QUESTION 1 Of 386 randomly selected medical students, 30 said that they planned to work in a rural community. Find a 95% confidence level to estimate the true proportion of all medical students who plan to work in a rural community QUESTION 2 33 packages are randomly selected from packages received by a parcel service. The sample has a mean weight of 11 pounds and a standard deviation of 2.6 pounds. What is the 95% confidence interval for the true mean weight of all packages received by the parcel service?_Round to nearest pound QUESTION 3 A light-bulb manufacture wants to determine how large a random sample the quality control department should take to be 99% confident that the sample mean will be within 20 hours of the population mean of all light-bulbs ( the margin of error). Determine the minimum sample size needed if we assume the population standard deviation is 250. QUESTION 4 A manufacturing company wants to use a sample to estimate the proportion of defective tennis balls that are produced by a machine and the company wants the margin of error to be within 3% (.03 ) of the population proportion of all defective tennis balls and to be 99% confident. There is no previous survey on this. What is the least number of tennis balls needed for the sample?

Q: A random group of 36 people in Toms River were asked “how much money did you spend at the grocery…

A: Given: Sample mean x¯=149.50 Sample size (n) = 36 Sample standard deviation (s) = 22.50 Confidence…

Q: In a survey, 8 people were asked how much they spent on their child's last birthday gift. The…

A: The sample size, i.e., the sampled number of people, n=8.The sample mean, x¯=$37.The sample standard…

Q: For 30 randomly selected Rolling Stones concerts the mean gross earnings is 2.59 million dollars.…

A:

Q: Find the margin of error at a 90% confidence level.

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

A:

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

A:

Q: Researchers from a certain country were interested in how characteristics of Uie spI environment…

A:

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: In a survey, 24 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: In a survey, 28 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: If a school district takes a random sample of 75 Math SAT scores and finds that the average is 433,…

A:

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: 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: A researcher wishes to determine the average take-home pay of a part-time college student. He takes…

A: The objective of this question is to calculate the 95% and 99% confidence intervals for the average…

Q: In a poll, 1200 adults in a region were asked about their online vs. in-store clothes shopping. One…

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Q: According the the U.S. Census bureau, the percentage of Santa Ana residents living in poverty is…

A: Given: p^1=0.177p^2=0.133n1=1200n2=1500

Q: In a poll, 1200 adults in a region were asked about their online vs. in-store clothes shopping. One…

A: Answer:- Given, Sample size, n = 1200  Sample proportion, p = 38% = 0.38  Confidence level = 95% =…

Q: If a school district takes a random sample of 80 Math SAT scores and finds that the average is 497,…

A:

Q: A marketing survey involves product recognition in New York and California. Of 558 New Yorkers…

A: The sample proportions are computed as follows:

Q: If a school district takes a random sample of 89 Math SAT scores and finds that the average is 497,…

A: From the provided information, Sample size (n) = 89 Sample mean (x̅) = 497 Population standard…

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: igh school soccer league there are a total of 256 players.

A: Given that Total players in high school soccer league (n) = 256 players The mean height of the…

Q: A survey of first time homeowners found that the sample mean annual income was $53,000. Assume that…

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

A: Givensample size(n)=25Mean(x)=38standard deviation(s)=19confidence interval(c)=98%

Q: I asked 45 randomly selected WSU college students how many US states they had visited. The students'…

A: The sample size is 45, sample mean is 3.8 and standard deviation is 1.5. The degrees of freedom is,…

Q: A doctor would like to estimate the mean difference in the number of hours of sleep for seniors in…

A: Random sample of 50 high school seniors  Random sample of 50 seniors from the colleges in his county…

Q: If a school district takes a random sample of 89 Math SAT scores and finds that the average is 494,…

A:

Q: A sample of 1,000 employees in an international company was taken to estimate the average…

A: Confidence interval of mean is given by, (X bar + - Z* sigma/√n) Where X bar is mean and sigma is…

Q: 26 randomly selected students took the calculas final. If the sample mean was 90. Construct a 95$…

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Q: 73 full time workers were surveyed on how many hours they spend watching television during the week.…

A: Given information, n=73x¯=3.2s=0.2 Confidence level is 95% significance level is 5% df=n-1=73-1=72 t…

Q: A survey of credit card holders revealed that Americans carried an average credit card balance of…

A: M1=$3900M2=$3300n1=400n2=450σ1=$880σ2=$810

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 marketing survey involves product recognition in New York and California. Of 558 New Yorkers…

A:

Q: In a survey, 31 people were asked how much they spent on their child's last birthday gift. The…

A:

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…

Q: At the local fish hatchery we take a random sample of 21 trout from one of the holding pens.  The…

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Q: A small business wanted to estimate the mean income from their customers. They took a sample of 256…

A: We have to find  the 90% confidence interval estimate for the population mean income of their…

Q: Sales personnel for City Distributors Inc. submit weekly reports listing the customer complaints…

A: Given Information: Sample size (n) = 8 Sample mean (x¯) = 8 Sample standard deviation (s) = 3.464 To…

Q: In a survey, 29 people were asked how much they spent on their child's last birthday gift. The…

A: Answer  Given  Confidence interval CI = 80%   α=1-0.80    =0.2 sample size [n] =29  degree of…

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.