n a random sample of 800 men aged 25 to 35 years, 24% said they live with one or both parents. In another sample of 850 women of the same age group, 18% said that they live with one or both parents. A. Construct a 95% confidence interval for the difference between the proportions of all men and all women aged 25 to 35 years who live with one or both parents. B. Test at the 2% significance level whether the two population proportions are different. C. Repeat the test of part b using the p-value approach.

Q: In a random sample of 100 men in the United States, 55 of them were married. Using a 95% confidence…

A: We have given that,  Sample size (n) = 100 and favorable cases (X) = 55  Then,  We will find the 95…

Q: 2. Out of a sample of 760 people, 367 own their homes.. Construct a 95% confidence interval for the…

A:

Q: Refer to the data set of 20 randomly selected presidents given below. Treat the data as a sample and…

A: Given that            sample size, n=20 Here, we have to construct a​ 95% confidence interval…

Q: A survey of several 8 to 11 year olds recorded the following amounts spent on a trip to the mall:…

A: The amounts spent on a trip to the mall is 19.36,29.95,7.82,8.36,21.07,14.06,22.22

Q: By law, all new cars must have airbags on both the driver's side and the passenger's side. There is…

A: Given that  n=55 , X=35

Q: Campus athletics is interested in the proportion of registered students living on campus who play…

A: We are given that 95% confidence interval is correctly calculated from the given data as (0.12,…

Q: Which of the following is thet-value for a 95% confidence level when the sample size is 20. O a.…

A:

Q: A statistics instructor randomly selected four bags of​ oranges, each bag labeled 10​ pounds, and…

A: From the provided information, Sample size (n) = 4 Confidence level = 95%

Q: A recent study of 1000 randomly chosen residents in each of two randomly selected states indicated…

A:

Q: n a survey of 500 students, 180 students said that they prefer classes on Mondays and Wednesdays…

A:

Q: . How many of the 3335 subjects used at least one prescription​ medication? b. Construct a​ 90%…

A: n = 3335 p = 0.804

Q: National Sporting Goods Association found that 34% of Americans surveyed had gone swimming during…

A: Given data,           P=34%=0.34           n=1200      Z- value at 95% confidence is ZC=1.96

Q: A telephone poll of 1,000 adult Americans was reported in an issue of Time Magazine. One of the…

A: (a) Given that, Critical values: By using the standard normal table of Z, the critical value at…

Q: In a survey of 3110 adults aged 57 through 85 years, it was found that 87.9% of them used at least…

A:

Q: In a poll of 29 randomly selected students showed that 29.2 % of the respondents believe in global…

A: Given,sample size(n)=29sample proportion(p^)=0.292

Q: 829 adult were surveyed in one city, and 51% of them are opposed to the use of the photo-cop for…

A: Given data: n = 829 p = 0.51

Q: A random sample of 200 BZU students shows that 80 of them have O blood type. Let p be the BZU…

A: There are totally 200 BZU students are sampled in which 80 of them have blood type. The variable p…

Q: The International Nanny Association (INA) reports that in a sample of 928 in- home child care…

A: Solution: A. Let X be the number of number of nannies who have passed the INA certification exam and…

Q: In a survey of 3015 adults aged 57 through 85 years, it was found that 85.6% of them used at least…

A:

Q: An apartment hunter is shopping for either a one-bedroom or two-bedroom apartment in different…

A: Here the new variable is  X= difference  = (Two bed- One bed)  Then mean = ΣXi/n X̄ = (200 + 290 +…

Q: A survey of several 9 to 11 year olds recorded the following amounts spent on a trip to the mall:…

A: From the provided information,Sample size (n) = 7Confidence level = 95%

Q: A study of 420,074 cell phone users found that 138 of them developed cancer of the brain or nervous…

A:

Q: We are interested in comparing the percentage of females versus males who expect their future to be…

A: Summarising the given information below. The number of total surveyed females and males:…

Q: A sample of 10 students took a class designed to improve their SAT math scores. Following are their…

A: Given data number of samples (n) = 10 Before (X) After (Y) d= (X - Y) d2 433 434 -1 1 478…

Q: A genetic experiment with peas resulted in one sample of offspring that consisted of 443 green peas…

A: Find z-critical value:…

Q: Suppose we are comparing two populations to see if they have the same mean. A sample from each…

A: Given data p-value = 0.0156 Null and Alternative Hypothesis Null Hypothesis H0: μ1 = μ2 Alternative…

Q: o test the effect of a physical fitness course on one's physical ability, the number of sit-ups that…

A: Solution:  Let x = Sit ups before  and y = Sit ups after  x y d=y-x (d-d) (d-d)2 34 49 15 7.1…

Q: Refer to the data set of 20 randomly selected presidents given below. Treat the data as a sample and…

A: Out of the given sample of size, n = 20 the number of presidents whose heights were more than their…

Q: A student at a local high school claimed that three- quarters of 17-year-old students in her high…

A: Information given below Hypothesis Ho: p=0.75 vs H1: p≠0.75  P value=0.9315  Level of significance…

Q: In a random sample of 315 people 90 of them were smokers. Based on the above sample test whether the…

A:

Q: 3469 adults aged 57 through 85​ years, it was found that 85.6​% of them used at least one…

A: It is given that sample size (n) is 3,469 and p hat value is 0.856.

Q: a. Interpret a contidence level of 94%. b. Construct and interpret a 94% confidenc interval to…

A: Sample proportion p = 775/1100 p = 0.7045   n = 1100

Q: Given the following data, find the LOWER endpoint for a 95% confidence interval for .…

A: Given data: Confidence level = 95% Sample size = 6

Q: An insurance company collects data on seat-belt use among drivers in a country. 90% confidence…

A: Givenn1=2000p1^=28%=0.28P2^=xn2=4821500=0.32α=1-0.90=0.10α2=0.05zc=zα2=z0.05=1.6449      (from z…