The average income for a county in Texas is reported in a study. The margin of error is 767.42. Use a confidence interval of 95%. What is the approximate standard deviation of the sample assuming n = 45? O 4,598.04 O 8,315.63 O 2,574.36 O 2,626.54

The average income for a county in Texas is reported in a study. The margin of error is 767.42. Use a confidence interval of 95%. What is the approximate standard deviation of the sample assuming n = 45? O 4,598.04 O 8,315.63 O 2,574.36 O 2,626.54

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

See similar textbooks

Related questions

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 random sample of twelve people, the mean driving distance to work was 21.7 miles and the…

Q: 2. Construct a 95% confidence interval for the population standard deviation of a random sample of…

A: given data sample size(n) = 15 sample standard deviation(s) = 12.7construct 95 % ci for σ.

Q: Is there a difference in the average age that people get married in the US and its close neighbor,…

A: Obtain the 95% confidence interval for the mean rating. The 95% confidence interval for the mean…

Q: Among various ethnic groups, the standard deviation of heights is known to be approximately three…

Q: Construct a confidence interval for the mean of the paired differences for the two populations with…

A: Solution: From the given information, n=32, d-bar=1.358, α=0.05 and sd=0.635.

Q: Among various ethnic groups, the standard deviation of heights is known to be approximately three…

A: Using Excel formula “=NORM.S.INV(1-(0.04/2))”, the critical value for 96% confidence level is…

Q: In a random sample of ten people, the mean driving distance to work was 19.2 miles and the…

Q: In a random sample of five people, the mean driving distance to work was 19.9 miles and the standard…

Q: In a random sample of twelve people, the mean driving distance to work was 21.5 miles and the…

A: Solution

Q: The average salary of 25 random employees in a certain financial sector was 3400 USD. The sample…

Q: In a random sample of five people, the mean driving distance to work was 19.9 miles and the…

A: Given Sample size, n = 5 Degree of freedom, n-1 = 4 Sample mean, x = 19.9 miles Sample standard…

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: In a random sample of seven people, the mean driving distance to work was 19.7 miles and the st and…

A: From provided information, random sample of 7 people is considered. The mean distance to work was…

Q: The monthly earnings of a group of business students are normally distributed with a standard…

A: Given that Margin of error =E =132 Population standard deviation =532 93% confidence level.

Q: In a random sample of six mobile devices, the mean repair cost was $75.00 and the standard…

A: Obtain the critical value: The level of confidence is 1- α = 0.99. The sample size is n = 6. The…

Q: stributed and use the t-distribution to find the margin of error and construct a 95% confidence…

A: df=n-1=10−1=9 The critical t-value for α=0.05 and df=9 degrees of freedom is tc=2.262. (from t…

Q: The average salary of 25 random employees in a certain financial sector was 3400 USD. The sample…

A: The Margin of error for 90% confidence interval for the population mean when the population standard…

Q: The average salary of 64 employee was 5400 and the sample standard deviation was 270. Find the 90%…

Q: Construct the requested confidence interval from the supplied information. Among a sample of 130…

A: Here we want to find the 95 percent confidence interval.

Q: given a standard deviation of + or - 0.023m at the 90% confidence level, what is the 68% error level…

A: The confidence level of the data is 90%. The standard deviation of the data with error percentage is…

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

Q: The mean time taken to design a house plan by 40 architects was found to be 19 hours with a standard…

A: Step 1:Given data :Sample size n=40degrees of freedom df=40−1=39Sample mean xˉ=19Sample standard…

Q: 7. Construct a 95% confidence interval for the population mean, μ. Assume the population has a…

A: CL=0.95 Sample size, n=20 Sample mean, x¯=3120 Sample sd, s=677

Q: In a random sample of eight cell phones, the mean full retail price was $468.50 and the standard…

A: The information provided in the question are as follows :- Sample size (n) = 8 Sample mean (x̄) =…

Q: In a random sample of five people, the mean driving distance to work was 22.5 miles and the standard…

Q: Kennedy High School reported that the mean score on an end of semester physics exam wass 508 with a…

A: Given, Mean score μ=508Standard deviation σ=121Sample size n=200

Q: Starting salaries of 55 college graduates who have taken a statistics course have a mean of $43,701…

A: Given information- Sample size, n = 55 Significance level, α = 0.01 Sample mean, x-bar = $43,701…

Q: In a random sample of five people, the mean driving distance to work was 23.2 miles and the standard…

A: Given,sample size(n)=5sample mean(x¯)=23.2standard deviation(s)=7.7degrees of…

