A researcher believes that children who attend elementary school in a rural setting differ in their amount of  physical activity than children who attend elementary school in an urban setting. The researcher collects a random sample from each population and records 63 of the 93 children from rural schools and 74 of the 88 children from urban schools participate in at least one hour of rigorous activity a day. Compute a 95% confidence interval for the difference in proportions of physical activity of rural and urban children.   a. Name of Interval: b. State the random variable(s) and the parameter(s) in words. c. State and check the assumptions for a hypothesis test/interval. d. Find the sample statistics and the confidence interval.

Q: Of 363 randomly selected medical students, 20 said that they planned to work in a rural community.…

A:

Q: A random sample of 12-ounce milkshakes from 11 fast-food restaurants had the following number of…

A: The sample mean (x̄) is calculated by adding up all the values in the sample and dividing by the…

Q: As part of the Pew Internet and American Life Project,, researchers conducted two surveys in late…

A: Given: n1= 800 ,x1= 584 n2= 2253 ,x2= 1059 At 95.0 % confidence level , z score= 1.96 Need to find…

Q: In a survey of 3005 adults aged 57 through 85 years, it was found that 81.7% of them used at least…

A: Given :  n = 3005 p̂=0.817

Q: knowing

A: Given: Total number of drivers = 383 Drivers who buckle up = 297 Proportion of drivers, p =…

Q: A pharmaceutical company is running tests to see how well its new drug lowers cholesterol. Twelve…

A: The objective of this question is to construct a 99% confidence interval for the true mean…

Q: a survey of 3088 adults aged 57 through 85​ years, it was found that 87.1​% of them used at least…

A: The formula for computing confidence interval is,  n is the sample size and P is the proportion.

Q: The average IQ score of an individual is 100. In a study based on a random sample of 1,000 high…

A:

Q: Insurance companies are interested in knowing the population percent of drivers who always buckle up…

A: Solution

Q: A surprising number of young adults (ages 19 to 25) still live in their parents' homes. A random…

A: Sample taken for ages 19 to 25 2253 men and 2629 women random sample In which 986 men and 923 women…

Q: A survey of 400 students at Red Rock College found that 180 planned to take summer classes. Find a…

A:

Q: Suppose a random sample of 142 graduates admissions officers were asked what role scores on…

A: Proportion is almost similar to the concept of probability. Proportion is a fraction of population…

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: 99% confidence interval for the proportion of the high school students who have been cyberbullied,…

A: Given that Sample size n = 14795 Number of high school students who have cyberbullied, X = 2219 The…

Q: In a survey of 1037 US adults ages 65 and over, 643 were concerned about getting the flu. A 99%…

A: Given,n=1037x=643p^=xnp^=6431037=0.6201-p^=1-0.620=0.380α=1-0.99=0.01α2=0.005Z0.005=2.58    (from…

Q: A study of 420,098 cell phone users found that 0.0324% of them developed cancer of the brain or…

A: sample size(n)=420098sample proportion()=0.0324%=0.000324

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: What proportion of the sample reported that they had been cyberbullied?

A: Givensample size(n)=14795No.of respondents say yes they had been cyberbullied(x1) = 2219No.of…

Q: A random variance survey of 495 adults found that 59 had diabetes. which of the following is a 98%…

A:

Q: What is the value of Z for this confidence interval?

A: Given,sample size(n)=14795no.of people who said yes=2219no.of people who said no =12576

Q: A pharmaceutical company is running tests to see how well its new drug lowers cholesterol. Eleven…

A: From the provided information, Sample size (n) = 11 Confidence level = 95%

Q: The career center at a large university surveyed a random sample of students who graduated the…

A: First, we need to calculate the proportions of engineering and business majors who were able to find…

Q: A surprising number of young adults (ages 19 to 25) still live in their parents' homes. A random…

A: Required conditions: n1>30 n2>30 n1*p1^>=10 n1*(1-p1^)>=10 n2*p2>=10 n2*p2^>=10

Q: andom sample of 1200 high school freshmen was taken and it was found that 559 of them were reading…

A: The population proportion (p) will be proportion of high school freshman who reads at or above their…

Q: An insurance company collects data on seat-belt use among drivers in a country. Of 1100 drivers…

A: We have given two age groups of drivers and an insurance company collects data on seat-belt use…

Q: A surprising number of young adults (ages 19 to 25) still live in their parents' homes. A random…

A:

Q: 7 out of 70 students said they were comfortable with “working in a workplace with a lot of…

A: Given that  57 out of 70 students said they were comfortable with “working in a workplace with a lot…

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

A:

Q: A study of 420,010 cell phone users found that 0.0319% of them developed cancer of the brain or…

A: Solution: Given information: n= 420010 Sample size  p^=0.0319100= 0.000319 Sample Proportion

Q: A medical study showed that 57 of 300 persons failed to recover from a given disease. Find a 95%…

A: Sample size: Number of successes: Confidence level:

Q: A high school counselor is interested in determining factors that contribute to high school drop out…

A: Solution; Given information: n1=41 Sample size of high school  graduatesn2= 38 Sample size of high…

Q: A university dean is interested in determining the proportion of students who receive some sort of…

A: Given that, X = 118 N = 200 99% confidence level

Q: Insurance companies are interested in knowing the population percent of drivers who always buckle up…

A: Givenx=316sample size(n)=391Confidence level=95%p^=xn=316391≅0.8082

Q: A survey of credit card holders revealed that Americans carried an average credit card balance of…

A: M1=$3900M2=$3300n1=400n2=450σ1=$880σ2=$810

Q: From a random sample of 64 high school seniors, the sample mean and standard deviation of the math…

A:

Q: Insurance companies are interested in knowing the popul drivers who always buckle up before riding…

A: Given data,No of people surveyed n=387who claims to buckle up is x=311Construct a 93 % CI?

Q: A surprising number of young adults (ages 19 to 25) still live in their parents' homes. A random…

A: Required conditions: n1>30 n2>30 n1*p1^>=10 n1*(1-p1^)>=10 n2*p2>=10 n2*p2^>=10

Q: An insurance company collects data on seat-belt use among drivers in a country. Of 1100 drivers…

A: Answer:-  Given, Sample size (30-39), n1 = 1100 Sample proportion (30-39), p1 = 0.25 Sample size (55…

Q: In a college 500 students were randomly selected for a survey to determine if they own a laptop. Of…

A: X: No. of students who own a laptop. Given:

Q: f 260 employees selected randomly from one company, 18.46% of them commute by carpooling. Construct…

A: Given data, n=260 p=18.46% =0.1846 z-value at 98% confidence is Zc=2.326

Q: A study of 420,098 cell phone users found that 0.0324% of them developed cancer of the brain or…

A: The question is about confidence interval.Given :Sample no. of cell phone users ( n ) = 420098Sample…

Q: In a study of the relationship between alcohol use and bicycling fatalities, the researcher…

A:

2-Proportion Test

In a 2-proportion test, two separate proportions are measured at the same time, and their similarity is evaluated. It is also known as a 2-proportion z-test, and a comparison of two different proportions is done in this method.