A study of 420,074 cell phone users found that 138 of them developed cancer of the brain or nervous system. Prior to this study of cell phone​ use, the rate of such cancer was found to be 0.0344​% for those not using cell phones. a. Use the sample data to construct a 95​% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system.

Q: Construct the requested confidence interval. In a Gallup Poll of 1005 individuals, 452 thought they…

A:

Q: During a flu vaccine shortage in the United States, less than a majority of people were able to…

A: Given data, n=2350 x=978 sample proportion p=xn=9782350=0.4162   z-value at 95% confidence is…

Q: A study of 420,055 cell phone users found that 137 of them developed cancer of the brain or nervous…

A: From the provided information, Sample size (n) = 420055 Out of which 133 developed cancer of the…

Q: In a study of 480 households, 168 households were found to have 4 cell phones. Find a 95% confidence…

A: Given: x=168 n=480 p^=xn=168480=0.35

Q: In a study of 200 accidents that required treatment in an emergency room, 80 occurred at work. Find…

A: Obtain the 90% confidence interval for the true proportion of accidents that occurred at work. The…

Q: In a survey of 1,200 social media users, 56% said it is not okay to be friends with your boss.…

A:

Q: In a survey of 950 randomly selected individuals, 17% answered yes to the question "Do you think the…

A: Sample size, n = 950Probability = 17% = 0.17 90% confidence level

Q: In a study of government financial aid for college​ students, it becomes necessary to estimate the…

A: formula for the sample size :

Q: In a random sample of 500 people eating lunchbat a cafeteria on various fridays, it was found that…

A: Given X=160 n=500 P=X/n

Q: The 2010 General Social Survey asked the question: “After an average work day, about how many hours…

A: a) There is 95% confidence that the proportion of number of hours spent relaxing or pursuing…

Q: A pew Poll conducted in 2013 surveyed a random sample of 2003 adult Americans, and 18% of them said…

A: The random variable X follows normal distribution. We have to construct 90% confidence interval.…

Q: In a random sample of 500 people eating lunch at a cafeteria on various Fridays, it was found that…

A: Given that Sample size n= 500  Favorable cases x=160 Sample proportion p^=x/n

Q: In 2017, the CDC administered the Youth Risk Behavior Survey (YRBS) to a random sample of 14,795…

A: The random variable cyber bullied follows normal distribution. The sample size is 14795. The number…

Q: A Gallup Youth Survey in 2001 found that of 501 randomly selected teenagers in the U.S., 271 of them…

A: Given Total number of teenagers selected = 501Number of teenagers said who get along with their…

Q: A survey found that out of 900 workers, 150 said they were interrupted by frank phone calls. Find…

A:

Q: In a survey of 50,00 people, 80% indicated that they listened to their favorite radio station…

A: Hern= 5000 P^ = 0.80 We have to construct 95% ci for p

Q: In 2017, the CDC administered the Youth Risk Behavior Survey (YRBS) to a random sample of 14,795…

A: We have given that,  Favorable cases (X) = 2219 and sample size   (n) = 14795. Then,  We will find…

Q: B. Construct a 95% confidence interval estimate of 4-4, the difference between the mean weight of…

A: Given Information: B) 95% confidence interval estimate of difference between the mean weight of…

Q: A &B

A: The provided data mentions the value of sample size, n is 928 and the value of x is 128. The…

Q: 1. A poll of 2,500 people shows that 50% approve of a smoking ban in bars and restaurants. What is…

A:

Q: In a survey of 7100 t.v views, 40% said they watch network news programs. Find the margin of error…

A: Solution: From the given information, p-hat=0.40 and n=7100. The confidence level is 0.95.

Q: An independent-measures research study uses a total of 18 participants to compare two treatment…

A:

Q: In a recent survey of US adults, 35 people were asked the question:   “Do you own a smart phone?”.…

A: Given data,n=35x=29Sample proportion p=xn=2935=0.83z-value at 90% confidence is Zc=1.645

Q: a. the incidence (%) of CHD among smokers: b. the incidence (%) of CHD among non-smokers: c. the RR…

A: Solution: a) The incidence rate of CHD among smokers here is computed as:= 180 / (3920 + 180) =…

Q: 9. A survey found that out of 900 workers, 150 said they were interrupted by frank phone calls. Find…

A:

Q: In a sample of 1901 patients who underwent a certain type of surgery, 22% experienced complications.…

A:

Q: An investigator computes a 95% confidence interval for a population mean on the basis of a sample of…

A: To find half of width, We need to calculate 4 times the sample Size.

