Answered: It was observed that 30 of the 100…

It was observed that 30 of the 100 randomly selected students smoked. Find the confidence interval estimate of 95% confidence level for \ pi ratio of smokers in the population.

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter13: Probability And Calculus

Section13.2: Expected Value And Variance Of Continuous Random Variables

Problem 10E

See similar textbooks

Related questions

Q: Out of 200 people sampled, 188 had kids. Based on this, construct a 95% confidence interval for the…

A: Given information- Confidence level, c = 95% Sample proportion is,

Q: Out of 200 people sampled, 62 had kids. Based on this, construct a 95% confidence interval for the…

Q: Out of 100 people sampled, 84 had kids. Based on this, construct a 95% confidence interval for the…

A: Introduction: The 100 (1 – α) % confidence interval for the population proportion, p is: Here, n is…

Q: Out of 600 people sampled, 402 had kids. Based on this, construct a 90% confidence interval for the…

Q: A survey of 120 female freshmen showed that 9 did not wish to work after marriage. Find the 99%…

Q: Out of 300 people sampled, 45 had kids. Based on this, construct a 90% confidence interval for the…

Q: A random sample of 300 of toy trains produced by a company are selected from a production line and…

A: Given Data: x=60 n=300 p^=xn=60300=0.2 z-value for 95% Confidence level: z=1.96

Q: A soft drinks manufacturer recently executed a taste test in two cities for a new fruit flavoured…

A: Obtain the 90% confidence interval for the difference between the true proportion of consumers in…

Q: A sociologist is studying the prevalence of crime in one major city. In a sample of 175 randomly…

A: Given data, n=175 x=54 sample proportion p=x/n=54/175 =0.31 z-value at 90% confidence is Zc=1.645

Q: Out of 400 people sampled, 32 had kids. Based on this, construct a 95% confidence interval for the…

A: Solution: Let X be the number of people had kids and n be the sample number of people. From the…

Q: Out of 200 people sampled, 96 had kids. Based on this, construct a 90% confidence interval for the…

A: Solution: Let X be the number of people had kids and n be the number of people sampled. From the…

Q: study done by researchers at a university concluded that 50% of all student athletes in this…

A: Provided information is, study done by researchers at a university concluded that 50% of all…

Q: Out of 600 people sampled, 282 had kids. Based on this, construct a 95% confidence interval for the…

A: Out of 600 people sampled, 282 had kids. Based on this, construct a 95% confidence interval for the…

Q: Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use…

Q: Out of 200 people sampled, 42 had kids. Based on this, construct a 95% confidence interval for the…

A: For 95% confidence intervalα=1-CI=1-0.95=0.05So, zα2=z0.052=z0.025=1.96

Q: According to a survey, 66% of all students in the U. S. own a laptop. For a simple random sample of…

A: It is given that, According to a survey, 66% of all students in the U. S. own a laptop i.e. the…

Q: A local health centre noted that in a sample of 400 patients, 80 were referred ta hospital. a.…

A: Given, n = 400 x = 80 Therefore, population proportion p^ = 80400 = 0.2 The critical…

Q: A recent study of 85 workers found that 61 people rode the bus to work each day. Find the 95%…

A: Given that A recent study of 85 workers found that 61 people rode the bus to work each day. Find…

Q: A poll of 1206 U.S. adults found that 51% regularly used Facebook as a news source. Find the…

Q: t of 500 people sampled, 180 had kids. Based on this, construct a 95% confidence interval for the…

A: Given data,n=500x=180construct a 95% confidence interval?

Q: In a random sample of 500 adults, 410 described themselves as vegetarians. Construct a 95%…

A: We have been provided with the following information about the adults who describe themselves as…

Q: A poll of 2977 U.S. adults found that 28% regularly used Facebook as a news source. Find the margin…

A: Solution: Given information: n = 2977 Sample size p^= 0.28 Sample proportion.

Q: Of 219 employees selected randomly from a company, 10% of them commute by carpooling. Construct a…

A: Given Data Sample Size, n = 219 Sample proportion = p̂= 0.10 Significance level, α…

Q: Out of 500 people sampled, 355 had kids. Based on this, construct a 90% confidence interval for the…

A: By using the given information, we have Out of 500 people sampled, 355 had kids. 90% confidence…

Q: Out of 200 people sampled, 178 had kids. Based on this, construct a 95% confidence interval for the…

A: We have given that, Favorable cases (X) = 178 and sample size (n) = 200 Then, We will find the…

Q: college performs a survey of 424 randomly chosen graduates to estimate the proportion of alumni who…

A: We have given that Sample size n =424 Favorable cases x =361 Sample proportion p^=x/n =361/424…

Q: A sociologist is studying the prevalence of crime in one major city. In a sample of 200 randomly…

A: Here we need to construct a 95% confidence interval for the proportion of people who have been…

Q: Out of 54 randomly selected patients of a local hospital who were surveyed, 42 reported that they…

A: Given that, The total patients n are 54 and the total patients who were satisfied x are 42.

Q: A poll of 2657 U.S. adults found that 86% regularly used Facebook as a news source. Find the margin…

A: Obtain the margin of error at the 90% level of confidence. The margin of error at the 90% level of…

Q: A survey of 200 applicants for police officer training showed that 40 of the applicants were women.…

Q: ut of 300 people sampled, 261 had kids. Based on this, construct a 99% confidence interval for the…

Q: Out of 600 people sampled, 192 had kids. Based on this, construct a 90% confidence interval for the…

A: Given data,n=600x=192sample proportion p=xn=192600=0.32z-value at 90% confidence is Zc=1.645

Q: Out of 400 people sampled, 304 had kids. Based on this, construct a 95% confidence interval for the…

Q: 300 people sampled, 120 had kids. Based on this, construct a 90% confidence interval for the true…

A: The value of z at a 90% confidence level is 1.645, which can be obtained by using the standard…

Q: Random sample of 200 patients in NYC were tested for the desease. With 95% confidence, the rate of…

A: We have to find margin of error.

