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Physicians' assistants The 2006 AAPA survey of the population of physicians' assistants who were working full time reported a mean annual income of \(\$ 84,396\) and standard deviation of \(\$ 21,975 .\) (Source: Data from 2006 AAPA survey [www.apa.org].) a. Suppose the AAPA had randomly sampled 100 physicians' assistants instead of collecting data for all of them. Describe the mean, standard deviation, and shape of the sampling distribution of the sample mean. b. Using this sampling distribution, find the \(z\) -score for a sample mean of \(\$ 80,000\) c. Using parts a and b, find the probability that the sample mean would fall within approximately \(\$ 4000\) of the population mean.

Short Answer

Expert verified
The sample mean is \(\$84,396\) with a standard deviation of \(\$2,197.50\), approximately normal. The \(z\)-score for \(\$80,000\) is \(-2.00\), and there's a 93.12% probability the sample mean is within \(\$4,000\) of \(\$84,396\).

Step by step solution

01

Understand the Sampling Distribution of the Sample Mean

For problem a, when you randomly sample 100 physicians’ assistants, the sampling distribution of the sample mean will have the following characteristics:- **Mean**: The mean of the sampling distribution (\( \bar{x} \)) is the same as the population mean, which is \(\\(84,396\).- **Standard Deviation**: The standard deviation of the sampling distribution (known as the standard error) is calculated by dividing the population standard deviation by the square root of the sample size. This gives us \(\frac{21,975}{\sqrt{100}} = \\)2,197.50\).- **Shape**: According to the Central Limit Theorem, the sampling distribution of the sample mean will be approximately normal because the sample size is large (\(n = 100\)).
02

Calculate the Z-Score for a Sample Mean of \(\$80,000\)

In problem b, to find the \(z\)-score of a sample mean of \(\\(80,000\), use the formula:\[z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}\]where \(\bar{x} = \\)80,000\), \(\mu = \\(84,396\), and \(\sigma_{\bar{x}} = \\)2,197.50\). Plug in the values:\[z = \frac{80,000 - 84,396}{2,197.50} \approx -2.00\]Thus, the \(z\)-score is \(-2.00\).
03

Determine the Probability of the Sample Mean within \(\$4,000\)

For problem c, find the probability that the sample mean would fall within \(\\(4,000\) of the population mean.1. The range of \(\\)4,000\) from the mean \(\\(84,396\) is \(\\)80,396\) to \(\\(88,396\).2. Convert these values to \(z\)-scores using the formula: - For \(\\)80,396\), \(z = \frac{80,396 - 84,396}{2,197.50} \approx -1.82\). - For \(\\(88,396\), \(z = \frac{88,396 - 84,396}{2,197.50} \approx 1.82\).3. Look up or use a calculator to find the probabilities associated with \(z = -1.82\) and \(z = 1.82\). These \(z\)-scores correspond to approximately 3.44% and 96.56%, respectively.4. The probability that the sample mean is within \(\\)4,000\) of the population mean is:\[P(-1.82 < z < 1.82) = 0.9656 - 0.0344 = 0.9312\]Thus, there is a 93.12% probability that the sample mean is within \(\$4,000\) of the population mean.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental concept in statistics. It tells us that when we take a large number of samples from a population, the distribution of the sample means will tend to be normal (bell-shaped), regardless of the population's original distribution. This theorem is crucial because it allows us to make inferences about a population even when the population distribution is unknown or non-normal.

In this exercise, the sample size is 100, which is considered large. According to the CLT, the shape of the sampling distribution of the sample means will be approximately normal. This means that we can use the properties of the normal distribution to calculate probabilities and make conclusions about the population mean.

By understanding the Central Limit Theorem, students can appreciate how statistical inference is possible even from non-normal populations. The larger the sample size, the more closely the sampling distribution will resemble a normal distribution.
Standard Error
The Standard Error (SE) measures how much sampling variability exists in the sample means. It is essentially the standard deviation of the sampling distribution. SE provides an estimate of how much the sample mean is expected to fluctuate from the true population mean.

In the exercise, the SE is calculated by dividing the population standard deviation by the square root of the sample size: \(SE = \frac{21,975}{\sqrt{100}} = 2,197.50\).

Key points about Standard Error:
  • The SE decreases as the sample size increases, meaning larger samples yield more precise estimates of the population mean.
  • It helps assess the reliability of the sample mean as an estimate of the population mean.
  • By knowing the SE, it becomes easier to compute confidence intervals and conduct hypothesis tests.
Understanding Standard Error is vital because it quantifies the margin of error when making inferences about a population from a sample.
Z-Score
A Z-Score is a measure that describes a value's position relative to the mean of a set of values, expressed in terms of standard deviations. In the context of a sampling distribution, a Z-Score tells us how many standard errors a sample mean is from the population mean.

