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Chapter 7: Problem 14

Education of the self-employed According to a recent Current Population Reports, the population distribution of number of years of education for self- employed individuals in the United States has a mean of 13.6 and a standard deviation of 3.0 . a. Identify the random variable \(X\) whose population distribution is described here. b. Find the mean and standard deviation of the sampling distribution of \(\bar{x}\) for a random sample of size 100 . Interpret the results. c. Repeat part b for \(n=400 .\) Describe the effect of increasing \(n\).

Short Answer

Expert verified
a. Random variable \(X\) is years of education. b. For \(n=100\), mean = 13.6, SD = 0.3. c. For \(n=400\), mean = 13.6, SD = 0.15. Increasing \(n\) decreases SD.

Step by step solution

01

Identify the Random Variable

The random variable \(X\) in this context represents the number of years of education for self-employed individuals in the United States.
02

Find Mean of the Sampling Distribution for n=100

The mean of the sampling distribution of \(\bar{x}\) is the same as the mean of the population. Hence, \(\mu_{\bar{x}} = \mu = 13.6\).
03

Calculate Standard Deviation for n=100

The standard deviation of the sampling distribution of \(\bar{x}\) is \(\frac{\sigma}{\sqrt{n}}\), where \(\sigma = 3.0\) and \(n = 100\). Calculate it as follows:\[\sigma_{\bar{x}} = \frac{3.0}{\sqrt{100}} = 0.3\]
04

Interpret Results for n=100

For a sample size of 100, the average number of years of education remains at 13.6, and the variability, or standard deviation, of the sample mean distribution decreases to 0.3, meaning the sample mean is more concentrated around the population mean.
05

Find Mean of the Sampling Distribution for n=400

The mean of the sampling distribution of \(\bar{x}\) remains \(13.6\) for any sample size \(n\), as it equals the population mean.
06

Calculate Standard Deviation for n=400

Use the formula \(\frac{\sigma}{\sqrt{n}}\) with \(n = 400\):\[\sigma_{\bar{x}} = \frac{3.0}{\sqrt{400}} = 0.15\]
07

Interpret Results for n=400

For a sample size of 400, the mean remains 13.6, but the standard deviation decreases to 0.15, indicating an even narrower distribution of the sample mean around the population mean.
08

Describe the Effect of Increasing n

Increasing the sample size \(n\) reduces the standard deviation of the sampling distribution \(\sigma_{\bar{x}}\), making the estimate of the population mean more precise.

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Key Concepts

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

Random Variable
In the context of sampling distribution, a **random variable** represents a numerical outcome of a specific event or experiment. To clarify this concept, consider that a random variable can take varied values based on an underlying probability distribution. For example, in our exercise, the random variable \( X \) corresponds to the number of years of education for self-employed individuals in the United States.
This variable is defined by its ability to capture different education levels within the self-employed group, each of which has a certain probability of occurrence. Such a representation allows us to statistically analyze the traits of this demographic, leading to insights about average education levels, variance, and perform further calculations utilizing these metrics.
Mean and Standard Deviation
The **mean** of a distribution gives us a central value, essentially summarizing the entire data into an average. It acts as a point of reference for understanding the general trend within the data. In the exercise, the mean education level of self-employed individuals is stated as 13.6 years.

For the purpose of a sampling distribution, the mean of the sample mean distribution is the same as the mean of the population, \( \mu_{\bar{x}} = 13.6 \). This consistent property allows us to make reliable estimates about a population based on samples.

The **standard deviation**, on the other hand, measures the dispersion or spread of data points relative to the mean. In this exercise, the population standard deviation is 3.0 years, telling us how diverse the education levels are around the average of 13.6 years.

When forming a sampling distribution, the standard deviation of the sample means, often referred to as the standard error \( \sigma_{\bar{x}} \), is calculated using \( \frac{\sigma}{\sqrt{n}} \), where \( n \) is the sample size. This measurement indicates the variability of the sample mean compared to the population mean, reducing as the sample size increases.
Sample Size Effect
The idea of **sample size effect** is important in understanding how estimates from samples relate to the true population parameters. A larger sample size typically results in more reliable and closer estimates to the actual population mean.
  • As shown in the exercise, with a sample size \( n = 100 \), the standard deviation of the sample mean distribution is 0.3.
  • In contrast, when the sample size increases to \( n = 400 \), the standard deviation becomes 0.15.

