A survey of 1060 randomly selected US teens ages 13-17 found that 605 of them say they have made a new friend online. (a) Find and interpret a 90% confidence interval for the proportion, p, of all US teens who have made a new friend online. (b) Give the best estimate for p and give the margin of error for the estimate. (c) Use the interval to determine whether we can be 90% confident that more than half of US teens have made a new friend online.

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Q: In another pre-election poll, a pollster finds that 438 out of 600 people polled favor, Smith.…

A: Given, X = 438 N = 600 The sample proportion is given by:

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Q: In a random sample of205people,147said that they watched educational television. Find…

A: Solution: Let X be the number of people who watched educational television and n be the sample…

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A: Provided, N=480 P=0.52 be the proportion of people who liked dogs. a=0.10

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A:

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Q: A cable company claims that 88% of its customers are fully satisfied with their service. A…

A: Given that  Sample size n =350 Favorable cases x =287 Sample proportion p^=x/n  =287/350 =0.82

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Q: Use the given degree of confidence and sample data to construct a confidence interval for the…

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Q: The figure to the right shows the results of a survey in which 999 adults from Country​ A, 1006…

A: Hey, since there are multiple subparts posted, we will answer the first three subparts. If you want…

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A:

Q: Wisconsin Public Radio wants to duplicate a survey conducted in 2011 that found that 68% of adults…

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A:

Q: According to some car loan statistics, 44% of Americans are relying on an auto loan to pay for their…

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Q: Construct a 99% confidence interval for the true proportion of all car license applications that…

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Q: A survey of 800 women shoppers found that 17% of them shop on impulse. what is 98% confidence…

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Q: f a Random Sample of 450 adults includes 155 who identify as happy, compute an 80% confidence…

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Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.