The USA Today claims that 44% of adults who access the Intenet read the international news online. You want to check the accuracy of their claim by surveying a random sample of 120 adults who access the Internet and asking theim if they read the international news online. Fifty- two adults respond "yes." • Verify that the sampling distribution of p can be approximated by the normal distribution. Use a 95% confidence interval to test the newspaper's claim. Use mathematics to explain your answer. If we were to increase the value of n from 120 to 200, how would the width of the confidence interval be effected?

Q: You measure 40 watermelons' weights, and find they have a mean weight of 76 ounces. Assume the…

A:

Q: ou measure 38 textbooks' weights, and find they have a mean weight of 43 ounces. Assume the…

A: Since we know population standard deviation, use z-distribution to find z-critical value. Find…

Q: y selected textbooks' weights, and find they have a mean weight of 78 ounces. Assume the population…

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A: It is given that confidence level is 0.90, standard deviation is 12 and margin of error E is 5.

Q: You mea sure 43 randomly selected textbooks' weights, and find they have a mean weight of 33 ounces.…

A: Given,   Sample size = 43 Sample mean = 33 Population standard deviation = 13.6   Since the…

Q: A study found that 77% of students report that their peers at least "sometimes" copy information…

A: Margin of error: Margin of error is a range of values that lie above and below the sample statistic…

Q: In a survey, 28 people were asked how much they spent on their child's last birthday gift. The…

A: Given: n=28 standard deviation s = $5 confidence level = 98%

Q: Suppose you use simple random sampling to select and measure 47 dogs' weights, and find they have a…

A: Given, mean is 78 and standard deviation is 13.5. Sample number is 47. We with use one sample z…

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

A:

Q: You measure 50 textbooks' weights, and find they have a mean weight of 71 ounces. Assume the…

A: We have given the following Sample mean = 71 Sample size = 50 Sample standard deviation = 3.8

Q: he average weight of 20 randomly selected cars was 3500 pounds. The sample standard deviation was…

A: given data, n=20df=n-1=20-1=19x¯=3500 pounds s=315 pounds CI=0.90α=1-0.90=0.10

Q: You measure 43 randomly selected textbooks' weights, and find they have a mean weight of 33 ounces.…

A: We have to find 95% confidence interval for the true population mean textbook weight.

Q: You measure 35 textbooks' weights, and find they have a mean weight of 36 ounces. Assume the…

A: Given information- Population standard deviation, σ = 9.3 ounces Sample size, n = 35 Sample mean,…

Q: students. You want to estimate the proportion p of students who never floss with 90% confidence and…

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Q: How large a sample is necessary to estimate the true proportion of unmarried Canadians aged 25-29…

A: Given data: Margin of error  = 5% = 0.05 Confidence level  = 0.90 To find: Sample size (n) = ?

Q: You measure 24 textbooks' weights, and find they have a mean weight of 53 ounces. Assume the…

A: Given that, x¯=53,n=24,σ=12.9 Critical value: By using the z-tables, the critical value at 5% level…

Q: You measure 45 watermelons' weights, and find they have a mean weight of 56 ounces. Assume the…

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Q: You measure 37 textbooks' weights, and find they have a mean weight of 59 ounces. Assume the…

A:

Q: Refer back to the pizza-delivery Try It exercise. The mean delivery time is 36 minutes and the…

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Q: If no previous study has been done, give the sample size we should use to find p with a margin of…

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Q: have a mean weight of 65 ounces. Assume the population standard deviation is 4.6 ounces. Based on…

A: Given Sample mean=65 Population standard deviation=4.6

Q: A runner wants to know her average mile running time. She records a sample of 25 running times. The…

A: Solution: Let X be the mile running time. From the given information, x-bar=6.5 minutes, s=0.7…

Q: In a survey, 28 people were asked how much they spent on their child's last birthday gift. The…

A:

Q: You measure 20 dogs' weights, and find they have a mean weight of 44 ounces. Assume the population…

A: Since we know population standard deviation, use z-distribution to find z-critical value. Find…

Q: You measure 50 textbooks' weights, and find they have a mean weight of 66 ounces. Assume the…

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Q: An administrator in charge of undergraduate education on a large campus wanted to estimate the…

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Q: We have developed a series of questions to measure job satisfaction of bus drivers in London,…

A:

Q: You measure 46 textbooks' weights, and find they have a mean weight of 34 ounces. Assume the…

A: given data population sd(σ) = 14.3n = 46x¯ = 3495% ci for mean (μ) = ?

Q: In a survey, 12 people were asked how much they spent on their child's last birthday gift. The…

A: Since population standard deviation is unknown, Use t-distribution to find t-critical value. Find…

Q: A credit counseling agency wants to estimate the mean amount of credit card debt held by Americans.…

A: Here we need to construct a 90% confidence interval for the population mean.

Q: A researcher wants to estimate the average amount of money a person spends on buying lottery tickets…

A: Given: Let X be the amount of money a person spends on buying lottery tickets each week Sample size…

Q: In a survey, 21 people were asked how much they spent on their child's last birthday gift. The…

A: Givensample size(n)=21Mean(x)=34standard deviation(s)=6confidence level=80%

Q: A recent article published by the Wall Street Journal said the Cincinnati airport had the most…

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Q: You measure 34 turtles' weights, and find they have a mean weight of 65 ounces. Assume the…

A:

Q: The American Wedding Study provides information related to weddings in the United States. Last year…

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

A: It is given that sample mean is 47 and the standard deviation is 12.