/*! 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 15 Let \(x\) denote the time (in mi... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(x\) denote the time (in minutes) that it takes a fifth-grade student to read a certain passage. Suppose that the mean value and standard deviation of \(x\) are \(\mu=\) 2 minutes and \(\sigma=0.8\) minute, respectively. a. If \(\bar{x}\) is the sample mean time for a random sample of \(n=9\) students, where is the \(\bar{x}\) distribution centered, and how much does it spread out about the center (as described by its standard deviation)? b. Repeat Part (a) for a sample of size of \(n=20\) and again for a sample of size \(n=100\). How do the centers and spreads of the three \(\bar{x}\) distributions compare to one another? Which sample size would be most likely to result in an \(\bar{x}\) value close to \(\mu\), and why?

Short Answer

Expert verified
The distribution centers for different sample sizes are all at \(\mu = 2\) minutes. The spreads for \(n=9\), \(n=20\), and \(n=100\) are respectively \(0.2667\) minutes, \(0.1789\) minutes, and \(0.08\) minutes. As the sample size increases, the spread decreases. Hence, a larger sample size such as \(n=100\) would be most likely to produce a sample mean close to the population mean.

Step by step solution

01

Calculation the distribution center and spread for n=9

The center of the distribution is the mean, which is given to be \(\mu = 2\) minutes. For a sample of size \(n=9\), the spread of distribution, or standard deviation, can be calculated as the standard error \(\sigma/ \sqrt{n}\). Here, \(\sigma = 0.8\) and \(n = 9\), so the standard error is \(0.8/ \sqrt{9} = 0.2667\) minutes.
02

Calculation the distribution center and spread for n=20

The center is still \(\mu = 2\) minutes. This time, the spread of distribution for a sample of size \(n=20\) can be calculated similarly as the standard error, which now is \(0.8/ \sqrt{20} = 0.1789\) minutes.
03

Calculation the distribution center and spread for n=100

Again, the center is \(\mu = 2\) minutes. The spread of distribution for a sample of size \(n=100\) is \(0.8/ \sqrt{100} = 0.08\) minutes.
04

Compare the centers and spreads for different sample sizes

The centers are the same for different sample sizes, all being \(\mu = 2\) minutes. The spread however becomes smaller as the sample size \(n\) increases. This shows that with a larger sample size, the sample mean is more likely to be close to the population mean as there's less variability.
05

Identify the sample size that would most likely result in an \(\overline{x}\) value close to \(\mu\)

Given the analysis in the previous step, the larger the sample size, the smaller the standard error. So, a sample size of \(n=100\) would be the most likely to result in an \(\overline{x}\) value close to \(\mu\), because it has the smallest spread among the three sample sizes discussed.

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Key Concepts

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

Sample Mean
When we talk about the sample mean, we refer to the average value calculated from a random sample drawn from a population. In our exercise, the sample mean is denoted by \(\bar{x}\), and it represents the average time taken by the sampled fifth-grade students to read a passage.

The sample mean serves as an estimate of the overall population mean, \(\mu\). Since individual readings might vary, the sample mean helps in getting a central value for the sample. As the sample size increases, the sample mean typically becomes a better estimate of the population mean. This is because larger samples tend to capture more of the population's variability. Due to this centrality, understanding the sample mean is critical for interpreting data collected from samples. It's important to remember that every random sample has its own sample mean, which can slightly differ from the population mean due to natural sampling variability. This leads us to consider the spread of the sample mean, which is where the concept of standard error comes into play.
Standard Error
The concept of standard error is a measure that tells us how much the sample mean might differ from the population mean. This is crucial in our exercise because it determines the spread of the sample mean around the population mean. The formula to calculate the standard error of the sample mean is \( \sigma / \sqrt{n} \), where \(\sigma\) is the population standard deviation, and \(n\) is the sample size.

For instance, when \(n=9\), the standard error is \(0.2667\) minutes. But as we increase the sample size to \(n=20\) or \(n=100\), the standard error reduces to \(0.1789\) and \(0.08\) minutes, respectively. The decreasing standard error with larger samples indicates less variability in the sample means. This highlights that with more observations, the estimate becomes more accurate, as the sample mean starts clustering closer to the population mean. Thus, standard error plays a key role in assessing how reliable our sample mean is in representing the population mean.
Distribution Spread
Distribution spread refers to how spread out the values of a dataset are. In the context of our exercise, it pertains to the variability or dispersion of the sample means around the population mean. The spread of the sample mean distribution is directly linked to the standard error. The smaller the standard error, the narrower the distribution spread of sample means.

