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Chapter 5: Problem 46

The inside diameter of a randomly selected piston ring is a random variable with mean value \(12 \mathrm{~cm}\) and standard deviation \(.04 \mathrm{~cm}\). a. If \(\bar{X}\) is the sample mean diameter for a random sample of \(n=16\) rings, where is the sampling distribution of \(\bar{X}\) centered, and what is the standard deviation of the \(\bar{X}\) distribution? b. Answer the questions posed in part (a) for a sample size of \(n=64\) rings. c. For which of the two random samples, the one of part (a) or the one of part (b), is \(\bar{X}\) more likely to be within \(.01 \mathrm{~cm}\) of \(12 \mathrm{~cm}\) ? Explain your reasoning.

Short Answer

Expert verified
\( \bar{X} \) for \( n=64 \) is more likely to be within 0.01 cm of 12 cm.

Step by step solution

01

Understand the Mean of the Sampling Distribution

The mean of the sampling distribution \( \bar{X} \) for any sample size \( n \) is the same as the mean of the population from which the samples are drawn. Therefore, for both \( n=16 \) and \( n=64 \), the mean of the sampling distribution \( \bar{X} \) is \( 12 \mathrm{~cm} \).
02

Calculate the Standard Deviation of the Sampling Distribution for n=16

The standard deviation of the sampling distribution, also known as the standard error, is found using the formula \( \frac{\sigma}{\sqrt{n}} \). For \( n=16 \):\[\text{Standard Deviation (SD)} = \frac{0.04}{\sqrt{16}} = \frac{0.04}{4} = 0.01 \mathrm{~cm}\]
03

Calculate the Standard Deviation of the Sampling Distribution for n=64

Similarly, for \( n=64 \):\[\text{Standard Deviation (SD)} = \frac{0.04}{\sqrt{64}} = \frac{0.04}{8} = 0.005 \mathrm{~cm}\]
04

Analyze Which Sample Likely Results in a Closer Mean to 12 cm

The smaller the standard deviation of \( \bar{X} \), the tighter the distribution is around the mean. For \( n=64 \), the standard deviation is smaller (0.005 cm compared to 0.01 cm for \( n=16 \)), indicating that \( \bar{X} \) is more likely to be close to the mean 12 cm when \( n=64 \). Therefore, \( \bar{X} \) for \( n=64 \) is more likely to be within 0.01 cm of 12 cm.

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Key Concepts

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

Mean of Sampling Distribution
When we talk about the mean of a sampling distribution, it's essential to understand that this mean is always calculated using the mean of the original population. Regardless of the sample size, the mean of the sampling distribution is identical to the population mean. Here, the population mean for the inside diameter of a piston ring is given as 12 cm.

This same mean applies to any sample mean, symbolized as \( \bar{X} \). Whether we collect a sample of 16 rings or 64 rings, the sampling distribution's mean remains at 12 cm.

This consistency is why we say the sampling distribution is centered at the population mean. It indicates that even though individual sample means vary, they hover around this population mean, providing an unbiased estimate as the sample size increases.
Standard Error
The standard error is a crucial concept in statistics and relates to the precision of the sample mean as an estimate of the population mean. It tells us how much sampling fluctuation we can expect.

The formula to calculate the standard error is \( \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation, and \( n \) is the sample size.

For the piston rings with a standard deviation of 0.04 cm:
  • When \( n = 16 \), the standard error becomes \( 0.01 \) cm.
  • When \( n = 64 \), this standard error reduces to \( 0.005 \) cm.
This shows that as the sample size increases, the standard error decreases, leading to a more precise estimate of the population mean.
Sample Size Effect on Variability
Sample size has a significant impact on the variability of the sampling distribution. Larger samples tend to provide more reliable estimates of the population mean because they reduce the effect of sample variability.

With a sample size of 16 rings, we calculated a standard error of 0.01 cm. For a sample of 64 rings, this standard error decreased to 0.005 cm.

