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Chapter 9: Problem 3

Find each \(X,\) given \(\hat{p}\) a. \(\hat{p}=0.60, n=240\) b. \(\hat{p}=0.20, n=320\) c. \(\hat{p}=0.60, n=520\) d. \(\hat{p}=0.80, n=50\) e. \(\hat{p}=0.35, n=200\)

Short Answer

Expert verified
a. 144, b. 64, c. 312, d. 40, e. 70

Step by step solution

01

Understand the Problem

The problem provides the sample proportion \( \hat{p} \) and sample size \( n \) for a set of examples. The goal is to calculate the expected number, \( X \), of successful outcomes in the sample, which can be found using the formula \( X = \hat{p} \times n \).
02

Calculate X for Part a

For the given \( \hat{p} = 0.60 \) and \( n = 240 \), calculate \( X \) using \( X = \hat{p} \times n \). Therefore, \( X = 0.60 \times 240 = 144 \).
03

Calculate X for Part b

For \( \hat{p} = 0.20 \) and \( n = 320 \), compute \( X \) as \( X = \hat{p} \times n = 0.20 \times 320 = 64 \).
04

Calculate X for Part c

Given \( \hat{p} = 0.60 \) and \( n = 520 \), use the formula to find \( X = \hat{p} \times n = 0.60 \times 520 = 312 \).
05

Calculate X for Part d

For \( \hat{p} = 0.80 \) and \( n = 50 \), calculate \( X = \hat{p} \times n = 0.80 \times 50 = 40 \).
06

Calculate X for Part e

With \( \hat{p} = 0.35 \) and \( n = 200 \), find \( X \) using \( X = \hat{p} \times n = 0.35 \times 200 = 70 \).

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Key Concepts

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

Sample Proportion
In statistics, understanding the concept of **sample proportion** is crucial. The sample proportion, typically denoted as \( \hat{p} \), represents the fraction of successes in a sample. For instance, in a survey where 150 out of 300 respondents prefer chocolate ice cream, the sample proportion \( \hat{p} \) is 0.50. It serves as an estimate of the true proportion (or probability) of success in the entire population.
  • The sample proportion enables statisticians to infer population characteristics.
  • It is calculated as \( \hat{p} = \frac{x}{n} \), where \( x \) represents the number of successes in the sample and \( n \) is the sample size.
  • This metric allows researchers to make predictions and test hypotheses.
Grasping sample proportions helps in solving problems related to population dynamics and surveys.
Sample Size
The **sample size**, indicated by \( n \), is the total number of observations or elements in a sample. It is a pivotal factor in research and impacts the reliability of the results. A larger sample size generally increases the accuracy of statistical analyses and reduces sampling error.
  • Larger samples give more reliable estimates of population parameters.
  • However, larger samples might require more resources and time to collect.
  • The selection of an adequate sample size depends on the study's objectives and resources.
For example, when calculating how many people in a city prefer electric cars, the researcher must decide on an appropriate sample size to confidently extrapolate findings to the entire population.
Probability
**Probability** is a fundamental concept in statistics. It quantifies the likelihood of an event occurring, ranging from 0 (impossibility) to 1 (certainty). In the context of sample proportions, probability represents the chance of a chosen sample exhibiting certain traits.
  • Probability helps in predicting future events based on existing data.
  • It enables statisticians to make informed decisions with limited information.
  • Sample proportion is a type of probability, detailing the likelihood of drawing a specific observation from a population.
For a deeper understanding, consider flipping a fair coin; the probability of getting heads is 0.5. Similarly, if 60% of a class votes for a particular candidate, the probability of randomly choosing a student who supports that candidate is 0.60.
Statistical Formulas
Statistical analysis relies heavily on various **formulas** that simplify complex calculations and provide insights into data trends. One such crucial formula involves finding the expected count of successes in a sample: \( X = \hat{p} \times n \).
  • This formula calculates the expected number of successes given a sample proportion and size.
  • Statistical formulas can be used across various fields to describe and predict outcomes.
  • They offer a structured way of resolving problems and drawing conclusions from data.
Utilizing these formulas accurately can help in making sense of real-world data. For example, if \( \hat{p}=0.60 \) and \( n=100 \), then the expected number of successful outcomes \( X \) would be 60, signifying that 60% of the sample exhibits a particular trait. Understanding these equations is vital for translating statistical jargon into actionable insights.

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Most popular questions from this chapter

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? $$ \begin{array}{ll}\hline & {\text { Women } \quad \text { Men }} \\ \hline \text { Sample size } & {250} & {300} \\ {\text { Number receiving aid }} & {200} & {180}\end{array} $$

Exam Scores at Private and Public Schools A researcher claims that students in a private school have exam scores that are at most 8 points higher than those of students in public schools. Random samples of 60 students from each type of school are selected and given an exam. The results are shown. At \(\alpha=0.05,\) test the claim. $$ \begin{array}{ll}{\text { Private school }} & {\text { Public school }} \\\ \hline \bar{X_{1}=110} & {\bar{X}_{2}=104} \\ {\sigma_{1}=15} & {\sigma_{2}=15} \\ {n_{1}=60} & {n_{2}=60}\end{array} $$

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. Improving Study Habits As an aid for improving students' study habits, nine students were randomly selected to attend a seminar on the importance of education in life. The table shows the number of hours each student studied per week before and after the seminar. At \(\alpha=0.10\), did attending the seminar increase the number of hours the students studied per week? $$ \begin{array}{l|ccccccccc}{\text { Before }} & {9} & {12} & {6} & {15} & {3} & {18} & {10} & {13} & {7} \\ \hline \text { After } & {9} & {17} & {9} & {20} & {2} & {21} & {15} & {22} & {6}\end{array} $$

Home Prices According to the almanac, the average sales price of a single- family home in the metropolitan Dallas/Ft. Worth/Irving, Texas, area is dollar 215,200. The average home price in Orlando, Florida, is dollar 198,000. The mean of a random sample of 45 homes in the Texas metroplex was dollar 216,000 with a population standard deviation of dollar 30,000 . In the Orlando, Florida, area a sample of 40 homes had a mean price of dollar 203,000 with a population standard deviation of dollar 32,500 . At the 0.05 level of significance, can it be concluded that the mean price in Dallas exceeds the mean price in Orlando? Use the \(P\) -value method.

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. Reading Program Summer reading programs are very popular with children. At the Citizens Library, Team Ramona read an average of 23.2 books with a standard deviation of \(6.1 .\) There were 21 members on this team. Team Beezus read an average of 26.1 books with a standard deviation of \(2.3 .\) There 23 members on this team. Did the variances of the two teams differ? Use \(\alpha=0.05 .\)
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