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The mean tuition cost at state universities throughout the United States is \(\$ 4260\) per year (St. Petersburg Times, December 11,2002 ). Use this value as the population mean and assume that the population standard deviation is \(\sigma=\$ 900 .\) Suppose that a random sample of 50 state universities will be selected. a. Show the sampling distribution of \(\bar{x}\) where \(\bar{x}\) is the sample mean tuition cost for the 50 state universities. b. What is the probability that the simple random sample will provide a sample mean within \(\$ 250\) of the population mean? c. What is the probability that the simple random sample will provide a sample mean within \(\$ 100\) of the population mean?

Short Answer

Expert verified
a) \(\bar{x} \sim N(4260, 127.28)\); b) 0.95; c) 0.576

Step by step solution

01

Understand the problem

We need to understand that we are dealing with the sampling distribution of the sample mean. Given parameters are the population mean \(\mu = \\(4260\), population standard deviation \(\sigma = \\)900\), and sample size \(n = 50\).
02

Find the standard error of the mean (SEM)

The standard error of the mean is calculated using the formula \( \text{SEM} = \frac{\sigma}{\sqrt{n}} \). Substituting \( \sigma = 900 \) and \( n = 50 \), we calculate \( \text{SEM} = \frac{900}{\sqrt{50}} \approx 127.28 \).
03

Sampling distribution of \( \bar{x} \)

The sampling distribution of the sample mean \( \bar{x} \) is normally distributed with mean \( \mu = 4260 \) and standard deviation given by the standard error, \( \text{SEM} \approx 127.28 \).
04

Probability of being within \(\$250\)

To find the probability that the sample mean is within \(\$250\) of the population mean, we calculate \(P(4260 - 250 < \bar{x} < 4260 + 250)\). The z-scores are calculated as \(z = \frac{250}{127.28} \approx 1.96\). Using the standard normal distribution, \(P(-1.96 < Z < 1.96)\) is approximately 0.95.
05

Probability of being within \(\$100\)

To find the probability that the sample mean is within \(\$100\) of the population mean, we calculate \(P(4260 - 100 < \bar{x} < 4260 + 100)\). The z-scores are \(z = \frac{100}{127.28} \approx 0.79\). Thus, \(P(-0.79 < Z < 0.79)\) is approximately 0.576.

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Key Concepts

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

Standard Error of the Mean
The standard error of the mean (SEM) is a crucial concept when dealing with sampling distributions. It tells us how much the sample mean is expected to fluctuate from the true population mean. In simpler terms, it measures the dispersion or spread of the sample means you might get if you took multiple samples from the same population. Think of it as giving you an idea of how accurate your sample mean is likely to be as an estimation of the population mean.

The SEM is calculated using the formula:
  • \( \text{SEM} = \frac{\sigma}{\sqrt{n}} \)
Where \( \sigma \) is the population standard deviation, and \( n \) is the sample size. As you can see, SEM depends on both the variability within the population and the size of your sample.

In the given exercise, with a population standard deviation of \\(900\, and a sample size of 50, the SEM is approximately \\)127.28\. This means that any sample mean we calculate from a sample of 50 universities is expected to vary by about \\(127.28\ from the true population mean of \\)4260\.
Population Mean
The population mean (\( \mu \)) is one of the most fundamental concepts in statistics. It refers to the average value of an entire population. For instance, in the context of the exercise, the population mean tuition cost across all state universities in the United States is given as \\(4260\.The population mean serves as the central point in the sampling distribution. When we draw samples and calculate their means, these means tend to cluster around the population mean. The beauty of the sampling distribution is that it helps us understand how close our sample mean is likely to be to this population mean.

In the exercise, the goal is to estimate how the sample mean of selected 50 state universities compares to the known population mean of \\)4260\. It's important to remember that the population mean is a fixed value, while individual sample means are variable. This fixed value provides a reference point, helping ensure the accuracy and reliability of statistical analyses when considering different samples.
Normal Distribution
Normal distribution, also known as the Gaussian distribution, is a continuous probability distribution that is symmetric around its mean. It follows the familiar bell curve shape and is characterized by two parameters: the mean (\( \mu \)) and the standard deviation (\( \sigma \)). When we talk about sampling distributions of sample means, under certain conditions, these distributions tend to be normal. This happens even if the underlying population distribution is not normal, according to the Central Limit Theorem. The theorem states that if you have a large enough sample size, the sampling distribution of the sample mean will be approximately normally distributed.In the exercise, the sample mean distribution for the tuition costs is normally distributed. This distribution has a mean equal to the population mean (\(\mu = 4260 \)) and a standard deviation equal to the standard error of the mean (\( SEM \approx 127.28 \)). This normality allows us to use the properties of the normal distribution to calculate probabilities (like finding how likely a sample mean falls within a specified range from the population mean). Thus, the normal distribution is fundamental in making inferences about population parameters based on sample statistics.

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Most popular questions from this chapter

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