This spread is key in understanding how much the sample mean can fluctuate each time we take a sample. A narrower distribution spread means that the sample means are consistently closer to the population mean. Analyzing different sample sizes in the exercise shows that increase in sample size results in a smaller distribution spread. This implies that the sample mean values will be less spread out, giving us higher confidence that the sample mean is a good estimator of the population mean. The spread gives us insight into the reliability and precision of our sample mean.
Population Mean
The population mean, denoted by \(\mu\), is the average of all possible observations from the whole population that we aim to study. In our exercise, \(\mu=2\) minutes, meaning that, on average, it takes two minutes for fifth-grade students to read the passage.

It represents the true center of the distribution if we were to measure every possible student in the population. However, practically speaking, it's usually not feasible to measure the entire population. Thus, we estimate it through samples. The sample mean serves as an estimate of the population mean and becomes more precise with larger sample sizes. Understanding the population mean is essential because it serves as a benchmark to evaluate how well our sample mean approximates the true average. With a known population mean, we can assess the accuracy of our sample estimates and understand the potential deviations in sample statistics.

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Most popular questions from this chapter

Suppose that a particular candidate for public office is in fact favored by \(48 \%\) of all registered voters in the district. A polling organization will take a random sample of 500 voters and will use \(\hat{p}\), the sample proportion, to estimate \(p\). What is the approximate probability that \(\hat{p}\) will be greater than \(.5\), causing the polling organization to incorrectly predict the result of the upcoming election?

Explain the difference between \(\sigma\) and \(\sigma_{\bar{x}}\) and between \(\mu\) and \(\mu_{\bar{x}}^{-\text {. }}\)

A manufacturing process is designed to produce bolts with a \(0.5\) -inch diameter. Once each day, a random sample of 36 bolts is selected and the bolt diameters are recorded. If the resulting sample mean is less than \(0.49\) inches or greater than \(0.51\) inches, the process is shut down for adjustment. The standard deviation for diameter is \(0.02\) inches. What is the probability that the manufacturing line will be shut down unnecessarily? (Hint: Find the probability of observing an \(\bar{x}\) in the shutdown range when the true process mean really is \(0.5\) inches.)

Consider the following population: \(\\{1,2,3,4\\} .\) Note that the population mean is $$ \mu=\frac{1+2+3+4}{4}=2.5 $$ a. Suppose that a random sample of size 2 is to be selected without replacement from this population. There are 12 possible samples (provided that the order in which observations are selected is taken into account): $$ \begin{array}{llllll} 1,2 & 1,3 & 1,4 & 2,1 & 2,3 & 2,4 \\ 3,1 & 3,2 & 3,4 & 4,1 & 4,2 & 4,3 \end{array} $$ Compute the sample mean for each of the 12 possible samples. Use this information to construct the sampling distribution of \(\bar{x}\). (Display the sampling distribution as a density histogram.) b. Suppose that a random sample of size 2 is to be selected, but this time sampling will be done with replacement. Using a method similar to that of Part (a), construct the sampling distribution of \(\bar{x}\). (Hint: There are 16 different possible samples in this case.) c. In what ways are the two sampling distributions of Parts (a) and (b) similar? In what ways are they different?

The thickness (in millimeters) of the coating applied to disk drives is one characteristic that determines the usefulness of the product. When no unusual circumstances are present, the thickness \((x)\) has a normal distribution with a mean of \(2 \mathrm{~mm}\) and a standard deviation of \(0.05 \mathrm{~mm}\). Suppose that the process will be monitored by selecting a random sample of 16 drives from each shift's production and determining \(\bar{x}\), the mean coating thickness for the sample. a. Describe the sampling distribution of \(\bar{x}\) (for a sample of size 16). b. When no unusual circumstances are present, we expect \(\bar{x}\) to be within \(3 \sigma_{\bar{x}}\) of \(2 \mathrm{~mm}\), the desired value. An \(\bar{x}\) value farther from 2 than \(3 \sigma_{\bar{x}}\) is interpreted as an indication of a problem that needs attention. Compute \(2 \pm 3 \sigma_{\bar{x}} .\) (A plot over time of \(\bar{x}\) values with horizontal lines drawn at the limits \(\mu \pm 3 \sigma_{x}^{-}\) is called a process control chart.) c. Referring to Part (b), what is the probability that a sample mean will be outside \(2 \pm 3 \sigma_{\bar{x}}^{-}\) just by chance (that is, when there are no unusual circumstances)? d. Suppose that a machine used to apply the coating is out of adjustment, resulting in a mean coating thickness of \(2.05 \mathrm{~mm}\). What is the probability that a problem will be detected when the next sample is taken? (Hint: This will occur if \(\bar{x}>2+3 \sigma_{\bar{x}}\) or \(\bar{x}<2-3 \sigma_{\bar{x}}\) when \(\left.\mu=2.05 .\right)\)

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