/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 A sample of 250 workers aged 16 ... [FREE SOLUTION] | 91Ó°ÊÓ

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A sample of 250 workers aged 16 and older produced an average length of time with the current employer ("job tenure") of 4.4 years with standard deviation 3.8 years. Construct a \(99.9 \%\) confidence interval for the mean job tenure of all workers aged 16 or older.

Short Answer

Expert verified
The 99.9% confidence interval is \([3.61, 5.19]\) years.

Step by step solution

01

Identify the given information

We are given a sample size ( ) of 250 workers, an average job tenure ( ) of 4.4 years, and a standard deviation ( ) of 3.8 years. Also, we have a confidence level of 99.9%.
02

Determine the appropriate distribution

Since the sample size is large (n = 250), we will use the normal distribution to construct the confidence interval. 99.9% confidence level corresponds to a critical value, which we will find using the z-table.
03

Find the critical value (z*)

For a 99.9% confidence level, the critical value ( ) can be found using a standard normal distribution table. The critical value is approximately 3.291.
04

Calculate the standard error (SE)

The standard error of the sample mean is calculated using the formula \( SE = \frac{\sigma}{\sqrt{n}} \). Substitute \( \sigma = 3.8 \) and \( n = 250 \) into the formula to get \( SE = \frac{3.8}{\sqrt{250}} \approx 0.24 \).
05

Calculate the confidence interval

The formula for the confidence interval is \( \bar{x} \pm z^* \times SE \). Plug in \( \bar{x} = 4.4 \), \( z^* = 3.291 \), and \( SE = 0.24 \) into the formula to get the confidence interval: \( 4.4 \pm 3.291 \times 0.24 \approx 4.4 \pm 0.79 \). This results in the interval: \( [3.61, 5.19] \).

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Key Concepts

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

Understanding Sample Size
In statistics, the term "sample size" refers to the number of observations or data points collected in a study, which is represented by the variable \( n \). It is a crucial element in determining the reliability and accuracy of your statistical inferences.
An appropriate sample size helps ensure that the sample data properly represents the larger population.
When the sample size is large, the estimated statistics (like the sample mean and variance) tend to approximate the true population parameters more closely. In our exercise, a sample size of 250 was used to estimate the average job tenure of all workers aged 16 and older. Because this is a large sample, we can use the normal distribution approximation to calculate the confidence interval effectively.
Choosing a larger sample size can reduce sampling error, enhancing the precision of your estimate. However, larger samples may be more costly or time-consuming to collect, so there's always a balance to strike between precision and resources.
Understanding Standard Error
The standard error quantifies the amount of variability in the sample mean from the true population mean. It is a critical component in the construction of confidence intervals.
The standard error is calculated using the formula \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
In our scenario, with a population standard deviation of 3.8 years and a sample size of 250, the standard error is approximately 0.24. This indicates that the average job tenure from different samples would deviate by about 0.24 years from the true population mean.
A smaller standard error means that our sample mean is more tightly clustered around the population mean, leading to narrower confidence intervals and more precise estimates. Lowering the standard deviation or increasing the sample size can help achieve a smaller standard error.
Understanding the Z-Score
The z-score is a standard score that indicates how many standard deviations an element is from the mean. When constructing confidence intervals, the z-score (represented as \( z^* \)) is used to find critical values based on the preferred level of confidence.
For example, a 99.9% confidence level corresponds to a critical z-score of approximately 3.291, as shown in the exercise. This value means that there is only a 0.1% probability that the true population mean falls outside the interval when constructing a 99.9% confidence interval.
You determine the critical z-score using a standard normal distribution table, which lists the cumulative probabilities of a standard normal distribution.
Utilizing the z-score, you can calculate the margin of error and ultimately the confidence interval. Higher confidence levels demand higher z-scores, resulting in wider intervals to ensure the true parameter is captured within the interval.

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Most popular questions from this chapter

A random sample is drawn from a population of unknown standard deviation. Construct a \(99 \%\) confidence interval for the population mean based on the information given. a. \(\quad n=49, x-17.1, s=2.1\) b. \(\quad n=169, x-=17.1, s=2.1\)

A government agency was charged by the legislature with estimating the length of time it takes citizens to fill out various forms. Two hundred randomly selected adults were timed as they filled out a particular form. The times required had mean 12.8 minutes with standard deviation 1.7 minutes. Construct a \(90 \%\) confidence interval for the mean time taken for all adults to fill out this form.

The designer of a garbage truck that lifts roll-out containers must estimate the mean weight the truck will lift at each collection point. A random sample of 325 containers of garbage on current collection routes yielded \(x-=75.3 \mathrm{lb}, \mathrm{s}=12.8 \mathrm{lb}\). Construct a \(99.8 \%\) confidence interval for the mean weight the trucks must lift each time.

Wildlife researchers tranquilized and weighed three adult male polar bears. The data (in pounds) are: 926,742,1,109 . Assume the weights of all bears are normally distributed. a. Construct an \(80 \%\) confidence interval for the mean weight of all adult male polar bears using these data. b. Convert the three weights in pounds to weights in kilograms using the conversion \(1 \mathrm{lb}=0.453 \mathrm{~kg}\) (so the first datum changes to \((\mathrm{g} 26)(0.453)-410)\). Use the converted data to construct an \(80 \%\) confidence interval for the mean weight of all adult male polar bears expressed in kilograms. c. Convert your answer in part (a) into kilograms directly and compare it to your answer in (b). This illustrates that if you construct a confidence interval in one system of units you can convert it directly into another system of units without having to convert all the data to the new units.

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