Construct a 90% confidence interval to estimate the true difference between the proportions of children who read at an early age when they are read to frequently compared to those who were read to less often, as described in the table of results. Let Population 1 be the children who were read to frequently and Population 2 be the children who were read to less often. Round the endpoints of the interval to three decimal places, if necessary.

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

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Q: Suppose the data in the table below was found from a case-control study.  What measure of…

A: Provided data in the table was found from a case-control study. We need to construct the 95%…

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Q: Suppose that the a college wants to have an idea of the average SAT score of the incoming freshman…

A: The confidence level is 0.90, the standard deviation is 310 and margin of error E is 50.

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

Q: Find a 95% confidence interval for the proportion of all U.S. adults with household incomes of at…

A: We need to construct the 95%95\%confidence interval for the population proportion. We have been…

Q: Example: The National Health Interview Survey for 2011-2014 estimates that 37%   of 18-24 year olds…

A: Solution-: Given: n=30,p=0.37,p^=0.45,α=0.05 We find 95% C.I. for the true population proportion and…

Q: The following table represents the probability of the number of cars owned by a college student.…

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

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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: 90% confidence interval is narrower than a 95% confidence interval using the exact same data.…

A: Conditions on confidence interval:   As the confidence level increases, width of the confidence…

Q: In 2010, the National Football league adopted new rules designed to limit head injuries. In a survey…

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Q: Pew Research Center survey asked a variety of science questions of independent rande public…

A: The value of z at 98% confidence level is 2.326, which can be obtained by using the standard normal…

Q: Construct a 99% confidence interval to assess whether this represents evidence that parents'…

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Q: In Oct. 2008, scientists tested cord blood samples for 20 newborn infants in USA. The scientists…

A: Given information- Sample size, n = 20 Now, degree of freedom is given by- Degree of freedom, df = n…

Q: A Food Marketing Institute found that 28% of households spend more than $125 a week on groceries.…

A: given data  p (spend more than $125) = p = 0.28 n = 235 P(p^<0.31) = ?

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: An issue of Time Style and Design reported on a poll conducted by Schulman Ronca & Bucuvalas Public…

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Q: arly-childhood-development studies indicate that the more often a child is read to from birth, the…

A: Answer  Given  X1 =43 n1 =91 x2 =51 n2 =81 Confidence interval =99%

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: In general, would you expect a 90% confidence interval on μ to be longer or shorter than a 95%…

A: Given It is given that the sample is same and CL is increased from 90% to 95%.

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: 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…

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Q: What is the effect on a confidence interval if one decreases both sample size and the level of…

A:

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

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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

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A: It is given that sample mean is 7.1 hours and the standard deviation is 5 hours.

Q: Come up with  your own data for a problem that would ask for a 90% confidence interval utilizing Z.…

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