Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 \( X = \hat{p} \times n = 0.35 \times 200 = 70 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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. Test Scores An instructor who taught an online statistics course and a classroom course feels that the variance of the final exam scores for the students who took the online course is greater than the variance of the final exam scores of the students who took the classroom final exam. The following data were obtained. At \(\alpha=0.05\) is there enough evidence to support the claim? $$ \begin{array}{cc}{\text { Online Course }} & {\text { Classroom Course }} \\\ \hline s_{1=3.2} & {s_{2}=2.8} \\ {n_{1}=11} & {n_{2}=16}\end{array} $$

Hospital Stays for Maternity Patients Health Care Knowledge Systems reported that an insured woman spends on average 2.3 days in the hospital for a routine childbirth, while an uninsured woman spends on average 1.9 days. Assume two random samples of 16 women each were used in both samples. The standard deviation of the first sample is equal to 0.6 day, and the standard deviation of the second sample is 0.3 day. At \(\alpha=0.01,\) test the claim that the means are equal. Find the \(99 \%\) confidence interval for the differences of the means. Use the \(P\) -value method.

Length of Hospital Stays The average length of "short hospital stays" for men is slightly longer than that for women, 5.2 days versus 4.5 days. A random sample of recent hospital stays for both men and women revealed the following. At \(\alpha=0.01,\) is there sufficient evidence to conclude that the average hospital stay for men is longer than the average hospital stay for women? $$ \begin{array}{lcc}{} & {\text { Men }} & {\text { Women }} \\ \hline \text { Sample size } & {32} & {30} \\ {\text { Sample mean }} & {5.5 \text { days }} & {4.2 \text { days }} \\ {\text { Population standard deviation }} & {1.2 \text { days }} & {1.5 \text { days }}\end{array} $$

Average Earnings for College Graduates The average earnings of year-round full-time workers with bachelor's degrees or more is dollar 88,641 for men and dollar 58,000 for women - a difference of slightly over dollar 30,000 a year. One hundred of each were randomly sampled, resulting in a sample mean of dollar 90,200 for men, and the population standard deviation is dollar 15,000 ; and a mean of dollar 57,800 for women, and the population standard deviation is dollar 12,800. At the 0.01 level of significance, can it be concluded that the difference in means is not dollar 30,000 ?

Working Breath Rate Two random samples of 32 individuals were selected. One sample participated in an activity which simulates hard work. The average breath rate of these individuals was 21 breaths per minute. The other sample did some normal walking. The mean breath rate of these individuals was 14. Find the 90 \(\%\) confidence interval of the difference in the breath rates if the population standard deviation was 4.2 for breath rate per minute.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free