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Chapter 9: Problem 4

Find each \(X,\) given \(\hat{p} .\) a. \(\hat{p}=0.24, n=300\) b. \(\hat{p}=0.09, n=200\) c. \(\hat{p}=88 \%, n=500\) d. \(\hat{p}=40 \%, n=480\) e. \(\hat{p}=32 \%, n=700\)

Short Answer

Expert verified
a: 72, b: 18, c: 440, d: 192, e: 224

Step by step solution

01

Understanding the Formula

To find the number of successful outcomes, \(X\), when given the sample proportion \(\hat{p}\) and the sample size \(n\), we use the formula \(X = \hat{p} \times n\). This formula is derived from the definition of a sample proportion, \(\hat{p} = \frac{X}{n}\), where \(\hat{p}\) is the proportion of successes, \(X\) is the number of successes, and \(n\) is the total number of trials.
02

Solving for Part a

Given \( \hat{p} = 0.24 \) and \( n = 300 \). Applying the formula, we have:\[ X = 0.24 \times 300 = 72 \].Thus, the number of successful outcomes is 72.
03

Solving for Part b

Given \( \hat{p} = 0.09 \) and \( n = 200 \). Applying the formula gives:\[ X = 0.09 \times 200 = 18 \].Therefore, there are 18 successful outcomes.
04

Solving for Part c

Given \( \hat{p} = 88\% = 0.88 \) and \( n = 500 \). Using the formula, we find:\[ X = 0.88 \times 500 = 440 \].Thus, there are 440 successful outcomes.
05

Solving for Part d

Given \( \hat{p} = 40\% = 0.40 \) and \( n = 480 \). Applying the formula yields:\[ X = 0.40 \times 480 = 192 \].So, there are 192 successful outcomes.
06

Solving for Part e

Given \( \hat{p} = 32\% = 0.32 \) and \( n = 700 \). Using the formula, we have:\[ X = 0.32 \times 700 = 224 \].This means there are 224 successful outcomes.

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Key Concepts

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

Successful Outcomes
Understanding successful outcomes is key when working with sample proportions in statistics. A successful outcome refers to the number of desired results or "successes" within a sample. For example, if you are looking at a survey where people's positive responses are successes, those who said "yes" represent the successful outcomes. The formula for finding the number of successful outcomes when you know the sample proportion (\(\hat{p}\)) and the sample size (\(n\)) is:
  • \(X = \hat{p} \times n\)
This tells us how many individuals or items in the sample have the characteristic we are interested in.
Sample Size
Sample size is an important concept in statistics as it refers to the total number of observations or trials in a sample. This could be the number of people surveyed, items tested, or repetitions of an experiment.A good sample size can affect the accuracy of the results. It is essential to have a sufficiently large sample size to ensure that the data accurately represents the population. When calculating successful outcomes, the sample size \(n\) is multiplied with the sample proportion \(\hat{p}\) to give \(X\), the number of successes. For example, if \(n = 300\) in a survey, this means you asked 300 individuals about their opinions.
Probability
Probability in statistics measures the likelihood of an event occurring. In the context of sample proportions, probability represents the ratio of successful outcomes to the total number of possible outcomes. For a sample size \(n\), the sample proportion \(\hat{p}\) gives us a probability of a success in that sample. This means that if \(\hat{p} = 0.24\), then there is a 24% chance that a randomly chosen individual from the sample will have the desired characteristic or outcome. Understanding probability helps in predicting outcomes and making decisions based on statistical data.
Statistics
Statistics is the study of collecting, analyzing, interpreting, and presenting data. It involves concepts like probability, averages, and sampling methods to derive meaningful conclusions from data. Using statistics, we can infer information about a large population based on data collected from a smaller sample. In problems involving sample proportions, statistics helps us determine how many successful outcomes are likely in a given sample. This information is crucial for making decisions in fields like market research, medicine, and social sciences.
Mathematics Education
Mathematics education involves teaching and learning concepts related to math, including how to use formulas and solve equations. Understanding sample proportions and their calculation is a vital part of statistics education. Effective mathematics education provides students with the tools to apply mathematical concepts to real-world problems. For instance, learning how to calculate successful outcomes using proportions enhances a student's ability to analyze and interpret data accurately. This skill is crucial not only for students pursuing higher education in mathematics but also for various real-life applications in everyday decision-making.

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Most popular questions from this chapter

Ages of Gamblers The mean age of a random sample of 25 people who were playing the slot machines is 48.7 years, and the standard deviation is 6.8 years. The mean age of a random sample of 35 people who were playing roulette is 55.3 with a standard deviation of 3.2 years. Can it be concluded at \(\alpha=0.05\) that the mean age of those playing the slot machines is less than those playing roulette?

For Exercises 7 through \(27,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Undergraduate Financial Aid A study is conducted to determine if the percent of women who receive financial aid in undergraduate school is different from the percent of men who receive financial aid in undergraduate school. A random sample of undergraduates revealed these results. At \(\alpha=0.01,\) is there significant evidence to reject the null hypothesis? $$ \begin{array}{ll}\hline & {\text { Women } \quad \text { Men }} \\ \hline \text { Sample size } & {250} & {300} \\ {\text { Number receiving aid }} & {200} & {180}\end{array} $$

For Exercises 2 through \(12,\) perform each of these steps. Assume that all variables are normally or approximately normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Pulse Rates of Identical Twins A researcher wanted to compare the pulse rates of identical twins to see whether there was any difference. Eight sets of twins were randomly selected. The rates are given in the table as number of beats per minute. At \(\alpha=0.01,\) is there a significant difference in the average pulse rates of twins? Use the \(P\) -value method. Find the \(99 \%\) confidence interval for the difference of the two. $$ \begin{array}{l|cccccccc}{\text { Twin } \mathbf{A}} & {87} & {92} & {78} & {83} & {88} & {90} & {84} & {93} \\ \hline \text { Twin B } & {83} & {95} & {79} & {83} & {86} & {93} & {80} & {86}\end{array} $$

For Exercises 9 through \(24,\) perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Noise Levels in Hospitals In a hospital study, it was found that the standard deviation of the sound levels from 20 randomly selected areas designated as "casualty doors" "was 4.1 dB A and the standard deviation of 24 randomly selected areas designated as operating theaters was 7.5 dB A. At \(\alpha=0.05,\) can you substantiate the claim that there is a difference in the standard deviations?

For Exercises 7 through \(27,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Animal Bites of Postal Workers In Cleveland, a random sample of 73 mail carriers showed that 10 had been bitten by an animal during one week. In Philadelphia, in a random sample of 80 mail carriers, 16 had received animal bites. Is there a significant difference in the proportions? Use \(\alpha=0.05 .\) Find the \(95 \%\) confidence interval for the difference of the two proportions.
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