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Chapter 7: Problem 8

A population consists of \(N=4\) numbers: 10,15,21,22 with a mean of \(\mu=17 .\) A random sample of \(n=2\) is selected in one of two ways: first, without replacement and second, with replacement. Use this information to answer the questions. How many possible random samples are available when sampling with replacement? List the possible samples. Find the sampling distribution for the sample mean \(\bar{x}\) and display it as a table and as a probability histogram.

Short Answer

Expert verified
a) 20.5 b) 21 c) 21.5 d) 22.5 Answer: c) 21.5

Step by step solution

01

Determine Possible Random Samples with Replacement

Since we are asked to select a random sample of \(n=2\) with replacement, each number gets replaced back into the population after it has been picked. Therefore, each of the \(N=4\) numbers can be picked twice for each of the two selections. The number of ways we can choose 2 items from a population of 4, with replacement, is given by \(_{N}C_n = C(4,2) = \frac{4!}{(2!(4-2)!)}=6\).
02

List the Possible Samples

Now we will write down all the possible sample combinations with replacement. These combinations are: 1. (10, 10) 2. (10, 15) 3. (10, 21) 4. (10, 22) 5. (15, 15) 6. (15, 21) 7. (15, 22) 8. (21, 21) 9. (21, 22) 10. (22, 22)
03

Determine the Sampling Distribution for the Sample Mean

To find the sampling distribution, we need to find the means for each possible sample combination. We calculate the mean for each sample and their corresponding probabilities: 1. (10, 10), mean: \(\frac{10 + 10}{2} = 10\) 2. (10, 15), mean: \(\frac{10 + 15}{2} = 12.5\) 3. (10, 21), mean: \(\frac{10 + 21}{2} = 15.5\) 4. (10, 22), mean: \(\frac{10 + 22}{2} = 16\) 5. (15, 15), mean: \(\frac{15 + 15}{2} = 15\) 6. (15, 21), mean: \(\frac{15 + 21}{2} = 18\) 7. (15, 22), mean: \(\frac{15 + 22}{2} = 18.5\) 8. (21, 21), mean: \(\frac{21 + 21}{2} = 21\) 9. (21, 22), mean: \(\frac{21 + 22}{2} = 21.5\) 10. (22, 22), mean: \(\frac{22 + 22}{2} = 22\)
04

Table and Probability Histogram

Now we create a table and a probability histogram that displays the different sample means and their corresponding probabilities. Sample Means | Probability -------------|------------ 10 | 1/10 12.5 | 1/10 15 | 1/10 15.5 | 1/10 16 | 1/10 18 | 1/10 18.5 | 1/10 21 | 1/10 21.5 | 1/10 22 | 1/10 In the histogram, the x-axis represents the sample means, and the y-axis represents their probabilities. For all the possible sample means, their probability is 1/10.

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Key Concepts

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

Understanding Probability and Statistics
Probability and statistics are branches of mathematics that deal with the analysis of random phenomena. Probability is the study of randomness and uncertainty, quantifying the likelihood of events to occur, while statistics involves collecting, analyzing, interpreting, presenting, and organizing data.

Concerning our exercise, we are dealing with a simple population comprising of four numbers with a known mean. Understanding how to determine the probability of different outcomes with random samples is a core application of these mathematical principles. To break down the process, we use the number of possible outcomes and calculate the individual and cumulative probabilities for a series of events鈥攍ike the mean of random samples.
Random Sampling with Replacement
Random sampling with replacement is a sampling technique where each member of the population may be selected more than once. In our scenario, we're choosing two numbers from our population of four. After a number is selected, it is 'replaced' back into the set, making it possible to pick it again.

Random sampling with replacement impacts the probability of selecting each combination. Since every member has an equal chance of being chosen every time, it results in more possible samples than sampling without replacement. Plus, the calculation of probabilities requires consideration for the fact that outcomes can repeat.
Calculating the Sample Mean
The sample mean, expressed as \( \bar{x} \) in statistics, is calculated by summing all the values in a sample and dividing by the number of observations. In our example with samples of size two, the mean is simply the sum of the two numbers divided by two.

To find the sampling distribution of the sample mean, we calculate the individual means of all possible samples. Since we are dealing with a finite population and clearly defined sample size, we can list and compute the sample mean for every possible pair. This approach is called the method of enumeration 鈥� a solid basis for learning before advancing to more complex sampling methods.
Interpreting a Probability Histogram
A probability histogram visually represents the sampling distribution. It is a graphical tool that helps us understand the behavior of the sample mean in repeated samples. Each bar corresponds to a possible sample mean, and its height reflects the probability of that mean occurring.

In the provided example, we plotted a probability histogram with the sample means on the x-axis and their probabilities on the y-axis. Since we sampled with replacement, each mean had an equal chance of 1/10. By learning to interpret such histograms, students can visually grasp the concept of distribution and the likelihood of various outcomes in a random set of data.

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Most popular questions from this chapter

The manager of a building supplies company randomly samples incoming lumber to see whether it meets quality specifications. From each shipment, 100 pieces of \(2 \times 4\) lumber are inspected and judged according to whether they are first (acceptable) or second (defective) grade. The proportions of second- grade \(2 \times 4\) s recorded for 30 shipments were as follows: \(\begin{array}{llllllllll}.14 & .21 & .19 & .18 & .23 & .20 & .25 & .19 & .22 & .17 \\ .21 & .15 & .23 & .12 & .19 & .22 & .15 & .26 & .22 & .21 \\ .14 & .20 & .18 & .22 & .21 & .13 & .20 & .23 & .19 & .26\end{array}\) a. Construct a control chart for the proportion of second-grade \(2 \times 4 \mathrm{~s}\) in samples of 100 pieces of lumber. b. Explain how the control chart can be of use to the manager of the building- supplies company.

Packages of food whose average weight is 450 grams with a standard deviation of 17 grams are shipped in boxes of 24 packages. If the package weights are approximately normally distributed, what is the probability that a box of 24 packages will weigh more than 11 kilograms?

A random sample of \(n\) observations is selected from a population with standard deviation \(\sigma=1 .\) Calculate the standard error of the mean \((S E)\) for the values of \(n\). $$ n=16 $$

A population consists of \(N=5\) items-two of which are considered "successes" \(\left(S,\right.\) and \(\left.S_{2}\right)\) and three of which are considered "failures" \(\left(F_{1}, F_{2},\right.\) and \(\left.F_{3}\right) .\) A random sample of \(n=2\) items is selected, without replacement. Use this information to answer the questions. For each of the samples in Exercise \(10,\) find the proportion of successes in the sample.

In a Fox News Poll \(^{3}\) conducted by Pulse Opinion Research in the state of Delaware, 1000 likely voters were asked to answer the following questions. \- All in all, would you rather have bigger government that provides more services or smaller government that provides fewer services? \- Do you agree or disagree with the following statement: The federal government has gotten totally out of control and threatens our basic liberties unless we clear house and commit to drastic change. \- Thinking about the health care law that was passed earlier this year, would you favor repealing the new law to keep it from going into effect, or would you oppose repealing the new law? Comment on the effect of wording bias on the responses gathered using this survey.
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