A random sample of 50 employees is chosen from a company. The average salary of the 50employees is $55,000 and the standard deviation is $5,000. Find a 99% confidence intervalfor the average salary of the entire company.O ($53,178.61, $56,821.38)O ($54,991.14, $55,008.86)($53,836.91, $56,163.09)O ($53,178.61, $56,821.39)

The confidence interval is the combined value of sample statistic and margin of error. It is…

Answered: A random sample of 50 employees is…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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In a random sample of 38 families, the average weekly food expense was $95.60 with a standard deviation of $22.50. the distribution of weekly food expenses is normally distributed. Find the 99% confidence interval for the average weekly food expense. Round to two decimal places.
Fill in the blank. A random sample of 25 roundtrip flights between Philadelphia and Dallas has an average airfare of $393.50 with a sample standard deviation of $50.30. The 95% confidence interval around this sample mean is Round to two decimal places as needed. ... A. ($372.74, $414.26) B. ($388.47, $398.53) OC. ($383.44, $403.56) O D. ($353.26, $433.74)
five samples of 15 minerals are taken from a population. The sample means for hardness of the minerals are 6.5, seven, seven, 6.5, and six. The standard deviation for the mineral is .745. Find the meaning of the sample means and the standard deviation of the sample means for the hardness
Construct the confidence interval for the population mean u. c=0.98, x=16.5, o=4.0, and n=60. A 98% confidence interval for u is (___,___). (Round to one decimal place as needed.
A random sample of 21 lunch orders at Noodles & Company showed a mean bill of $13.54 with a standard deviation of $6.45. Find the 98 percent confidence interval for the mean bill of all lunch orders. (Round your answers to 4 decimal places.) The 98% confidence interval is from _____to______
Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 399 drivers and find that 285 claim to always buckle up. Construct a 83% confidence interval for the population proportion that claim to always buckle up. Use interval notation, for example, [1,5]
Use the Student's t distribution to find tc for a 0.95 confidence level when the sample is 17. (Round your answer to three decimal places.)
Use the Student's t distribution to find tc for a 0.90 confidence level when the sample is 25. (Round your answer to three decimal places.)
A Hollywood film producer is working on a new action film. She wants to estimate the true mean length of action films that have appeared in theatres during the past decade. She randomly selects 30 films and records the run time of each. The average time was 119 minutes with a standard deviation of 21.5 minutes. Construct a 95% confidence interval that estimates the true mean length of action films over the past decade. 3. Do. Calculate the 95% confidence interval. (97.5, 140.5) (110.83, 127.16) (110.97, 127.03) (110.58, 127.43)
Use the Student's t distribution to find tc for a 0.95 confidence level when the sample is 3. (Round your answer to three decimal places.)
In a sample of 100 college students, 25 of them own a car. Calculate the 95% confidence interval for the true population of college students who own a car. Choose the closest one. A. (0.165132, 0.334868) B. (0.231725, 0.452302) C. (0.25, 0.75) D. (0.1, 0.9)
A random sample of 1000 NAU students in Flagstaff found 470 NAU students who drink coffee daily. Find a 95% confidence interval for the true percent of NAU students in Flagstaff who drink coffee daily. Express your results to the nearest hundredth of a percent. Answer ___to____%

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