What does this imply? As the sample size grows:
  • The sampling distribution becomes narrower.
  • The estimates of the sample mean are less scattered around the population mean.
  • There's an increased likelihood of the sample mean being very close to the actual population mean.
Smaller standard error at a larger sample size, like 64, signifies less variability, indicating that the sample mean is more likely to be within 0.01 cm of the 12 cm population mean compared to a sample of 16. By understanding this, you can see the importance of considering sample size in statistical analysis.

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Most popular questions from this chapter

Consider a system consisting of three components as pictured. The system will continue to function as long as the first component functions and either component 2 or component 3 functions. Let \(X_{1}, X_{2}\), and \(X_{3}\) denote the lifetimes of components 1,2 , and 3 , respectively. Suppose the \(X_{i}\) s are independent of one another and each \(X_{i}\) has an exponential distribution with parameter \(\lambda\). a. Let \(Y\) denote the system lifetime. Obtain the cumulative distribution function of \(Y\) and differentiate to obtain the pdf. [Hint: \(F(y)=P(Y \leq y)\); express the event \(\\{Y \leq y\\}\) in terms of unions and/or intersections of the three events \(\left\\{X_{1} \leq y\right\\},\left\\{X_{2} \leq y\right\\}\), and \(\left\\{X_{3} \leq y\right\\}\).] b. Compute the expected system lifetime.

In cost estimation, the total cost of a project is the sum of component task costs. Each of these costs is a random variable with a probability distribution. It is customary to obtain information about the total cost distribution by adding together characteristics of the individual component cost distributions-this is called the "roll-up" procedure. For example, \(E\left(X_{1}+\cdots+X_{n}\right)=E\left(X_{1}\right)+\cdots+E\left(X_{n}\right)\), so the roll-up procedure is valid for mean cost. Suppose that there are two component tasks and that \(X_{1}\) and \(X_{2}\) are independent, normally distributed random variables. Is the roll-up procedure valid for the 75 th percentile? That is, is the 75 th percentile of the distribution of \(X_{1}+X_{2}\) the same as the sum of the 75 th percentiles of the two individual distributions? If not, what is the relationship between the percentile of the sum and the sum of percentiles? For what percentiles is the roll-up procedure valid in this case?

In an area having sandy soil, 50 small trees of a certain type were planted, and another 50 trees were planted in an area having clay soil. Let \(X=\) the number of trees planted in sandy soil that survive 1 year and \(Y=\) the number of trees planted in clay soil that survive 1 year. If the probability that a tree planted in sandy soil will survive 1 year is \(.7\) and the probability of 1-year survival in clay soil is .6, compute an approximation to \(P(-5 \leq X-Y \leq 5\) ) (do not bother with the continuity correction).

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of \(1.2\). a. If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 9 pins is at least 51 ? b. What is the (approximate) probability that the sample mean hardness for a random sample of 40 pins is at least 51 ?

A rock specimen from a particular area is randomly selected and weighed two different times. Let \(W\) denote the actual weight and \(X_{1}\) and \(X_{2}\) the two measured weights. Then \(X_{1}=\) \(W+E_{1}\) and \(X_{2}=W+E_{2}\), where \(E_{1}\) and \(E_{2}\) are the two measurement errors. Suppose that the \(E_{i} \mathrm{~s}\) are independent of one another and of \(W\) and that \(V\left(E_{1}\right)=V\left(E_{2}\right)=\sigma_{E^{2}}^{2}\). a. Express \(\rho\), the correlation coefficient between the two measured weights \(X_{1}\) and \(X_{2}\), in terms of \(\sigma_{W}^{2}\), the variance of actual weight, and \(\sigma_{X}^{2}\), the variance of measured weight. b. Compute \(\rho\) when \(\sigma_{W}=1 \mathrm{~kg}\) and \(\sigma_{E}=.01 \mathrm{~kg}\).
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