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According to a 2018 Money magazine article, Maryland has one of the highest per capita incomes in the United States, with an average income of $$\$ 75,847$$. Suppose the standard deviation is $$\$ 32,000$$ and the distribution is right-skewed. A random sample of 100 Maryland residents is taken. a. Is the sample size large enough to use the Central Limit Theorem for means? Explain. b. What would the mean and standard error for the sampling distribution? c. What is the probability that the sample mean will be more than $$\$ 3200$$ away from the population mean?

Short Answer

Expert verified
Central Limit Theorem applies as the sample size is greater than 30. The mean for the sampling distribution would be \(\$ 75847\) while the standard error is \(\$ 3200\). The probability that the sample mean will be over \(\$ 3200\) away from the population mean can be calculated by taking \(P(Z < -1) + P(Z > 1)\) with the standard normal distribution table.

Step by step solution

01

Analyze the Application of Central Limit Theorem

Based on the Central Limit Theorem (CLT), if the sample size is greater than or equal to 30, it's large enough to apply the CLT. In this case, the sample size is 100, thus the sample satisfies the condition, and we can use the Central Limit Theorem for means.
02

Calculate the Mean and Standard Error

For a sampling distribution, the mean is the same as the population mean, that is, \(\$ 75,847\). The standard error (SE) of the mean is the standard deviation of the population divided by the square root of the sample size. So, SE = \(\$ 32,000\)/sqrt(100) = \(\$ 3200\). Therefore, the mean is \(\$ 75847\) and standard error is \(\$ 3200\).
03

Find the probability that sample mean differ from population mean by more than \(\$ 3200\)

To calculate this, we need to calculate the Z-score first and then use that to find the probability. The Z-score for a value X is given by \((X - Mean)/SE\). Here, X is \(\$ 3200\) away from the mean, so we have two Z-scores, one for \(\$ 75847 - 3200\) and second one for \(\$ 75847 + 3200\). We are looking for the probability that the sample mean differ from the population mean by more than \(\$ 3200\), so it is \(P(Z < -1) + P(Z > 1)\). Using the standard normal distribution table, we can find each of these probabilities, and add them together.

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Key Concepts

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

Sampling Distribution
When we select a sample from a larger population and calculate its mean, we are creating one instance of what is known as a sampling distribution.

But what happens if we take another sample, and another, each of the same size, and independently from the same population? We'd end up with a collection of sample means, which, when plotted as a histogram, tends to form what we call a sampling distribution.

This distribution has its own mean and standard deviation, and for a large number of samples, it will resemble a normal distribution, regardless of the shape of the population distribution, thanks to the Central Limit Theorem (CLT). This is crucial because it allows us to make inferences about the population from which the samples are drawn, using the properties of the normally distributed sampling distribution.
Standard Error
The standard deviation of the sampling distribution is what we refer to as the standard error (SE).

This quantity measures the dispersion of the sample means around the population mean. In our exercise, the standard error gives us an idea of the variability we can expect from the means of different samples of Maryland residents' incomes.

To calculate the SE, we use the formula: \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation and \( n \) is the sample size. In the case of Maryland incomes, the SE helps us understand whether a sample mean is within expected ranges of fluctuation or if it's an outlier.
Probability Distribution
A probability distribution defines the likelihood of each possible outcome within a random variable's set of outcomes.

There are many types of probability distributions, each describing the probabilities associated with different phenomena, such as binomial distribution for number of successes in a series of yes/no experiments, or uniform distribution when every outcome is equally likely.

The normal distribution, which is one of the most commonly encountered in statistics, is particularly important here as the distribution of sample means (for sufficiently large sample sizes) can be described by a normal distribution due to the CLT. This probability distribution enables us to calculate probabilities associated with the sample means relative to the population mean, a crucial step in many statistical inference procedures.
Z-score
The Z-score is a statistical measure that describes a value's relationship to the mean of a group of values.

Expressed in terms of standard deviations, a Z-score tells us how many standard deviations away from the mean a particular value is. The Z-score is calculated using the formula: \( Z = \frac{(X - \mu)}{SE} \), where \( X \) is the value, \( \mu \) is the mean, and SE is the standard error.

In our exercise, calculating the Z-scores for the values \( \$75,847 \pm \$3,200 \) and comparing them with standard normal distribution tables gives us the probabilities that the sample mean will be more than \( \$3,200 \) away from the population mean. These scores are crucial in determining the likeliness of obtaining a sample with a given characteristic purely by chance.

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Most popular questions from this chapter

Drivers in Alaska drive fewer miles yearly than motorists in any other state. The annual number of miles driven per licensed driver in Alaska is 9134 miles. Assume the standard deviation is 3200 miles. A random sample of 100 licensed drivers in Alaska is selected and the mean number of miles driven yearly for the sample is calculated. (Source: 2017 World Almanac and Book of Facts) a. What value would we expect for the sample mean? b. What is the standard error for the sample mean?

A fast-food chain advertises that the mediumsize serving of French fries weighs 135 grams. A reporter took a random sample of 10 medium orders of French fries and weighed each order. The weights (in grams) were \(111,124,125,156,127,134\), \(135,136,139,141\). Assume the population distribution is Normal. (Source: soranews24.com) a. Test the hypothesis that the medium servings have a population mean different from 135 grams. Use a significance level of \(0.05\). b. Construct a \(95 \%\) confidence interval for the population mean. How does your confidence interval support your conclusion in part a? Do you think the consumers are being misled about the serving size? Explain.

A \(95 \%\) confidence interval for the ages of the first six presidents at their inaugurations is \((56.2,59.5)\). Either interpret the interval or explain why it should not be interpreted.

The undergraduate grade point average (GPA) for students accepted at a random sample of 10 medical schools in the United States was taken. The mean GPA for these accepted students was \(3.75\) with a standard error of \(0.06\). The distribution of undergraduate GPAs is Normal. (Source: Accepted.com) a. Decide whether each of the following statements is worded correctly for the confidence interval. Fill in the blanks for the correctly worded one(s). Explain the error for the ones that are incorrectly worded. i. We are \(95 \%\) confident that the sample mean is between ____ \(-\) and ____. ii. We are \(95 \%\) confident that the population mean is between ____. iii. There is a \(95 \%\) probability that the population mean is between ____ and ____. b. Based on your confidence interval, would you believe that the population mean GPA is \(3.80\) ? Why or why not?

The mean age of all 2550 students at a small college is \(22.8\) years with a standard deviation is \(3.2\) years, and the distribution is right-skewed. A random sample of 4 students' ages is obtained, and the mean is \(23.2\) with a standard deviation of \(2.4\) years. a. \(\mu=? \quad \sigma=? \quad \bar{x}=? \quad s=?\) b. Is \(\mu\) a parameter or a statistic? c. Are the conditions for using the CLT fulfilled? What would be the shape of the approximate sampling distribution of many means, each from a sample of 4 students? Would the shape be right-skewed, Normal, or left-skewed?

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