researcher takes a random sample of 100 Americans that have never had a pet and a separate random sample of 100 Americans that are current or previous pet owners. She is interested in determining whether there is a difference in the proportion of individuals who believe in the afterlife among these two groups. She observes that 76 of the "never pet owners" and 84 of the "ever pet owners" believe in the afterlife. Determine the margin of error for a confidence interval for the difference in the two appropriate proportions at a 95% confidence level. a. The assumptions necessary to perform the calculation are not satisfied. b. 0.0837 c. 0.0719

Q: In a survey of 650 randomly selected U.S. males ages 18-64, 396 say they have gone to the dentist in…

A: Given,n=650x=396p^=xnp^=396650=0.6091-p^=1-0.609=0.391α=1-0.90=0.1α2=0.05Z0.05=1.645     (from…

Q: There are 3000 beads of various colors in a jar. In a sample of 125 beads, 40% were red. Calculate…

A:

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: When Mendel conducted his famous genetics experiment with peas, one sample of offspring consisted of…

A: Obtain the 95% confidence interval estimate of the percentage of yellow peas. The 95% confidence…

Q: In a random sample of 100 customers, the items that customers purchased were tracked. It was found…

A: Given,sample size(n)=100sample proportion(p^)=0.171-p^=1-0.17=0.83α=1-0.90=0.1α2=0.05Z0.05=1.645

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

A: It is given that, x is 311 and n is 400.

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

A: Given data, n=260 p= 18.46% = 0.1846   The z value at 90% confidence is 1.64

Q: A recent year was randomly picked from 1985 to the present. In that year, there were 2051 Hispanic…

A: Given, x1 = 2051 n1 = 12328 x2 = 321 n2 = 2441 α = 0.1       To find, 90% confidence interval for…

Q: A random sample of 230 people living in Seattle found that 83 of them had a pet. A random sample of…

A: Solution: Given information: n1= 230 Sample size of people iving in Seattle n2= 172  Sample size of…

Q: In a sample of 100,40 said they preferred Pepsi to Coke. A 95% confidence interval for the…

A:

Q: A Pew Internet Project Data Memo presented data comparing adult gamers with teen gamers with respect…

A: There are two independent samples which are adult gamers and teen gamers. We have to construct 95%…

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: In a random sample of 500 people aged 20-24, 22% were smokers. In a random sample of 450 people aged…

A: We have to find confidence interval...

Q: A researcher was interested in knowing how many people in the city supported a new tax. She sampled…

A:

Q: A poll of 827 students at Alpha state college found that 61% of those polled preferred a quarter…

A: It is given that the proportion of the students who preferred a quarter system to a trimester…

Q: In a sample of 100, 40 said they preferred Pepsi to coke. A 95% confidence interval for the…

A:

Q: If n=41n=41, ¯xx¯=36, and s=9s=9, find the margin of error for a 80% confidence level.

A: Answer - Given, Sample size n = 41,    x  = 36    and s = 9               80% confidence interval…

Q: A researcher wanted to check whether the percentage of Canadian adults with obesity has decreased in…

A: Number of Canadian adults in 2005= 75 Number of Canadian adults in 2009 = 120   significance level =…

Q: A department chair wanted to know whose students performed better on a common pass/fail final exam:…

A: p1 :  proportion of students passed in smith's  class p2 :  proportion of students passed in Jones's…

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: 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: 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 a survey of 2255 randomly selected US adults (age 18 or older), 1787 of them use the Internet…

A: We want to calculate the 90% confidence interval for each of the following proportions

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: one with a 95% confidence level and the other with a 99.7% confidence level. Which interval is…

A: Confidence interval is probability, which refers that population parameter will fall within set of…

Q: qualities as options, one of which is "communicator", as the most important quality for an effective…

A: It is given that  Favourable cases, X = 121 Sample size, n = 275 Sample proportion, p^ = X/n…

Q: 324 out of a simple random sample of 400 students hated their online math course. Find a 99%…

A: Given: Number of students that hated the online math course (y) = 324 Number of students (n) = 400…

Q: Suppose we observed 15 successes in a random sample of 60 subjects. What is the 99% confidence…

A:

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: a newspaper article stated that the us supreme court received 812 letters from around the country on…

A: To find the 95% confidence interval for the true proportion of all American adults who feel that…

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

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Q: In a sports poll, 142 of 500 randomly selected Americans indicated that they consider themselves to…

A: Given that, In a sports poll, 142 of 500 randomly selected Americans indicated that they consider…

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

A: