/*! 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 103 A specific labor union is becomi... [FREE SOLUTION] | 91Ó°ÊÓ

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A specific labor union is becoming defensive about gross absenteeism by its members. The union decided to check this by monitoring a random sample of its members. The union leaders had always claimed that in a typical month, \(95 \%\) of its members are absent less than 10 hours per month. Use the data to respond to this claim. Use a one-sided tolerance limit and choose the confidence level to be \(99 \%\). Be sure to interpret what you learned from the tolerance limit calculation. The number of members in this sample was 300 . The number of hours absent was recorded for each of the 300 members. The results were \(x=6.5\) hours and \(s=2.5\) hours.

Short Answer

Expert verified
Using the given data, a one-sided tolerance limit calculation shows that there's a 99% chance that a member selected at random would be absent less than the calculated threshold value of hours in a month. This value should then be used to validate the claim made by the union leaders.

Step by step solution

01

Determine the Standard Error of the Mean.

The first step is to calculate the standard error of the mean (SEM). The formula for SEM is \(SEM=\frac{s}{\sqrt{n}}\), where \(s\) is the standard deviation and \(n\) is the sample size. Using the provided values, plug 2.5 for \(s\) and 300 for \(n\) to get \(SEM=\frac{2.5}{\sqrt{300}}\).
02

Find the Value of Z

Next, find the Z value for the 99% confidence level. A Z value is a measure of how many standard deviations an element is from the mean. For a one-sided test at the 99% confidence level, the Z value is 2.33. This value can be obtained from a standard Z table.
03

Calculate the Tolerance Limit

Now, calculate the one-sided tolerance limit using the formula \(TL = x + (Z * SEM)\), where \(x\) is the mean, \(Z\) is the Z value, and \(SEM\) is the standard error of the mean. Plug in 6.5 for \(x\), 2.33 for \(Z\), and the value calculated in step 1 for \(SEM\), to get the tolerance limit.
04

Interpret the Tolerance Limit.

The calculated tolerance limit is a threshold value. If it is less than 10, this would mean that there's a 99% chance that a member selected at random would be absent less than this threshold value of hours in a month. Compare the tolerance limit with the claim made by the union leaders and draw a valid conclusion.

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Key Concepts

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

Tolerance Limits
A tolerance limit helps us determine a range in which we expect a certain percentage of the population to fall. In the exercise, this concept is crucial to verify the union's claim. To do this, a one-sided tolerance limit is calculated. This involves finding the maximum number of hours that 99% of union members are absent, based on the sampled data. The one-sided aspect here means we're only considering an upper bound.The formula used is \( TL = x + (Z \times SEM) \), where \( x \) is the mean of the sample, \( Z \) is the Z-score corresponding to the desired confidence level, and \( SEM \) is the Standard Error of the Mean. In simple terms, if this calculated limit is found to be less than 10 hours, it supports the union's claim that 95% of their members are absent less than 10 hours per month. This tolerance limit calculation acts as a statistical tool to help understand and verify claims about large populations based on smaller samples.
Standard Error of the Mean
The Standard Error of the Mean (SEM) is essential for understanding how much the sample mean could vary from the actual population mean. It essentially measures the variability of the sample mean, finding how bulletproof it is in estimating the real average.The SEM is calculated using the formula \( SEM = \frac{s}{\sqrt{n}} \), where \( s \) stands for the standard deviation and \( n \) represents the sample size. In our exercise, the sample's standard deviation \( s \) is 2.5, and the sample size \( n \) is 300.A smaller SEM value indicates that the sample mean is a more precise estimate of the population mean. In other words, it assures us of the accuracy and reliability of our sample mean calculations while making tolerance limit decisions.
Confidence Interval
A confidence interval gives us a range of values in which we expect the true population parameter (like the mean) to lie. It provides a more comprehensive insight than a single estimate, offering a degree of certainty about where the true value likely falls. For the given problem, the Z value for a 99% confidence level is used to find the tolerance limit. This Z value reflects how many standard deviations away a point can lie from the mean with the corresponding level of confidence. By employing the confidence interval concept, we calculate and interpret statistical tolerance limits. Thus, we can make evidence-based statements regarding the overall absenteeism of the union members with a specified level of confidence.
One-Sided Test
A one-sided test in statistics is applied when the research or claim focuses on a specific direction of effect. In simpler terms, it checks if a population parameter (like the mean) is either greater than or less than a specified value, but not both. In the exercise, a one-sided test is utilized to verify if 99% of union members are absent less than a specified threshold (10 hours). The threshold's one-sided nature means we're only considering it as an upper bound. That differentiates it from a two-sided test, which would check both above and below this threshold. Hence, applying a one-sided test is particularly useful when the research question asks about exceeding or not exceeding a certain value, helping us draw more focused conclusions based on our data and hypothesis.

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Most popular questions from this chapter

A certain geneticist is interested in the proportion of males and females in the population that have a certain minor blood disorder. In a random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder. Compute a \(95 \%\) confidence interval for the difference between the proportion of males and females that have the blood disorder.

A random sample of 100 automobile owners shows that, in the state of Virginia, an automobile is driven on the average 23,500 kilometers per year with a standard deviation of 3900 kilometers. Assume the distribution of measurements to be approximately normal. (a) Construct a \(99 \%\) confidence interval for the average number of kilometers an automobile is driven annually in Virginia. (b) What can we assert with \(99 \%\) confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23.500 kilometers per year?

A random sample of size \(n_{1}=25\) taken from a normal population with a standard deviation \(\sigma_{1}=5\) has a mean \(\bar{x}_{1}=80 .\) A second random sample of size \(n_{2}=36,\) taken from a different normal population with a standard deviation \(\sigma_{2}=3,\) has a mean \(x_{2}=75 .\) Find a \(94 \%\) confidence interval for \(\mu_{1}-\mu_{2}\).

An alternative form of estimation is accomplished through the method of moments. The method involves equating the population mean and variance to the corresponding sample mean \(\overline{\mathcal{T}}\) and sample variance \(\mathrm{s}^{2}\) and solving for the parameter, the result being the moment estimators. In the case of a single parameter, only the means are used. Give an argument that in the case of the Poisson distribution the maximum likelihood estimator and moment estimators are the same.

An automotive company is considering two types of batteries for its automobile. Sample information on the life of the battery is being used. Twenty batteries of type \(A\) and twenty batteries of type \(B\) are being used. The summary statistics are \(X A=32.91\), \(x_{B}=30.47, s_{A}=1.57,\) and \(s_{B}-1.74 .\) Assume the data on each battery are normally distributed and assume \(\sigma_{A}=\sigma_{B}\) (a) Find a \(95 \%\) confidence interval on \(\mu_{A}-\mu_{B}\). (b) Draw some conclusion from (a) that provides some insight into whether \(A\) or \(B\) should be adopted.

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