Provide an appropriate response. A random sample of 40 students has a mean annual earnings of $3120 and a standard deviation of $577. Construct the confidence interval for the population mean, µ atthe 94% confidence level. O($2910, $3330) ($4812, $5342) ($2948, $3291) ($1210, $2110)

Provide an appropriate response. A random sample of 40 students has a mean annual earnings of $3120 and a standard deviation of $577. Construct the confidence interval for the population mean, µ atthe 94% confidence level. O($2910, $3330) ($4812, $5342) ($2948, $3291) ($1210, $2110)

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Q: Of 500 random people surveyed, 72 were smokers. Find the 95% confidence interval. SE=0.0157

Q: The results of a study of 268 students who paid a private tutor to help them improve their scores on…

Q: e: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the…

A: Sample size n =47 Sample mean=109 Population standard deviation =32

Q: A random sample of 54 four-year-olds attending day care centers provided a yearly tuition average of…

Q: Listed below are the blood pressure measurements taken from the right and left arm of 5 patients.…

A: From the provided information, The mean and standard deviation of the right arm data can be obtained…

Q: A survey of 81 individuals finds that the average amount they spend on entertainment each month is…

A: Given n=81, Xbar=145,s=54 We want to find 95% CI for population mean

Q: confidence interval

Q: 15. A random sample of 100 visitors to a popular theme park spent an average of $142 on the trip…

A: SOLUTION: Given average x=142 standard deviation σ=47.5 random sample n=100

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

A: we have given that sample size n=21 ,xbar=30.9 ,s=12.7 margin of error= E = tcritical . s/sqrt (n)

Q: The previous mean amount of time spent using the lab computer was 61 minutes. Find a 95% confidence…

A: confidence interval with outliers: sample size (n) = 12 sample mean (x̄) = Σx/n = 812/12 =…

Q: A statistics practitioner took a random sample of 56 observations from a population whose standard…

A: We have given that, Sample size (n) = 56, population standard deviation (σ) = 35 and sample mean…

Q: Among those taking the Graduate Record Exam, 18 people are randomly selected. This group had a mean…

A: The point estimate for the population mean for the GRE test is 550. The margin of error is, Thus,…

Q: A statistics practitioner took a random sample of 48 observations from a population whose standard…

A: sample size(n)=48 population standard deviation(σ)=25 Mean(x)=105

Q: Salaries of 11 college graduates who took a statistics course in college have a mean of $78,782 and…

A: Given that, n=11, x=$78782, σ=$2000. The 100(1-α)% confidence interval for the population mean is…

Q: 9. 9. In a survey, 25 people were asked how much they spent on their child's last birthday gift.…

Q: C. Find a 95% confidence interval for the average life time of all the television sets produced by a…

Q: The Wechsler Adult Intelligence Scale (WAIS) is an IQ test. The scores on the WAIS for the 20 to 34…

Q: estimate the mean age of all students currently enrolled. In a random sample of 20 students the mean…

A: Solution

Q: Salaries of 25 college graduates who took a statistics course in college have a mean of $87,169 and…

A: Introduction: The 100 (1 – α) % confidence interval for the population standard deviation, σ, is the…

Q: A statistics practitioner took a random sample of 43 Note: For each confidence interval, enter your…

A: n = 43 standard deviation = 27 mean = 104 95% confidence interval for mean is given by x±zα2*σn 104…

Q: A statistics practitioner took a random sample of 55 observations from a population whose standard…

A: It is given that, Sample mean = 104. Population standard deviation, sigma is 31. The sample size, n…

Q: atistics practitioner took a random sample of 49 observations from a population whose standard…

A: Sample size n =49 Sample mean=103 Population standard deviation =25

Q: Construct a confidence interval of the population proportion at the given level of confidence. X =…

Q: In a survey, 19 people were asked how much they spent on their child's last birthday gift. The…

A: From the provided information, Sample size (n) = 19 Sample mean (x̄) = $46 Sample standard deviation…

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: In a survey, 16 people were asked how much they spent on their child's last birthday gift. The…

A: We have given that, Sample mean (x̄) = 50.2, sample size (n) = 16 and standard deviation (s) = 11.2…

Q: An existing inventory for a test measuring self-esteem indicates that the scores have a standard…

A: Given Information Sample size(n) = 70 Sample mean = 60 Population standard deviation = 8…

Q: . Estimate the mean with an approximate 95% confidence interval. The magazine estimates the true…

A: Let X be the random variable such that salary of the magazine's subscriber x̅ : Sample mean of…

Q: Zoologists investigated the reproductive traits of spider mites with an infection. Male and female…

A: Hello! As you have posted more than 3 sub parts, we are answering the first 3 sub-parts. In case…

Q: You measure 37 textbooks' weights, and find they have a mean weight of 73 ounces. Assume the…