Q: In a random sample of six people, the mean driving distance to work was 24.3 miles and the…

Q: Suppose that you wanted to estimate the population mean amount spent eating out per week by…

A: The question is about sample size estimation Given: Sample std. deviation = 200 dollars Margin of…

Q: In a random sample of eight people, the mean driving distance to work was 24.2 miles and the…

Q: For the first six months of the year, an analyst records the monthly closing stock price (in $) for…

A: 1) Mean is given as, Mean=655+58+...+62=56.83 Std deviation is given as, s=n−1∑(Xi−Xˉ)2=4.07 2)…

Q: for a sample of 57 that has a mean of 6200 and a standard deviation of 254, calculate the 95%…

Q: or a sample of 121 that has a mean of 6540 and a standard deviation of 780, calculate the 90%…

Q: The speed of greyhounds is normally distributed with a standard deviation of 0.15 mph. How many dogs…

A: Population standard deviation()=0.15Margin of error(E)=0.05

Q: The mean number of travel days per year for salespeople employed by three hardware distributors…

Q: In a random sample of ten people, the mean driving distance to work was 23.2 miles and the standard…

A: Given,sample size(n)=10sample mean(x¯)=23.2standard deviation(S)=7.4degrees of…

Q: In a random sample of 60 refrigerators, the mean repair cost was $157.08. Assume the population…

A: Given that,The sample size is refrigerators.The sample mean is The population standard deviation is…

Q: 83 students are asked in a survey how many hours they work during a week. Their responses have a…

Q: The annual earnings of 14 randomly selected computer software engineers have a sample standard…

Concept explainers

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.

Question

The average income for a county in Texas is reported in a study. The margin of error is 767.42. Use a confidence
interval of 95%. What is the approximate standard deviation of the sample assuming n = 45?
O 4,598.04
O 8,315.63
O 2,574.36
O 2,626.54

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step 3

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 3 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Point Estimation, Limit Theorems, Approximations, and Bounds$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, algebra and related others by exploring similar questions and additional content below.

Similar questions

A sample of 22 public high school students spent an average of 7 hours studying per week. Assuming the distribution of hours spent studying is normally distributed. Find a 95% confidence interval for the mean number of hours spent studying by public high school students.? Assume the standard deviation is 1.3 hours.
In a random sample of twelve people, the mean driving distance to work was 18.8 miles and the standard deviation was 4.9 miles. Assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 90% confidence interval for the population mean μ. Interpret the results.
Salaries of 46 college graduates who took a statistics course in college have a mean, x, of $61,400. Assuming a standard deviation, σ, of $19,045 , construct a 90% confidence interval for estimating the population mean.
The average salary of 25 random employees in a certain financial sector was 3400 USD. The sample standard deviation was 540 USD. We are interested in the 90% confidence interval for the population mean. What is the value of the margin of error? Select one: O a. 153.99 O b. 92.39 с. 184.79 O d. 123.19
A group of 49 randomly selected students has a mean age of 22.4 years. Assume the population standard deviation is 3.8. Construct a 90% confidence interval for the population mean. Include a statement interpreting your answer.
In a recent sample of 89 used cars sales costs, the sample mean was $6,225 with a standard deviation of $3,151. Assume the underlying distribution is approximately normal. Which distribution should you use for this problem? Construct a 95% confidence interval for the population mean cost of a used car.
In a random sample of 5 mobile devices, the mean repair cost was $60.00 and the standard deviation was $12.00. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a 95% confidence interval for the population mean. Interpret the results.
In a random sample of eleven people, the mean driving distance to work was 18.1 miles and the standard deviation was 7.6 miles. Assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 99% confidence interval for the population mean μ. Interpret the results.
A study of a local city is being conducted to find out the average number of people that use public transportation every day. The sample mean should be within 11 people of the mean people using public transportation. The standard deviation is 100. What is the sample size necessary for a 90% confidence level? O 15 O 223 O 224 O 14.95
In a sample of 100 individuals, the average height is 170 cm and the standard deviation is 15 cm. Using a 95% confidence level, what is the margin of error for the mean height of the population?
In a random sample of twelve people, the mean driving distance to work was 21.3 miles and the standard deviation was 7.3 miles. Assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 99% confidence interval for the population mean μ. Interpret the results. Identify the margin of error. (Rol square miles miles per hour miles s needed.)
A survey of 73 randomly selected homeowners finds that they spend a mean of $74 per month on home maintenance. Construct a 95% confidence interval for the mean amount of money spent per month on home maintenance by all homeowners. Assume that the population standard deviation is $16 per month. Round to the nearest cent.