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A newborn baby has extremely low birth weight (ELBW) if it weighs less than 1000 grams. A study of the health of such children in later years examined a random sample of 219 children who had been born with ELBW. Their mean weight at birth was \(x=810\) grams. This sample mean is an unbiased estimator of the mean weight \(\mu\) in the population of all ELBW babies. This means that a. in many samples from this population, the mean of the many values of \(x\) will be equal to \(\mu\). b. as we take larger and larger samples from this population, \(x\) will get closer and closer to \(\mu\). c. in many samples from this population, the many values of \(x\) will have a distribution that is close to Normal.

Short Answer

Expert verified
Option a is correct: mean of sample means equals population mean.

Step by step solution

01

Understanding the Problem

We are assessing which of the options (a, b, c) best describes the property of an unbiased estimator of the population mean. The key focus here is the behavior of the sample mean in relation to the population mean.
02

Evaluate Option a

Option a states: 'in many samples from this population, the mean of the many values of \(x\) will be equal to \(\mu\).' This statement aligns with the definition of an unbiased estimator, which implies that the expected value of the sample mean equals the population mean.
03

Evaluate Option b

Option b states: 'as we take larger and larger samples from this population, \(x\) will get closer and closer to \(\mu\).' This describes the Law of Large Numbers, which is related to convergence but not directly a characterization of an unbiased estimator.
04

Evaluate Option c

Option c states: 'in many samples from this population, the many values of \(x\) will have a distribution that is close to Normal.' This relates to the Central Limit Theorem about the distribution form, but does not directly reflect a property about being unbiased.
05

Determine the Correct Option

Based on the evaluations, option a accurately describes the property of an unbiased estimator. Therefore, for an unbiased estimator of \(\mu\), in many samples from this population, the mean of the values of \(x\) equals \(\mu\).

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Key Concepts

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

Unbiased Estimator
An unbiased estimator is a statistical term that piques the interest of many when understanding estimations and averages. The core idea here is that an estimator is called 'unbiased' if the average of its estimated values equals the parameter it is estimating, over many samples. Imagine you're a baker. If every batch of cookies you make consistently averages to the recipe's exact sugar quantity, your method of measuring is an unbiased estimator. In statistical language, if you are estimating a population parameter like the mean weight of newborns with extremely low birth weight (ELBW), your estimator – in this case, the sample mean – should, on average, hit the target across numerous samples. This concept of unbiased property assures us that we're not systematically off the mark when we take repeated samples.
Sample Mean
The sample mean is a fundamental concept in statistics. It is essentially the average of all data points in your sample. For example, if we take 219 newborns with ELBW and calculate their average birth weight, we get the sample mean. It summarizes the central tendency of that sample and provides insight into the population from which the sample is drawn. The sample mean serves as a useful tool because it acts as an unbiased estimator of the population mean. This means that over many samples, the average of the sample means will be equal to the population mean. Calculating the sample mean involves adding up all data points and dividing by the number of observations, expressed as \( \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \), where \(n\) is the number of observations. This straightforward computation is key in various statistical analyses.
Population Mean
The term 'population mean' refers to the average of an entire population. In the context of ELBW newborns, it would be the true average birth weight if we consider every single ELBW newborn globally. While it's often impractical to measure every member of a population, if we could, the population mean \( \mu \) would be that definitive average. Unlike the sample mean, which varies depending on the sample we draw, the population mean is a fixed value. However, it’s usually unknown, which is why sample means are used as estimators. Understanding the difference between these means is crucial since it helps in making sense of statistics and how they relate to the entire population versus just a sample.
Law of Large Numbers
The Law of Large Numbers (LLN) is a vital principle in probability and statistics. It states that as the size of a sample increases, the sample mean will tend to get closer to the population mean. If you imagine flipping a fair coin, initially the outcomes may fluctuate. However, with more flips, the proportion of heads or tails will likely settle towards the expected probability (50% for heads or tails). Within statistical analyses, as you increase your sample size, the impact of random variations diminishes. For ELBW newborns, aggregating data from increasingly larger samples would lead the sample mean weights towards the true population mean. This principle underscores why larger datasets often provide more reliable estimations. It’s a practical reminder that patience pays off, as more data can reveal more accurate insights.
Central Limit Theorem
The Central Limit Theorem (CLT) is one of the cornerstones of statistical theory. It explains why the distribution of sample means will typically approximate a normal distribution, even if the original data isn’t normally distributed. As per the CLT, given a sufficiently large sample size, the sample means from a population will shape into a normal distribution around the true population mean. This magical transformation occurs because we're averaging out the individual differences. For instance, if we consider the weights of ELBW newborns, their individual weights might not follow a normal distribution. However, if we take numerous samples and compute their means, the distribution of these means would approximate normality. This is powerful as it allows for the application of various statistical tests and methods that assume normality, aiding in inference making. It's a reassuring principle for statisticians, allowing them to make generalizations about populations even when dealing with naturally skewed data.

