A major metropolitan newspaper selected a simple random sample of 1,600 readers from their list of 100,000 subscribers. They asked whether the paper should increase its coverage of local news. Forty percent of the sample wanted more local news. What is the 99% confidence interval for the proportion of readers who would like more coverage of local news? Select one: a. 0.37 to 0.43 b. 0.30 to 0.50 c. 0.35 to 0.45 d. 0.32 to 0.48

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Q: Calculate the lower 95% confidence interval for the population mean. (round to the nearest tenth)…

A: given data n = 31σ2 = 15∴σ = 15 = 3.873x¯ = 5695% ci for μ = ?

Q: Use the sample data and confidence level given below to complete parts  (c).   A research institute…

A: It is given that, the number of respondents said yes (x) is 587 and the sample size (n) is 912.  The…

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Q: A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll,…

A: Given: n= 1031 ,x= 545 need to find point estimation for p.

Q: The percentage of high school graduates applying to multiple colleges has been increasing over the…

A: Givenn=374x=157p^=xn=157374=0.41981-p^=1-0.4198=0.5802α=1-0.90=0.1α2=0.05Z0.05 =1.645

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A: Given data: n = 1300  p̂ = 0.21. Confidence level = 95 %

Q: Use the sample data and confidence level given below to complete parts (a) through (d). A research…

A:

Q: In a survey of 3700 TV viewers, 834 said they watch network news programs. Find the margin of error…

A: Solution

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Q: A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll,…

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Q: In a random sample of 1003 US adults, 32% said that national identity is strongly tied to birth…

A: It is given that, 32% of the 1003 adults said that national identity is strongly tied to birth…

Q: Suppose we observed 120 successes in a random sample of 300 subjects. What is the 98% confidence…

A:

Q: A student polls his school to see if students in the school district are for or against the new…

A: Given: n=600 and x=480. Then,p^=xn=480600=0.8

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A: From the provided information Sample size (n) = 29, People who recommend x = 9 The sample proportion…

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A: Given  e=Margin of error=8%, level of significance ɑ=0.10

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A:

Q: Of 600 students sampled, 480 said they hoped to own a house someday. With 95% confidence, what is…

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Q: Determine the sample size for a on-line survey with a variability of 30% and a 95% confidence…

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A:

Q: A survey of 26 individuals found that they spent an average of $39.21 on exercise attire for the…

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A: Given find the multiplier for 60% confidence

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Q: A new fertilizer was applied to the soil of 222 bean plants. 28% showed increased growth. Find the…

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A: Solution:There is no estimates available, hence the sample proportion is assumed to be 0.50.The…

Q: In order to make the decision for one business, your team constructed a 90% confidence interval, the…

A: For 90% confidence interval, the width of that confidence interval is 500.We have to reduce the…

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A:

Q: a survey of 183 randomly selected gun owners, it was found that 67 of them said they owned a gun…

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Q: Mr. Price calculates a 95% confidence interval for the difference between the proportion of Aaron…

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Q: (a) Find a 95% confidence interval for p1 - P2. (Use 3 decimal places.)

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A: Given that  Margin of error =E=6.50%=0.0650 99% confidence.

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Q: Use the sample data and confidence level given below to complete parts (a) through (d). A research…

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A: Given Data Lower limit,LL = 12.5 upper limit,UL = 18.5

Q: In a survey of 1036 adults, a poll asked, "Are you worried or not worried about having enough money…

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Q: A poll of 1021 U.S. adults found that 46% regularly used Facebook as a news source. Find the…

A: Given, n = 1021 p^=0.46 q^ = 1 - p^ = 1 - 0.46 = 0.54 α=0.1

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.