12 A study is conducted between high school students and college students to compare their proficiency atwriting computer programs for microcomputers. For this study, the researchers wish to compare themean time (in minutes) of high school students to write an error-free program (x) with the mean time(in minutes) of college students to write an error-free program (uy). Assuming normality of bothpopulations, compute a 90% confidence interval estimate for (μx - μy).Data taken from two independent samples were summarized as follows:StatisticsHigh SchoolCollege84Mean time70Standard deviation1012Sample size1010

Answered: Image /qna-images/answer/acd0206a-c74c-45a4-8b1e-6907e3fce9f0.jpg

Answered: (PX) time (in minutes) of college…

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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A 30 month study is conducted to determine the difference in rates of accidents per month between two departments in an assembly plant. If a sample of 12 from the first department averaged 12.3 accidents per month with a standard deviation of 3.5, and a sample of 9 from the second department averaged 7.6 accidents per month with standard deviation of 3.4, find: The t-value for a 95% confidence interval is:
An electrical engineer wishes to compare the mean lifetimes of two types of transistors in an application involving high-temperature performance. A sample of 60 transistors of type A were tested and were found to have a mean lifetime of 1827 hours and a standard deviation of 168 hours. A sample of 180 transistors of type B were tested and were found to have a mean lifetime of 1658 hours and a standard deviation of 225 hours. Find a 95% confidence interval for the difference between the mean lifetimes of the two types of transistors.
During a natural gas shortage, a gas company randomly sampled residential gas meters in order to monitor daily gas consumption. On a particular day, a sample of 100 meters showed a sample mean of 250 cubic feet and a sample standard deviation of 50 cubic feet. Provide a 90% confidence interval estimate of the mean gas consumption for the population.
A random sample of 4040 adults with no children under the age of 18 years results in a mean daily leisure time of 5.135.13 hours, with a standard deviation of 2.462.46 hours. A random sample of 4040 adults with children under the age of 18 results in a mean daily leisure time of 4.254.25 hours, with a standard deviation of 1.711.71 hours. Construct and interpret a 9090% confidence interval for the mean difference in leisure time between adults with no children and adults with children left parenthesis mu 1 minus mu 2 right parenthesisμ1−μ2. Let mu 1μ1 represent the mean leisure hours of adults with no children under the age of 18 and mu 2μ2 represent the mean leisure hours of adults with children under the age of 18. The 90%confidence interval for left parenthesis mu 1 minus mu 2 right parenthesisμ1−μ2 is the range from hours to hours. (Round to two decimal places as needed.)
A large multi-branch bank is affiliated with both the MasterCard and Visa credit cards. For a random sample of 100 MasterCard cardholders, it is observed that the average month-end account balance is R682.58 with a sample standard deviation of R307.05. For a random sample of 100 Visa cardholders, the average month-end account balance is R550.55 with a sample standard deviation of R265.29. The 90% confidence interval for the mean difference between the month-end account balances for the two types of credit card holders (rounded to three decimals) is:
A random sample of 4040 adults with no children under the age of 18 years results in a mean daily leisure time of 5.995.99 hours, with a standard deviation of 2.292.29 hours. A random sample of 4040 adults with children under the age of 18 results in a mean daily leisure time of 4.414.41 hours, with a standard deviation of 1.661.66 hours. Construct and interpret a 9090% confidence interval for the mean difference in leisure time between adults with no children and adults with children left parenthesis mu 1 minus mu 2 right parenthesisμ1−μ2. Let mu 1μ1 represent the mean leisure hours of adults with no children under the age of 18 and mu 2μ2 represent the mean leisure hours of adults with children under the age of 18. The 9090% confidence interval for left parenthesis mu 1 minus mu 2 right parenthesisμ1−μ2 is the range from nothing hours to nothing hours. (Round to two decimal places as needed.)
A study was conducted to estimate hospital costs for car accident victims who woreseat belts. Suppose that 46 randomly selected cases had an average of $15150and a standard deviation of $8500. If the average hospital costs for those that didnot wear seatbelts was $19000, construct a 90% confidence interval to see if theaverage costs are significantly different.
A 30 month study is conducted to determine the difference in rates of accidents per month between two departments in an assembly plant. If a sample of 12 from the first department averaged 12.3 accidents per month with a standard deviation of 3.5, and a sample of 9 from the second department averaged 7.6 accidents per month with standard deviation of 3.4. The 95% confidence interval between the difference in rates between the two departments (round off to three decimals) is:
A random sample of 4040 adults with no children under the age of 18 years results in a mean daily leisure time of 5.015.01 hours, with a standard deviation of 2.492.49 hours. A random sample of 4040 adults with children under the age of 18 results in a mean daily leisure time of 4.044.04 hours, with a standard deviation of 1.891.89 hours. Construct and interpret a 9595% confidence interval for the mean difference in leisure time between adults with no children and adults with children left parenthesis mu 1 minus mu 2 right parenthesisμ1−μ2. Let mu 1μ1 represent the mean leisure hours of adults with no children under the age of 18 and mu 2μ2 represent the mean leisure hours of adults with children under the age of 18. The 95% confidence interval for left parenthesis mu 1 minus mu 2 right parenthesisμ1−μ2 is the range from hours to hours.
From a research survey, it was found that a random sample of 18 elementary graduates of an exclusive private school has a mean NEAT score of 90 with a standard deviation of 16, while a random sample of 24 elementary graduates of a public school showed a mean NEAT score of 85 with a standard deviation of 12. Is there a significant difference in the NEAT performance of these two groups of elementary graduates? a. Use the traditional method, and b. The P-value method c. Construct a 95% confidence interval for the difference in mean.
A wildlife study is designed to find the mean weight of salmon caught by an Alaskan fishing company. find the minimum sample size needed in order to construct a 95% confidence interval for the mean weight of such Salmon, to within 0.20 pounds. Assume that standard deviation 2.10.
To study the effect of curing temperature on shear strength of a certain rubber compound, 80 specimens were cured at 150°C and 95 were cured at 130°C. The specimens cured at 150°C had an average shear strength of 620 psi, with a standard deviation of 20 psi. Those cured at 130°C had an average shear strength of 750 psi, with a standard deviation of 30 psi. Find a 95% confidence interval for the difference between the mean shear strengths of specimens cured at the two temperatures.

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