4. A newspaper reports on the topics that teenagers most want to discuss with their parents. The findings, the results of a poll, showed that 46% would like more discussion about the family's financial situation, 37% would like to talk about school, and 30% would like to talk about religion. These and other percentages were based on a national sampling of 505 teenagers. Estimate the proportion of all teenagers who want more family discussions about religion. Use a 95% confidence level.

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A: Given information Error = 4% = 0.04 Confidence level = 0.95

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A: The formula for the confidence interval is , CI=p^± Z*p^1-p^n Where,  p^=the proportion in sample…

Q: Suppose the 95% confidence interval for a population mean is (20.5, 35.5). What is the margin of…

A:

Q: Suppose we want to estimate the proportion of teenagers (aged 13-18) who are lactose intolerant. If…

A: Given information Error = 2% = 0.02 Confidence level = 0.95

Q: In a survey conducted by the Gallup organization September 6-9, 2012, 1,017 adults were asked, "In…

A: Given information: Bar chart displaying the proportion of individuals with positive opinions about…

Q: Suppose we want to estimate the proportion of teenagers (aged 13-18) who are lactose intolerant. If…

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Q: If n = 200 and p^ = 0.64, find the margin of error at a 95% confidence level.

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Q: The. Nielsen television ratings used a random sample of 2,000 homes and found that 45% of the homes…

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Q: In a survey conducted by the Gallup organization September 6-9, 2012, 1,017 adults were asked, "In…

A: Given that  Total= 1017 Find Frequency table

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A: Given that there are 533 teenagers and the percentage of more family discussions about school is 37%…

Q: 11. A television station surveys 200 customers and finds that 23% of customers watch the evening…

A: Margin of error = z*√(pq/n) Where z= z value            n=no of customer            q=1-p…

Q: If 73.3% of the 45 students in our class are under the age of 20, find the 95% confidence interval…

A: The provided information are:

Q: How large is the sample if the confidence level is 97%, Ɛ = 2.5%  and ρ = .35?

A:

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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Q: If n=12, ¯xx¯(x-bar)=44, and s=5, find the margin of error at a 99% confidence level

A: Given: margin of error for a 99% confidence interval. n=12 x¯(x-bar)=44 s=5

Q: Suppose we want to estimate the proportion of teenagers (aged 13-18) who are lactose intolerant. If…

A: Given information Error = 3% = 0.03 Confidence level = 0.95

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A: Given that Elly and Drew work together to collect the data to estimate the percentage of their…

Q: If n=200 and ˆpp^ (p-hat) =0.39, find the margin of error at a 99% confidence level

A: Find z-critical value:…

Q: 2, A recent survey asked San Diego residents the following question: "If you had been aware that…

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Q: A magazine provided results from a poll of 1000 adults who were asked to identify their favorite…

A: Given Information :  A magazine provided results from a poll of 1000 adults who were asked to…

Q: If n=13, = 42, and s=7, find the MARGIN OF ERROR at a 99% confidence level

A:

Q: A magazine provided results from a poll of 2000 adults who were asked to identify their favorite…

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Q: If n=12, ¯xx¯(x-bar)=48, and s=6, find the margin of error at a 98% confidence level

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Q: 3. Aresearcher wants to estimate the daily expenses of college students. He wants a 99% confidence…

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Q: In a survey conducted by the Gallup organization September 6-9, 2012, 1,017 adults were asked, "In…

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Q: 19) A political survey had a margin of error of 4.25%. If the surveyor used a 90% confidence level…

A: Assume p-hat as 0.5, since no preliminary estimate is available.

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Q: (c) If the confidence levels were 98% rather than 80%, would the margin of error be larger or…

A: The confidence levels are 98% and 80%.

Q: A magazine provided results from a poll of 500 adults who were asked to identify their favorite pie.…

A: Solution: Given information: n= 500 Sample size Here we need to interpret the meaning of each…

Q: (d) Use the combined data of 80 students, construct a 95% confidence interval estimate of the…

A: Given that  Total students=n=40+40=80 Number of pass students=X=28+32=60

Q: If n=10, ¯xx¯(x-bar)=44, and s=8, find the margin of error at a 80% confidence level

A:

Q: 26.4, 226.9, 227.1, and 228.6.  Student B:  229.5, 226.1, 230.7, 223.8, and 227.5

A: For student A: Sample size, n=5 228.3, 226.4, 226.9, 227.1, 228.6…

Q: A magazine provided results from a poll of 1500 adults who were asked to identify their favorite…

A: We have to find out given answes ..

Q: If n=17, (x-bar)=44, and s=8, find the margin of error at a 95% confidence level

A: The sample mean is 44 and standard deviation is 8.