Suppose that a safety group surveyed 1,200 drivers. Among those​ surveyed,78​% said that careless or aggressive driving was the biggest threat on the​ road, and 33​% said that cell phone usage by other drivers was the driving behavior that annoyed them the most. Based on these data and assuming that the sample was a simple random​ sample, construct and interpret a 99​% confidence interval estimate for the true proportion in the population of all drivers who are annoyed by cell phone users. The confidence interval estimate is _____ to _____. ​(Round to three decimal places as needed. Use ascending​ order.)

Q: A poll reported that 67.7% of adults in a certain country believe that organic produce is better for…

A:

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: Thirteen out of the 137 randomly selected faculty members who were surveyed at a community college…

A: Obtain the 90% confidence interval for the proportion of all faculty member at the community college…

Q: Several years ago, 39% of parents who had children in grades K-12 were satisfied with the quality of…

A: I have used Excel function = NORMSINV() for finding critical value

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: Construct a 99.8% confidence interval for the proportion of women who work at home who are college…

A: Given,n=519x=165p^=xnp^=165519=0.318The point estimation for the proportion is 0.318 .

Q: A learn-to-type software program claims that it can improve your typing skills. To test the claim…

A: We make the following table for the difference from the given information  Typing Speeds (in Words…

Q: a) Construct a 95% confidence interval for the slope.

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Q: For a sample of 120 businesses, the average profit they generate in a month is $54,000 with a…

A: We have given that,  Average or sample mean (x̄) = 54000, standard deviation (s) = 8800 and sample…

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

A: (a)given that,possible chances (x)=410size(n)=603success rate ( p )= x/n = 0.6799point of estimate =…

Q: 1. A survey of 224 children of ages 9 to 17 was conducted shortly after 1986 space-shuttle disaster.…

A:

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: *** a. Use the sample data to construct a 90% confidence interval estimate of the percentage of cell…

A: A study of 420,041 cell phone users found that 0.0314% of them developed cancer of the brain or…

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: 1. In a recent poll 62 out of 149 randomly chosen students at Allen High School say they plan to buy…

A:

Q: In a survey, 13 people were asked how much they spent on their child's last birthday gift. The…

A: We have to construct ci.

Q: In a survey, 45 people were asked how much they spent on their last partners birthday gift. The…

A: Given information- Sample size, n = 45 Significance level, α = 0.10 Sample mean, x-bar = $90 Sample…

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: A study of 420,078 cell phone users found that 132 of them developed cancer of the brain or nervous…

A: given that,possible chances (x)=132size(n)=420078success rate ( p )= x/n = 0.000314point of estimate…

Q: A recent study randomly selected women from various countries around the world and collected…

A:

Q: Construct a 99% confidence interval for the proportion of those aged 65 and over who have sleep…

A: Given

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

A:

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

A:

Q: #4. A college president commissioned a study to find the proportion of students enrolled at her…

A:

Q: A medical researcher claims that smoking can result in the wrinkled skin around the eyes. The…

A: Given Information: Smokers: n1=150 x1=95 Nonsmokers: n2=250 x2=105 Confidence level is 95%

Q: GROUP 2 You work for a consumer advocate ogency and want to find the mean repair cost of a washing…

A: According to the provided information, Mean (x̅) = 100 Standard deviation (σ) = 17.50 Sample size…

Q: In a survey, 24 people were asked how much they spent on their child's last birthday gift. The…

A: From the available information, Sample size, n =24 Sample mean, x = 31 Sample standard deviation, s…

Q: An economist at the Florida Department of Labor and Employment needs to estimate the unemployment…

A: An economist at the Florida Department of Labor and Employment needs to estimate the unemployment…

Q: 1. In a recent study, the researchers want to find the information about the percentage of adults…

A: Given datan = 3026proportion of people who prefer to get their news online = 0.38

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

A: (a)given that,possible chances (x)=410size(n)=603success rate ( p )= x/n = 0.6799point of estimate =…

Q: Ten different families are tested for the number of gallons of water a day they use before and after…

A: From the provided information, The difference between before and after are as follow: Before…

Q: . A sociologist develops a test to measure attitudes about public transportation, and 20 randomly…

A: Answer - A sociologist develops a test to measure attitudes about public transportation,…

Q: A SURVEY WAS CONDUCTED IN 2010 AND 2020. THE 2010 RESULTS SUMMARIZED THE RESPONSES OF 1250…

A: Confidence interval are used to measure the range where the population proportion lies. It is…

Q: A researcher examined the extent of cooperation among kindergarten children. She unobtrusively…

A:

Q: 1. Online News In a Harris poll of 2036 adults, 40% said that they prefer to get their news online.…

A: Given that the sample size is n=2036 and sample proportion, p^=0.40. Confidence level=95%…

Q: A summary at the beginning of the report states "All drivers want some kind of workplace benefit."

A: Here given Students who participated in a UCLA Labor Summer Research Program in 2017 surveyed 260…

Q: survey, 30 people were asked how much they spent on their child's last birthday gift. The results…

A: From the available information, Sample size, n=30 Sample mean,  Sample standard deviation, s=$19.…

Q: company manufactures a certain over-the-counter drug. The company samples 80 pills and finds that…

A: We have given that  Sample size n = 80 Sample mean = 325.5 Standard deviation s=10.3 90%confidence…

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A:

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

A: Given :-   randomly selects 200 students and finds that 118 of them are receiving financial aid…

Average

In mathematics, everyone must have heard the term average in one place or other. Almost all types of calculation require taking out an average. In everyday life, the term is used for denoting a certain amount that can be said to be a typical quantity for multiple people or objects. In other words, if a group of people possesses a certain object, then the average gives us an idea regarding the quantity of the object possessed by each person. However, this is solely based on mathematical calculation. It is important to note that most people will not strictly have the average value in a group.