey was 4.5 percentage points with \(95 \%\) confidence. Explain what this means.

Katrina wants to estimate the proportion of adult Americans who read at least 10 books last year. To do so, she obtains a simple random sample of 100 adult Americans and constructs a \(95 \%\) confidence interval. Matthew also wants to estimate the proportion of adult Americans who read at least 10 books last year. He obtains a simple random sample of 400 adult Americans and constructs a \(99 \%\) confidence interval. Assuming both Katrina and Matthew obtained the same point estimate, whose estimate will have the smaller margin of error? Justify your answer.

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