To compute the Z-Score for a sample mean, you use the formula:
\[z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}\
\]where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, and \(\sigma_{\bar{x}}\) is the standard error.

For example, in this exercise, the Z-Score for a sample mean of \(\$80,000\) comes out to be \(-2.00\). This means the sample mean is 2 standard errors below the population mean.

Z-Scores are incredibly useful because:
  • They allow for the comparison of scores from different distributions by standardizing values.
  • They help determine probabilities and make decisions in hypothesis testing.
  • A negative Z-Score indicates a value below the mean, while a positive Z-Score indicates a value above the mean.
By mastering Z-Scores, you can better understand data deviations and their significance.
Probability
Probability is the branch of mathematics that deals with the likelihood of different outcomes. In the context of sampling distributions, probability helps us understand the chances of observing a particular sample mean, or a range of sample means, in relation to the population mean.

In the exercise, we calculate the probability that the sample mean is within \(\\(4,000\) of the population mean. We use the Z-Scores corresponding to \(\\)80,396\) and \(\$88,396\) to find these probabilities using standard normal distribution tables or calculators.

Here are some useful insights about probability in this context:
  • Z-Scores correspond to points on the standard normal distribution, which helps us determine probabilities.
  • The area under the curve between two Z-Scores gives the probability that a value falls within that range.
  • Understanding these probabilities allows us to make informed predictions and decisions based on sample data.
By grasping the concept of probability, students can analyze the likelihood of various sampling outcomes and draw meaningful conclusions about a population.

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Most popular questions from this chapter

Restaurant profit? Jan's All You Can Eat Restaurant charges \(\$ 8.95\) per customer to eat at the restaurant. Restaurant management finds that its expense per customer, based on how much the customer eats and the expense of labor, has a distribution that is skewed to the right with a mean of \(\$ 8.20\) and a standard deviation of \(\$ 3\). a. If the 100 customers on a particular day have the characteristics of a random sample from their customer base, find the mean and standard deviation of the sampling distribution of the restaurant's sample mean expense per customer. b. Find the probability that the restaurant makes a profit that day, with the sample mean expense being less than \$8.95. (Hint: Apply the central limit theorem to the sampling distribution in part a.)

Household size According to the 2010 U.S. Census Bureau Current Population Survey (www.census .gov/population/www/socdemo/hh-fam/cps2010.html), the average number of people in family households which contain both family and non-family members is 4.43 with a standard deviation of \(2.02 .\) This is based on census information for the population. Suppose the Census Bureau instead had estimated this mean using a random sample of 225 homes. Suppose the sample had a sample mean of 4.2 and standard deviation of 1.9 . a. Identify the random variable \(X .\) Indicate whether it is quantitative or categorical. b. Describe the center and variability of the population distribution. What would you predict as the shape of the population distribution? Explain. c. Describe the center and variability of the data distribution. What would you predict as the shape of the data distribution? Explain. d. Describe the center and variability of the sampling distribution of the sample mean for 225 homes. What would you predict as the shape of the sampling distribution? Explain.

Random variability in baseball A baseball player in the major leagues who plays regularly will have about 500 atbats (that is, about 500 times he can be the hitter in a game) during a season. Suppose a player has a 0.300 probability of getting a hit in an at-bat. His batting average at the end of the season is the number of hits divided by the number of at-bats. This batting average is a sample proportion, so it has a sampling distribution describing where it is likely to fall. a. Describe the shape, mean, and standard deviation of the sampling distribution of the player's batting average after a season of 500 at-bats. b. Explain why a batting average of 0.320 or of 0.280 would not be especially unusual for this player's yearend batting average. (That is, you should not conclude that someone with a batting average of 0.320 one year is necessarily a better hitter than a player with a batting average of \(0.280 .)\)

Playing roulette \(\quad\) A roulette wheel in Las Vegas has 38 slots. If you bet a dollar on a particular number, you'll win \(\$ 35\) if the ball ends up in that slot and \(\$ 0\) otherwise. Roulette wheels are calibrated so that each outcome is equally likely. a. Let \(X\) denote your winnings when you play once. State the probability distribution of \(X\). (This also represents the population distribution you would get if you could play roulette an infinite number of times.) It has mean 0.921 and standard deviation 5.603 . b. You decide to play once a minute for 12 hours a day for the next week, a total of 5040 times. Show that the sampling distribution of your sample mean winnings has mean \(=0.921\) and standard deviation \(=0.079 .\) c. Refer to part b. Using the central limit theorem, find the probability that with this amount of roulette playing, your mean winnings is at least \(\$ 1,\) so that you have not lost money after this week of playing. (Hint: Find the probability that a normal random variable with mean 0.921 and standard deviation 0.079 exceeds \(1.0 .\) )

What is a sampling distribution? How would you explain to someone who has never studied statistics what a sampling distribution is? Explain by using the example of polls of 1000 Canadians for estimating the proportion who think the prime minister is doing a good job.

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