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Chapter 7: Problem 11

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.

Short Answer

Expert verified
Answer: The proportions of successes in the samples are 1.0, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.0, 0.0, and 0.0.

Step by step solution

01

List all possible samples of size 2

(Select two items from the given population) We have a total of 5 items, where 2 are successes and 3 are failures. Therefore, the possible samples of size 2 are: 1. \(S_1, S_2\) 2. \(S_1, F_1\) 3. \(S_1, F_2\) 4. \(S_1, F_3\) 5. \(S_2, F_1\) 6. \(S_2, F_2\) 7. \(S_2, F_3\) 8. \(F_1, F_2\) 9. \(F_1, F_3\) 10. \(F_2, F_3\)
02

Determine the proportion of successes in each sample

Out of the total items in each sample, count the number of successes. 1. Sample 1: (\(S_1\), \(S_2\)) - 2 successes out of 2 items; proportion of successes: \(\frac{2}{2} = 1\) 2. Sample 2: (\(S_1\), \(F_1\)) - 1 success out of 2 items; proportion of successes: \(\frac{1}{2} = 0.5\) 3. Sample 3: (\(S_1\), \(F_2\)) - 1 success out of 2 items; proportion of successes: \(\frac{1}{2} = 0.5\) 4. Sample 4: (\(S_1\), \(F_3\)) - 1 success out of 2 items; proportion of successes: \(\frac{1}{2} = 0.5\) 5. Sample 5: (\(S_2\), \(F_1\)) - 1 success out of 2 items; proportion of successes: \(\frac{1}{2} = 0.5\) 6. Sample 6: (\(S_2\), \(F_2\)) - 1 success out of 2 items; proportion of successes: \(\frac{1}{2} = 0.5\) 7. Sample 7: (\(S_2\), \(F_3\)) - 1 success out of 2 items; proportion of successes: \(\frac{1}{2} = 0.5\) 8. Sample 8: (\(F_1\), \(F_2\)) - 0 successes out of 2 items; proportion of successes: \(\frac{0}{2} = 0\) 9. Sample 9: (\(F_1\), \(F_3\)) - 0 successes out of 2 items; proportion of successes: \(\frac{0}{2} = 0\) 10. Sample 10: (\(F_2\), \(F_3\)) - 0 successes out of 2 items; proportion of successes: \(\frac{0}{2} = 0\) Therefore, the proportions of successes in the samples are: 1. 1.0 2. 0.5 3. 0.5 4. 0.5 5. 0.5 6. 0.5 7. 0.5 8. 0.0 9. 0.0 10. 0.0

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Key Concepts

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

Proportion of Successes
When analyzing data from a random sample, one common measure is the 'proportion of successes'. This is particularly useful when considering events with two possible outcomes, such as 'success' or 'failure'.

To calculate the proportion of successes, you simply count the number of successful outcomes and divide by the total number of observations in the sample. In our exercise, for example, if a sample contains 2 successes out of 2 items, the proportion of successes would be \(\frac{2}{2} = 1\) or 100%. If there is 1 success and 1 failure in the sample, the proportion of success is \(\frac{1}{2}=0.5\) or 50%.

Understanding and calculating this proportion is vital in fields such as statistics, epidemiology, and social sciences, where it is necessary to measure the prevalence or frequency of certain outcomes within a given set of data.
Random Sample
The term 'random sample' refers to a subset of data taken from a larger set, where each member of the larger set has an equal chance of being included in the sample. Ensuring randomness is crucial for the representativity of the sample, making it possible to infer conclusions about the entire population from the sample data.

In the context of our exercise, a random sample without replacement means that once an item is selected, it cannot be chosen again. This affects the probabilities for subsequent selections since the population from which we are sampling decreases in size. For instance, once a 'success' has been selected in our first draw, it leaves a different ratio of 'successes' to 'failures' for the second draw, hence changing the probability of drawing a 'success' or a 'failure'.
Combinatorics
Combinatorics is a branch of mathematics dealing with the counting, arrangement, and combination of objects. It plays a key role in probability and statistics, as seen in our exercise, where we use combinatorial methods to list all possible samples from a population.

In probability sampling without replacement, combinatorics helps determine the total number of possible samples that can be taken. For example, if we have 5 items and we want to take a sample of 2 without replacement, we would use combinations to find out how many unique pairs we can create. This is often given by the binomial coefficient \(C(n, k)\), where 'n' is the total number of items and 'k' is the number of items to choose. Here, we calculate it as \(C(5, 2)\), which gives 10 possible samples, as listed in the step-by-step solution to our exercise.

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