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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 cm of \(12 \mathrm{~cm}\) ? Explain your reasoning.

Short Answer

Expert verified
The distribution is centered at 12 cm, with the sample size of 64 being more likely to be within 0.01 cm of the mean.

Step by step solution

01

Understand the Problem

We need to determine the mean and standard deviation of the sampling distribution of the sample mean diameter for different sample sizes and compare the distributions.
02

Definition and Calculation for Sample Size n=16

The mean of the sampling distribution of the sample mean \( \bar{X} \) is the same as the population mean, \( \mu = 12 \, \text{cm} \). The standard deviation of the sampling distribution, known as the standard error (SE), is calculated using the formula: \[ \text{SE} = \frac{\sigma}{\sqrt{n}} \]where \( \sigma = 0.04 \, \text{cm} \) is the population standard deviation and \( n = 16 \).So the standard error is:\[ \text{SE} = \frac{0.04}{\sqrt{16}} = \frac{0.04}{4} = 0.01 \, \text{cm} \]
03

Definition and Calculation for Sample Size n=64

For \( n = 64 \), the mean of the sampling distribution is still \( \mu = 12 \, \text{cm} \). The standard error is now calculated with:\[ \text{SE} = \frac{0.04}{\sqrt{64}} = \frac{0.04}{8} = 0.005 \, \text{cm} \]
04

Comparison of Sampling Distributions

For \( n = 16 \), the standard error is \( 0.01 \, \text{cm} \) and for \( n = 64 \), the standard error is \( 0.005 \, \text{cm} \). A smaller standard error means the sample mean is more likely to be close to the population mean. So, \( \bar{X} \) in part (b) is more likely to be within 0.01 cm of 12 cm because the standard error is smaller.

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Key Concepts

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

Standard Error
The standard error (SE) plays a crucial role when assessing the reliability of a sample mean as an estimate of the population mean. It measures how spread out the sample means are around the true population mean. This provides insight into the variability and consistency of our sample data.
To calculate the standard error, we use the formula:\[\text{SE} = \frac{\sigma}{\sqrt{n}}\]
Here, \( \sigma \) represents the population standard deviation, and \( n \) is the sample size.
  • As \( n \) increases, \( \sqrt{n} \) increases, leading to a smaller SE.
  • A small SE indicates that our sample mean is likely to be a good estimate of the population mean.
In this way, the SE helps us understand the level of precision we can expect when inferring from sample data.
Sample Mean
The sample mean, denoted as \( \bar{X} \), is an average of a set of measurements taken from a sample of the entire population. It is our best estimate of the population mean \( \mu \).
For practical purposes:
  • It is calculated by summing all measurements in a sample and dividing by the number of observations.
  • The sample mean serves as a point estimate, giving us an idea about the central tendency of the data observed in a sample.
In statistical inferences, the sample mean's reliability as an estimator improves with larger sample sizes, as indicated by reductions in standard error.
Standard Deviation
Standard deviation (SD) is a key statistical measure that describes the amount of variability or dispersion in a set of data points around the mean. In this context, for any random variable such as the inside diameter of piston rings:
  • The population standard deviation \( \sigma = 0.04 \) cm indicates typical deviations from the mean diameter of 12 cm.
  • Importantly, this value is used to derive the standard error, by quantifying the signal's noise in our sample mean estimates.
The role of SD becomes critical when predicting the likely variation of individual observations from the calculated mean.
Population Mean
The population mean, symbolized by \( \mu \), represents the true average of a certain characteristic of an entire population. In our piston ring example, the population mean is given as 12 cm.
Recognizing properties of the population mean is fundamental:
  • It serves as a parameter around which sample data give us valuable insights, through the mean and variability of the samples.
  • Sample means tend to converge on \( \mu \) as the number of observations \( n \) increases, due to the Law of Large Numbers.
For inference, knowing the population mean allows us to understand how representative the sample data is of the actual population.

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Most popular questions from this chapter

The time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal distribution with mean value \(10 \mathrm{~min}\) and standard deviation \(2 \mathrm{~min}\). If five individuals fill out a form on one day and six on another, what is the probability that the sample average amount of time taken on each day is at most 11 min?

The number of parking tickets issued in a certain city on any given weekday has a Poisson distribution with parameter \(\mu=50\). What is the approximate probability that a. Between 35 and 70 tickets are given out on a particular day? [Hint: When \(\mu\) is large, a Poisson rv has approximately a normal distribution.] b. The total number of tickets given out during a 5 -day week is between 225 and 275 ?

Manufacture of a certain component requires three different machining operations. Machining time for each operation has a normal distribution, and the three times are independent of one another. The mean values are 15,30 , and 20 min, respectively, and the standard deviations are 1,2 , and \(1.5 \mathrm{~min}\), respectively. What is the probability that it takes at most 1 hour of machining time to produce a randomly selected component?

Suppose that you have ten lightbulbs, that the lifetime of each is independent of all the other lifetimes, and that each lifetime has an exponential distribution with parameter \(\lambda\). a. What is the probability that all ten bulbs fail before time \(r\) ? b. What is the probability that exactly \(k\) of the ten bulbs fail before time \(t\) ? c. Suppose that nine of the bulbs have lifetimes that are exponentially distributed with parameter \(\lambda\) and that the remaining bulb has a lifetime that is exponentially distributed with parameter \(\theta\) (it is made by another manufacturer). What is the probability that exactly five of the ten bulbs fail before time \(t\) ?

The joint probability distribution of the number \(X\) of cars and the number \(Y\) of buses per signal cycle at a proposed left-turn lane is displayed in the accompanying joint probability table. \begin{tabular}{cc|ccc} \(p(x, y)\) & & 0 & 1 & 2 \\ \hline & 0 & \(.025\) & \(.015\) & \(.010\) \\ & 1 & \(.050\) & \(.030\) & \(.020\) \\ & 2 & \(.125\) & \(.075\) & \(.050\) \\ \(x\) & 3 & \(.150\) & \(.090\) & \(.060\) \\ & 4 & \(.100\) & \(.060\) & \(.040\) \\ & 5 & \(.050\) & \(.030\) & \(.020\) \end{tabular} a. What is the probability that there is exactly one car and exactly one bus during a cycle? b. What is the probability that there is at most one car and at most one bus during a cycle? c. What is the probability that there is exactly one car during a cycle? Exactly one bus? d. Suppose the left-turn lane is to have a capacity of five cars, and that one bus is equivalent to three cars. What is the probability of an overflow during a cycle? e. Are \(X\) and \(Y\) independent rv's? Explain.
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