This reduction in standard deviation, also known as standard error, causes the sample mean to cluster more tightly around the population mean, indicating less fluctuation and greater precision.

Hence, increasing the sample size not only reduces the sampling error but also enhances the confidence in reflecting the true population characteristics. It is a fundamental idea in statistics that validates stronger and more refined insights through expansive data collection efforts.

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Most popular questions from this chapter

Basketball shooting In college basketball, a shot made from beyond a designated arc radiating about 20 feet from the basket is worth three points, instead of the usual two points given for shots made inside that arc. Over his career, University of Florida basketball player Lee Humphrey made \(45 \%\) of his three-point attempts. In one game in his final season, he made only 3 of 12 three-point shots, leading a TV basketball analyst to announce that Humphrey was in a shooting slump. a. Assuming Humphrey has a \(45 \%\) chance of making any particular three-point shot, find the mean and standard deviation of the sampling distribution of the proportion of three-point shots he will make out of 12 shots. b. How many standard deviations from the mean is this game's result of making 3 of 12 three-point shots? c. If Humphrey was actually not in a slump but still had a \(45 \%\) chance of making any particular three-point shot, explain why it would not be especially surprising for him to make only 3 of 12 shots. Thus, this is not really evidence of a shooting slump.

Comparing pizza brands \(\quad\) The owners of Aunt Erma's Restaurant plan an advertising campaign with the claim that more people prefer the taste of their pizza (which we'll denote by A) than the current leading fast-food chain selling pizza (which we'll denote by \(\mathrm{D}\) ). To support their claim, they plan to randomly sample three people in Boston. Each person is asked to taste a slice of pizza \(A\) and a slice of pizza \(D\). Subjects are blindfolded so they cannot see the pizza when they taste it, and the order of giving them the two slices is randomized. They are then asked which pizza tastes better. Use a symbol with three letters to represent the responses for each possible sample. For instance, ADD represents a sample in which the first subject sampled preferred pizza \(A\) and the second and third subjects preferred pizza \(\mathrm{D}\) a. Identify the eight possible samples of size \(3,\) and for each sample report the proportion that preferred pizza \(A\). b. In the entire Boston population, suppose that exactly half would prefer pizza \(\mathrm{A}\) and half would prefer pizza D. Explain why the sampling distribution of the sample proportion who prefer Aunt Erma's pizza, when \(n=3,\) is \begin{tabular}{cc} \hline Sample Proportion & Probability \\ \hline 0 & \(1 / 8\) \\ \(1 / 3\) & \(3 / 8\) \\ \(2 / 3\) & \(3 / 8\) \\ 1 & \(1 / 8\) \\ \hline \end{tabular} c. In part b, we can also find the probabilities for each possible sample proportion value using the binomial distribution. Use the binomial with \(n=3\) and \(p=0.50\) to show that the probability of a sample proportion of \(1 / 3\) equals \(3 / 8 .\) (Hint: This equals the probability that \(x=1\) person out of \(n=3\) prefer pizza A. It's especially helpful to use the binomial formula when \(p\) differs from \(0.50,\) since then the eight possible samples listed in part a would not be equally likely.)

Exam performance An exam consists of 50 multiplechoice questions. Based on how much you studied, for any given question you think you have a probability of \(p=0.70\) of getting the correct answer. Consider the sampling distribution of the sample proportion of the 50 questions on which you get the correct answer. a. Find the mean and standard deviation of the sampling distribution of this proportion. b. What do you expect for the shape of the sampling distribution? Why? c. If truly \(p=0.70,\) would it be very surprising if you got correct answers on only \(60 \%\) of the questions? Justify your answer by using the normal distribution to approximate the probability of a sample proportion of 0.60 or less.

Multiple choice: Standard deviation Which of the following is not correct? The standard deviation of a statistic describes a. The standard deviation of the sampling distribution of that statistic. b. The standard deviation of the sample data measurements. c. How close that statistic falls to the parameter that it estimates. d. The variability in the values of the statistic for repeated random samples of size \(n\).

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 \(1.0 .\) )
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