Q: Out of 200 people sampled, 132 had kids. Based on this, construct a 90% confidence interval for the…

A: Given Data Sample Size, n = 200 Number of successes, x = 132.0 Significance level,…

Q: Out of 100 people sampled, 14 had kids. Based on this, construct a 95% confidence interval for the…

A: Given data, x=14 n=100 p=x/n= 14/100= 0.14 The z value at 95% confidence is Z=1.96

Q: Out of 500 people sampled, 415 had kids. Based on this, construct a 95% confidence interval for the…

Q: During a period of 11 years, of 1700 people that were selected for grand jury duty, 30% of them were…

A: For large sample size population proportion will follow normal dustribution as per central limit…

Q: A random sample of 200 employees at a company shows that 84 drive US made automobiles (the rest…

Q: A poll of 2612 U.S. adults found that 15% regularly used Facebook as a news source. Find the margin…

A: The margin of error is defined as MOEγ=zγ×σn,γ is the confidence level,σ is the sample standard…

Q: Out of 100 people sampled, 48 had kids. Based on this, construct a 90% confidence interval for the…

Q: Out of 100 people sampled, 68 had kids. Based on this, construct a 95% confidence interval for the…

A: It is given that:

Q: A study done by researchers at a university concluded that 65% of all student athletes in a country…

A: We have calculate 95% confidence level margin of error for the proportion.

Q: A medical researcher has conducted a study that involved 200 hypertension patients and 180…

Q: A poll of 2338 U.S. adults found that 71% regularly used Facebook as a news source. Find the…

A: Given information Total US adults = 2338 Proportion of Adults, who used facebook = 0.71 Level of…

Q: urvey of 200 college students revealed that 160 of them eat dessert daily. construct a 95%…

A: Given that A survey of 200 college students . Sample size (n) = 200 Students revealed that eat…

Q: Out of 200 people sampled, 84 had kids. Based on this, construct a 95% confidence interval for the…

A: Correct answer is 0.351 <p< 0.492

Q: In a random sample of 296 people, 179 said that they watched educational television. Find the upper…

Q: Out of 600 people sampled, 120 had kids. Based on this, construct a 99% confidence interval for the…

A: Given p^=sample proportion of kids in the sample of people=120600n=sample size=600Confidence…

Concept explainers

Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.

Question

It was observed that 30 of the 100 randomly selected students smoked. Find the confidence interval estimate of 95% confidence level for \ pi ratio of smokers in the population.

Definition Definition Method in statistics by which an observation’s uncertainty can be quantified. The main use of interval estimating is for describing a range that is made by transforming a point estimate by determining the range of values, or interval within which the population parameter is likely to fall. This range helps in measuring its precision.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step 3

VIEW

Step by step

Solved in 3 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Point Estimation, Limit Theorems, Approximations, and Bounds$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, probability and related others by exploring similar questions and additional content below.

Similar questions

Also, find the 95% confidence interval for the mean time required to graduate for all college graduates!
A random survey of 70 male sophomores showed 16 preferred off campus living and out of 63 female sophomores, 36 preferred off campus living. Find a 95% confidence interval for the difference between the genders.Use 4 decimals
In a survey, a random sample of teenagers were asked whether they have a sibling. The resulting confidence interval for the proportion of teenagers who have a sibling is (.61,.91). What is the sample proportion, p^?
In a random sample, 136 of * persons given a flu vaccine experienced some discomfort. Construct a - % confidence interval for the true proportion of persons who will experience some discomfort from the vaccine. 350 94
In a survey of 1250 U.S. adults, 363 said that their favorite sport to watch is football. The 95% confidence interval for the proportion of adults in the United States who say that their favorite sport to watch is football is_
A random sample of accountants were asked whether they double-check their calculations on a tax form. The resulting confidence interval for the proportion of accountants who double-check their work is (0.89,0.97). What is the sample proportion, p^?
In a survey of 6000 women, 3431 say they change their nail polish once a week. Construct a 99% confidence interval for the population proportion of women who change their nail polish once a week.
A random sample of 300 high school students in a particular town showed that 165 smoke on a regular basis. Find the 95% confidence interval estimating the population percentage for smokers at this high school.
We wish to estimate what percent of adult residents in a certain county are parents. Out of 500 adult residents sampled, 245 had kids. Based on this, construct a 99% confidence interval for the proportion of adult residents who are parents in this county. If n = 560 and ˆpp^ = 0.6, construct a 95% confidence interval.
Decide if the following statement is true or false and explain your reasoning: Suppose you found a sample proportion p^p^and built a 95% and 99% confidence interval around it. If the the population proportion pp is in actually the 95% confidence interval, then it must also be in the 99% confidence interval.
Construct a 95% confidence interval for the average score on the final exam for all students with 5 absences.
A survey collected data on attitudes toward the quality of customer service in retail stores. The survey found that 270 of the 750 American adults sampled felt that customer service is better today than it was two years ago. Find a 94% confidence interval for the proportion of American adults who feel customer service is better today than it was two years ago. p^±z⋅CFp^(1−p^)n−−−−−√p^±z⋅CFp^(1−p^)n Note: Since the population is filled with words, μμ and σσ dont exist. We are trying to estimate a proportion here. What is p^p^ ? ["", "", "", ""] What is the value for z? ["", "", "", ""] What is CF? ["", "", "", ""] What is n?