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For the class of \(2010,\) the average score on the Mathematics portion of the SAT (Scholastic Aptitude Test) is 516 with a standard deviation of 116. Find the mean and standard deviation of the distribution of mean scores if we take random samples of 100 scores at a time and compute the sample means.

Short Answer

Expert verified
The mean of the distribution of sample means is 516 and the standard deviation of the distribution of sample means is 11.6.

Step by step solution

01

Calculate the mean of the sampling distribution

The mean of the sampling distribution (\(\mu_{\bar{x}}\)) is simply the mean of the population. Therefore, \(\mu_{\bar{x}} = \mu = 516\). So, the mean of the distribution of sample means is 516.
02

Calculate the standard deviation of the sampling distribution

The standard deviation of the sampling distribution (\(\sigma_{\bar{x}}\)) is given by the formula \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the standard deviation of the population and n is the sample size. Substituting given values, we have \(\sigma_{\bar{x}} = \frac{116}{\sqrt{100}} = \frac{116}{10} = 11.6.\) Therefore, the standard deviation of the sampling distribution is 11.6.

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Key Concepts

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

Mean of Sample Means
The mean of sample means, often referred to as the expected value of the sampling distribution, is a cornerstone concept in sampling distribution theory. To determine this, we use the mean of the population from which samples are drawn. In our exercise, the population mean (\( \mu \) is 516. This value doesn't change when we calculate the average of multiple samples from this population. Here’s why it remains 516:
  • The expected value of the sample means (\( \mu_{\bar{x}} \) should reflect the true mean of the whole population.
  • Repeatedly drawing multiple samples of the same size will fluctuate around this population mean, but their average will eventually converge to it.
By understanding this, even if each of those samples gives slightly different averages due to random variability, overall, they are centered around the population mean. Thus, finding the mean of sample means is straightforward: it is equal to the population mean, here, 516.
Standard Deviation of Sample Means
The standard deviation of the sample means, commonly known as the standard error, provides insight into the variability or spread of sample means around the population mean. It tells us how much the average of a sample of observations is expected to vary from sample to sample, given the sample size.The formula to calculate this standard deviation is:\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \] where \( \sigma \) is the population standard deviation, and \( n \) is the sample size. In our context:
  • \( \sigma \), the population standard deviation, is 116.
  • \( n \), the number of observations within each sample, is 100.
Plugging in these numbers, the detailed calculation follows:\[ \sigma_{\bar{x}} = \frac{116}{\sqrt{100}} = \frac{116}{10} = 11.6 \] This number, 11.6, highlights that the sample means are expected to typically vary by about 11.6 units from the true population mean.
Central Limit Theorem
The Central Limit Theorem (CLT) is a foundational principle in statistics that explains why the means of samples drawn from a population tend to follow a normal distribution, even if the original population distribution is not normal, given a sufficiently large sample size.Key aspects of the Central Limit Theorem include:
  • Sample Size: The theorem holds as long as the sample size is large enough, typically \( n \geq 30 \), making the shape of the sampling distribution approximately normal.
  • Universality: Regardless of how skewed or non-normal the original population distribution is, the distribution of the sample mean approaches normality as the sample size increases.
  • Application: This characteristic allows statisticians to make inferences about the population mean using the properties of the normal distribution (mean and standard deviation) derived from sample data.
In our exercise, with a sample size of 100, which is quite large, the CLT assures us that the distribution of these sample means will be normal. This is crucial for making reliable statistical predictions and analyses based on the sample data.

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Most popular questions from this chapter

Random samples of the given sizes are drawn from populations with the given means and standard deviations. For each scenario: (a) Find the mean and standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) (b) If the sample sizes are 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 50 from Population 1 with mean 3.2 and standard deviation 1.7 and samples of size 50 from Population 2 with mean 2.8 and standard deviation 1.3

We see in the AllCountries dataset that the percent of the population that is elderly (over 65 years old) is 17.0 in Austria and 15.9 in Denmark. Suppose we take random samples of size 200 from each of these countries and compute the difference in sample proportions \(\hat{p}_{A}-\hat{p}_{D}\), where \(\hat{p}_{A}\) represents the proportion of the sample that is elderly in Austria and \(\hat{p}_{D}\) represents the proportion of the sample that is elderly in Denmark. Find the mean and standard deviation of the differences in sample proportions.

If random samples of the given sizes are drawn from populations with the given proportions: (a) Find the mean and standard error of the distribution of differences in sample proportions, \(\hat{p}_{A}-\hat{p}_{B}\) (b) If the sample sizes are 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 300 from population \(A\) with proportion 0.15 and samples of size 300 from population \(B\) with proportion 0.20

Use StatKey or other technology to generate a bootstrap distribution of sample means and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample standard deviation as an estimate of the population standard deviation. Mean price of used Mustang cars online (in \(\$ 1000\) s) using the data in MustangPrice with \(n=25,\) \(\bar{x}=15.98,\) and \(s=11.11\)

Is a Normal Distribution Appropriate? In Exercises 6.13 and \(6.14,\) indicate whether the Central Limit Theorem applies so that the sample proportions follow a normal distribution. In each case below, does the Central Limit Theorem apply? (a) \(n=500\) and \(p=0.1\) (b) \(n=25\) and \(p=0.5\) (c) \(n=30\) and \(p=0.2\) (d) \(n=100\) and \(p=0.92\)

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