Find and interpret a 95% confidence interval for the proportion p of US 8th-grade students who drink alcohol. (b) Give the best estimate for p and give the margin of error for the estimate.

Q: Several years ago, 50% of parents who had children in grades K-12 were satisfied with the quality of…

A: Given claim: 50% of the parents who had children in grades K-12 were satisfied with the quality of…

Q: A manager needs to know the proportion of employees who have insurance. He selects 200 employees and…

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Q: Several years ago, 36% of parents who had children in grades K-12 were satisfied with the quality of…

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Q: 2. A recent study of 25 students showed that they spent on average of $18.35 for gasoline per week.…

A: Given  x̄ = 18.35 n=25 s=3 degree of freedom = df = n -1 = 25-1 = 24 significance level α = 0.10…

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: Several years ago, 33% of parents who had children in grades K-12 were satisfied with the quality of…

A: The formula for the confidence interval for the proportion is, Here, confidence level is 0.90. Use…

Q: standardized achievement tests to nationwide samples of 17-year-olds in school. One year, the tests…

A: Sample size (n)=6000Sample proportion (p)=0.952We have to find 95% confidence interval for the…

Q: cording to a report in USAToday, more and more parents are helping ndom sample of 40 young adults…

A: The formula to construct confidence interval is, p^±Zα/2*p^(1-p^)n The formula for margin of error…

Q: A developmental psychologist would like to determine how much fine motor skill changes for children…

A: In statistical inference, there are two types of estimation,  Point estimation and Interval…

Q: How do you solve? ref W9Q7.

A: For Women nw=10 α=1-0.90=0.10 Degrees of freedom is obtained as- df=10-1=9 The mean and standard…

Q: ) Find the 99% confidence interval (CI) and margin of error (ME) used to estimate the population…

A: Given data,n=175p=72%=0.72z-value at 99% confidence is Zc=2.576

Q: A​ 95% confidence interval for the proportion of U.S. adults who subscribe to some sort of streaming…

A: In the above question, we are given,  a confidence interval is

Q: Sales of trumpets in a day is modeled as Normal. A random sample of 36 days is looked at. The sample…

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Q: A health insurance firm wants to estimate the proportion of a population that would receive the flu…

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Q: 20. OxyContin The drug OxyContin (oxycodone) is used to treat pain, but it is dangerou because it is…

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Q: Family Closeness. A study conducted among college students evaluated the connection between drug use…

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Q: Several years ago, 43% of parents who had children in grades K-12 were satisfied with the quality of…

A: Given  Several years ago, 43% of parents who had children in grades K-12 were satisfied with the…

Q: Estimate 95% confidence interval for the proportion of household experiencing burglary in 2010."

A: 25.8 households per 1,000 were the victims of home burglary. These data are obtained from a survey…

Q: n a survey, 25 people were asked how much they spent on their child's last birthday gift. The…

A: Givensample size(n)=25Mean(x)=38standard deviation(s)=19confidence interval(c)=98%

Q: Suppose you calculate a 95% confidence interval for a proportion to to be: 0.39 to 0.78. Determine…

A: Given,   Lower bound = 0.39​ Upper bound = 0.78   Here, we use to determine the margin of error.

Q: (a) Construct and interpret a 95% confidence interval for the mean price paid for coffee in 2015 by…

A: given   n =105       x¯  = 3.42     σ  =0.75 a) confidence interval  limits are    [x¯-   αZ/2 (σ/n)…

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…

Q: The Nellie Mae organization conducts an extensive annual study of credit card usage by college…

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Q: A statistics instructor was interested in the proportion of all Kent State students who live on…

A: The proportion that students at Kent State live on campus is obtained below:From the given…

Q: The number ofunhealthy days based on the AQI (Air Quality Index)for a random sample of metropolitan…

A: Use EXCEL Procedure for mean and standard deviation for the given samples. Follow the instruction to…

Q: We plan to sample parents in order to construct a 99% confidence interval for the proportion who…

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Q: Several years​ ago, 42​% of parents who had children in grades​ K-12 were satisfied with the quality…

A: Let p be the population proportion of parents satisfied with the quality of education the students…

Q: An arborist is interested in determining the mean diameter (in centimeters) of mature white oak…

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Q: A study was being done to estimate the proportion of students who receive financial aid. If the…

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Q: Several years ago, 43% of parents who had children in grades K-12 were satisfied with the quality of…

A:

Q: Sleep apnea: Sleep apnea is a disorder in which there are pauses in breathing during sleep. People…

A: Sample size  n = 433Number of people aged 65 and over had sleep apnea(x) = 107

Q: Several years ago, 46% of parents who had children in grades K-12 were satisfied with the quality of…

A: Given  p=0.46, n=1055, X=488

Q: several years ago, 48% of parents who had children in grades K-12 were satisfied with the quality of…

A: Given that  Sample size n =1035 Favorable cases x =497 Sample proportion p^=x/n  =497/1035 =0.4802

Q: An insurance company selected samples of clients under 18 years of age and over 18 and recorded the…

A: Given Information: Under age of 18  n1=480 x1=160 Over age of 18 n2=500 x2=140