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The All Countries dataset includes land area, in square kilometers, for all 213 countries in the world. The mean land area for all the countries is \(608,120 \mathrm{sq} \mathrm{km}\) with standard deviation 1,766,860 . For samples of size \(50,\) what percentage of sample means will be less than 400,000 sq km? What percentage will be greater than \(900,000 ?\)

Short Answer

Expert verified
The percentages of samples less than 400,000 and greater than 900,000 would depend on the calculated z-scores and resulting probabilities, which are based on the population mean, sample size, and population standard deviation. The final values can be found by following the detailed steps.

Step by step solution

01

Calculate z-scores

The formula to calculate a z-score is \(Z = (X - \mu) / \sigma\), where X is the sample mean, \(\mu\) is the population mean and \(\sigma\) is the population standard deviation divided by the square root of the sample size. Here, the sample size \(n = 50\), the population mean \(\mu = 608120\), and the population standard deviation \(\sigma = 1766860\). Let's apply this formula for \(X = 400000\) and \(X = 900000\). The resultant z-scores will show how many standard deviations away our sample means are from the population mean.
02

Look up the p-value in the Z-Table

The Z-table displays the cumulative probability of a standard normal distribution up until a given Z-score. The calculated z-values from the first step will be used here to determine the area under the curve to the left of these points. In other words, the table gives us the probability that a value falls within the interval \(-\infty, Z\). We need to find the percentages less than 400,000 and greater than 900,000, corresponding to the calculated Z-values.
03

Calculate the required percentages

The probabilities obtained in step 2 represent percentage of sample means will be less than 400,000 sq km and less than 900,000 sq km. To get the percentage of sample means greater than 900,000, subtract the obtained p-value from 1 and express it as a percent.

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Key Concepts

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

Z-Score Calculation
Understanding the z-score is crucial for students who are learning about statistics, particularly when dealing with sampling distributions. The z-score is a measure of how many standard deviations an element is from the mean. This concept allows us to standardize different data points so they can be accurately compared.

To calculate a z-score, the formula is: \[ Z = \frac{(X - \mu)}{\sigma} \] where:
  • \(X\) is the sample mean,
  • \(\mu\) is the population mean, and
  • \(\sigma\) is the population standard deviation.
When dealing with sample sizes greater than 1, as in the given exercise, we adjust the standard deviation by dividing it by the square root of the sample size (\(\sqrt{n}\)), which gives us the standard error. This process helps us find out how far from the population mean a sample mean is likely to be. The further away the Z-score, the less likely the sample mean is a reflection of the population mean.
Standard Deviation
Standard deviation is a critical statistic that tells you how spread out the values in a data set are around the mean. It is a measure of variability that indicates the typical distance between the values in the dataset and their mean. A small standard deviation means that the values tend to be close to the mean, whereas a large standard deviation indicates that the values are spread out over a wider range.

In the context of the exercise, the standard deviation is a very large number compared to the mean, indicating that the land areas of countries differ greatly. It is important to note that, when we calculate the z-scores for samples, we use the standard error instead of the standard deviation of the population. The standard error takes into account the sample size, which is done by dividing the standard deviation by the square root of the sample size. This adjustment is pivotal because it reflects the anticipated variability in sample means rather than individual data points.

Remembering this distinction between standard deviation and standard error is crucial when working out problems involving sample statistics.
Normal Distribution
The normal distribution, also known as the Gaussian distribution, is a symmetrical, bell-shaped distribution that describes the expected probability distribution of many natural phenomena, including averages of samples. In statistics, we often assume that the means of samples of a population are normally distributed, especially with a large number of samples or a large sample size, according to the Central Limit Theorem.

This concept is essential to understanding the exercise because it justifies using the Z-table to interpret z-scores. The normal distribution is centered around the population mean, and it has a standard deviation that defines its width. Areas under the normal distribution curve represent probabilities. For example, about 68% of the data lies within one standard deviation from the mean, and 95% lie within two standard deviations in a perfect normal distribution.

In practice, using the normal distribution allows us to calculate the likelihood of occurrence of different sample means, such as in the textbook problem where we're looking for the percentage of sample means less than 400,000 or greater than 900,000 square kilometers. This is profoundly useful for students and professionals who must make inferences about population parameters based on sample statistics.

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Most popular questions from this chapter

In Exercises 6.159 and \(6.160,\) situations comparing two proportions are described. In each case, determine whether the situation involves comparing proportions for two groups or comparing two proportions from the same group. State whether the methods of this section apply to the difference in proportions. (a) Compare the proportion of students who use a Windows-based \(\mathrm{PC}\) to the proportion who use a Mac. (b) Compare the proportion of students who study abroad between those attending public universities and those at private universities. (c) Compare the proportion of in-state students at a university to the proportion from outside the state. (d) Compare the proportion of in-state students who get financial aid to the proportion of outof-state students who get financial aid.

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Standard Error from a Formula and a Bootstrap Distribution In Exercises 6.19 to \(6.22,\) use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of peanuts in mixed nuts, with \(n=100\) and \(\hat{p}=0.52\)

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion: (a) Find the mean and standard error of the distribution of sample proportions. (b) If the sample size is large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 30 from a population with proportion 0.27

Use a t-distribution. Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the endpoints of the t-distribution with \(2.5 \%\) beyond them in each tail if the samples have sizes \(n_{1}=15\) and \(n_{2}=25\)

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