A: From the provided information, Sample size (n) = 37 Sample mean (x̅) = 73 Population standard…

Q: The average chef at a restaurant earns $17 an hour with a deviation of $1.75. What are the chances…

Q: A toy manufacturer wants to know how many new toys children buy each year. A sample of 242 children…

Q: Time (min.) 78 70 73 79 68 136 72 60 50 8 61 72

A: Given, 78 70 73 79 68 136 72 60 50 8 61 72

Q: Construct a 90% confidence interval for the mean when x̅=50 and there are 31 observations. Suppose…

Q: A statistics practitioner took a random sample of 47 observations from a population whose standard…

A: Given that Sample size n =47 Sample mean =97 Population standard deviation =32

Q: The housing market has recovered slowly from the economic crisis of 2008. Recently, in one large…

A: given data sample size (n) = 49sample mean ( x¯ ) = 8929sample standard deviatio (s) = 117199% ci…

Q: A statistics practitioner took a random sample of 59 observations from a population whose standard…

A: sample size(n)=59 Population standard deviation(σ)=32 Mean(x)=95

Q: The annual earnings of 13 randomly selected computer software engineers have a sample standard…

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

Q: A statistics practitioner took a random sample of 43 observations from a population whose standard…

A: We have given that, sample mean (x̄) = 109, standard deviation (σ) = 30 and sample size (n) = 43

Q: A sample of 30 days of sales were taken. The mean of those 30 days of sales were $10,000 with a…

A: Given values, sample size (n) = 30 Mean = 10,000 Standard deviation = 2000 z value for 95%…

Q: In a survey, 373 people were asked how much they spent on their child's last birthday gift. The…

A: Givensample size(n)=373Mean(x)=36.85standard deviation(σ)=19.05Confidence level = 80%

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

Provide an appropriate response.
A random sample of 40 students has a mean annual earnings of $3120 and a standard deviation of $577. Construct the confidence
interval for the population mean, µ atthe 94% confidence level.
O($2910, $3330)
($4812, $5342)
($2948, $3291)
($1210, $2110)

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

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 4 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, statistics and related others by exploring similar questions and additional content below.

Similar questions

in a test of a lifelong brand of tires , a survey revealed that the mean life was 35,000 mikes and the standard deviation was 7083 mikes. if the shape of the population distribution is known to be normal, approximately what percentage of the tires lasts less than 20834 miles ?
An existing inventory for a test measuring self-esteem indicates that the scores have a standard deviation of 8. A psychologist gave the self-esteem test to a random sample of 70 individuals, and their mean score was 69. Construct a 99% confidence interval for the true mean of all test scores. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. What is the lower limit of the 99% confidence interval? What is the upper limit of the 99% confidence interval?
A statistics practitioner took a random sample of 42 observations from a population whose standard deviation is 30 and computed the sample mean to be 102. Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits. A. Estimate the population mean with 95% confidence. Confidence Interval = B. Estimate the population mean with 90% confidence. Confidence Interval = C. Estimate the population mean with 99% confidence. Confidence Interval =
A statistics practitioner took a random sample of 50 observations from a population whose standard deviation is 30 and computed the sample mean to be 109. Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits. A. Estimate the population mean with 95% confidence. Confidence Interval = B. Estimate the population mean with 90% confidence. Confidence Interval = C. Estimate the population mean with 99% confidence. Confidence Interval =
A statistics practitioner took a random sample of 55 observations from a population whose standard deviation is 28 and computed the sample mean to be 110. Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits. A. Estimate the population mean with 95% confidence. Confidence Interval =| B. Estimate the population mean with 90% confidence. Confidence Interval = C. Estimate the population mean with 99% confidence. Confidence Interval =
In a survey, 28 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $45 and standard deviation of $14. Construct a confidence interval at a 99% confidence level. Give your answers to one decimal place. ________+________
In a survey, 21 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $30.9 and standard deviation of $12.7. Estimate how much a typical parent would spend on their child's birthday gift (use a 90% confidence level). Give your answers to 3 decimal places. Express your answer in the format of a + E. $ 30.9 V + $ 4.770
Salaries of 25 college graduates who took a statistics course in college have a mean of $79,350 and a standard deviation of $1,280. Construct a 98% confidence interval for estimating the population standard deviation. Enter the upper bound of the confidence interval. (Round your answer to nearest whole dollar.)
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.
need help!
A random sample of 64 shoppers showed that they spend an average of $27.31 per visit at the Sunday Mornings Bookstore. The standard deviation of the population is $2.14. Find the 90% confidence interval of the true mean. Please include all work showing how you found your answer.
In a survey, 19 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $44.7 and standard deviation of $16.8. Estimate how much a typical parent would spend on their child's birthday gift (use a 98% confidence level). Give your answers to 3 decimal places. Express your answer in the format of E. $ + $