Q: Of 380 randomly selected medical students, 21 said that they planned to work in a rural community.…

A: Note- Here I am taking the sample size as 380 because then only the answer will match with the given…

Q: Researchers investigating the new Johnson & Johnson COVID Vaccine selected a random sample of 200…

A:

Q: In a random sample of 184 college students, 97 had part-time jobs. Find the margin of error for the…

A: Given : Sample Size, n = 184 Number of successes, x = 97 confidence level,c=0.99

Q: In a survey funded by the UW school of medicine, 750 of 1000 adult Seattle residents said they did…

A: Find z-critical value:…

Q: A Pew Research Center survey of 1,857 internet users found that 1,124 people agreed that online…

A: Given: CI=95% p^=11241875=0.599 n=1875

Q: A poll of 2500 people shows that 50% approve of a smoking ban in bars and restaurants in Kentucky.…

A: Given, n = 2500 p = 50% = 0.50 At 95% Confidence level, the value of Z is 1.96

Q: 1. A recent survey showed that from a sample of 500 packages delivered by a Postal Service, 480 were…

A: We have given that  Sample size n= 500 , favorable cases x=480 Sample proportion p^=x/n=…

Q: A survey of 503 women shoppers found that 46% of them shop on impulse. What is the 99% confidence…

A: Given : Sample Size, n = 503 confidence level,c=0.99

Q: in a study of government financial aid for college​ students, it becomes necessary to estimate the…

A:

Q: In a random sample of 250 students who enrolled in Science, 150 took notes during classes. Find an…

A: Given that n=250,x =150,alpha =0.05 Find an approximate 95% confidence interval for the population…

Q: in a recent survey of 1000 adults ages 18 to 40, 34% said they had no credit cards. find the 95%…

A: The sample size n=1000. The sample proportion is 0.34. The confidence level is 95%. The area between…

Q: A marketing study with a random sample of 400 results in 160 subjects reacting positively to the…

A:

Q: a) Using a sample size of 75, construct a 95% confidence interval estimate of the mean birth weight…

A: here given ,  A study has been conducted to verify the mean of birth weight at Apple County. From…

Q: A farmer wants to test the effectiveness of a pest control method in allowing strawberry blooms to…

A: Solution: Let X be the number of blooms that yield marketable strawberries and n be the sample…

Q: Among 1815 randomly selected high school students surveyed by the Centers for Diseases Control, 300…

A: Given : X = 300 n = 1815 p^ = xn    = 3001815    = 0.1653

Q: 2. Estimate the proportion of WCU students who have type B blood using 95% confidence interval.…

A: Given , n=100x=24p^=xn=24100=0.241-p^=1-0.24=0.76α=1-0.95=0.05α2=0.025Z0.025=1.96

Q: The Associated Press (October 9, 2002) reported that in a survey of 4722 American youngsters aged 6…

A: Obtain the 99% confidence level for the proportion of all American youngsters who are seriously…

Q: An airline transports over 10 000 passengers daily, and the airline is curious what proportion of…

A: Critical value: Using Excel formula "=NORM.S.INV(1-(0.1/2))", the z-value is obtained as 1.645.

Q: In the SQF dataset, the sample proportion of the those stopped by police who are Black is 53%. The…

A:

Q: A survey found that out of 800 workers, 150 said they were interrupted by frank phone calls. Find…

A: Given,n=800x=150p^=xnp^=150800=0.18751-p^=1-0.1875=0.8125α=1-0.95=0.05α2=0.025Z0.025=1.96

Q: 2. A survey found that out of 900 workers, 150 said that they were interrupted by frank phone calls.…

A: In question, We have given the number of favorable cases 150 and sample size 900 then we'll find 95%…

Q: In a survey of 2480 golfers, 15% said they were left-handed. The survey's margin of error was 3%.…

A: We want to find the CI for population proportion (p)

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.