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Most popular questions from this chapter

The Medical College Admission Test. Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate the mean score \(\mu\) of those who took the MCAT on your campus, you will obtain the scores of an SRS of students. The scores follow a Normal distribution, and from published information, you know that the standard deviation of scores for all MCAT takers is 10.6. Suppose that (unknown to you) the mean score of those taking the MCAT on your campus is \(500.0\). a. If you choose one student at random, what is the probability that the student's score is between 495 and 505 ? b. You sample 25 students. What is the sampling distribution of their average score \(x\) ? c. What is the probability that the mean score of your sample is between 495 and 505 ?

Larger Sample, More Accurate Estimate. Suppose that, in fact, the total cholesterol level of all men aged 20-34 follows the Normal distribution with mean \(\mu=182\) milligrams per deciliter (mg/dL) and standard deviation \(\sigma=37 \mathrm{mg} / \mathrm{dL}\). a. Choose an SRS of 100 men from this population. What is the sampling distribution of \(x\) ? What is the probability that \(x\) takes a value between 180 and 184 \(\mathrm{mg} / \mathrm{dL}\) ? This is the probability that \(x\) estimates \(\mu\) within \(\pm 2 \mathrm{mg} / \mathrm{dL}\). b. Choose an SRS of 1000 men from this population. Now what is the probability that \(x\) falls within \(\pm 2 \mathrm{mg} / \mathrm{dL}\) of \(\mu\) ? The larger sample is much more likely to give an accurate estimate of \(\mu\).

The Law of Large Numbers Made Visible. Roll two balanced dice and count the total spots on the up-faces. The probability model appears in Example 12.5 (page 277). You can see that this distribution is symmetric with 7 as its center, so it's no surprise that the mean is \(\mu=7\). This is the population mean for the idealized population that contains the results of rolling two dice forever. The law of large numbers says that the average \(x\) from a finite number of rolls tends to get closer and closer to 7 as we do more and more rolls. a. Click "More dice" once in the Law of Large Numbers applet to get two dice. Click "Show \(\mu_{x}\) " to see the mean 7 on the graph. Leaving the number of rolls at 1 , click "Roll dice" three times, recording each roll. How many spots did each roll produce? What is the average for the three rolls? You see that the graph displays at each point the average number of spots for all rolls up to the last one. This is exactly like Ejgure 15.1. b. Click "Reset" to start over. Set the number of rolls to 100 and click "Roll dice." The applet rolls the two dice 100 times. The graph shows how the average count of spots changes as we make more rolls. That is, the graph shows \(x\) as we continue to roll the dice. Sketch (or print out) the final graph. c. Repeat your work from part (b). Click "Reset" to start over, then roll two dice 100 times. Make a sketch of the final graph of the mean \(x\) against the number of rolls. Your two graphs will often look very different. What they have in common is that the average eventually gets close to the population mean \(\mu=7\). The law of large numbers says that this will always happen if you keep on rolling the dice.

The Bureau of Labor Statistics announces that last month it interviewed all members of the labor force in a sample of 60,000 households; \(3.5 \%\) of the people interviewed were unemployed. The boldface number is a a. sampling distribution. b. statistic. c. parameter.

Detecting the Emerald Ash Borer. The emerald ash borer is a serious threat to ash trees. A state agriculture department places traps throughout the state to detect the emerald ash borer. When traps are checked periodically, the mean number of ash borers trapped is only \(2.2\), but some traps have many ash borers. The distribution of ash borer counts is finite and strongly skewed, with standard deviation \(3.9 .\) a. What are the mean and standard deviation of the average number of ash borers \(x\) in 50 traps? b. Use the central limit theorem to find the probability that the average number of ash borers in 50 traps is greater than 3.0.

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