A certain counselor wishes to compare the mean IQ scores of two different social groups. A random sample of 12group IQ scores1showed an average of85and astandard deviation offiftewhile a random sample chosen independently ofelegroup IQ scores2showed an average of 108and a standard deviation of13. Assumingthat the populations of IQ scores for each of the groups follow a normal distribution and that the variances of these populations are equal, construct aconfidence interval for the90%forμ₁ −μ₂, the difference between the meanµ₁group IQ scoresland the meanμgroup IQ scores2. Then find the lower limit andthe upper limit of the confidence interval of the90%.-Carry out intermediate calculations using at least three decimal places. Round answers to at least two decimal places. (If necessary, you can refer to a list offormulas.)Lower limit:Upper limit:XŚ

Given that for group 1: mean (M1) = 85, standard deviation (s1) = 15 and n1 = 12group 2: mean (M2) =…

Answered: A certain counselor wishes to compare…

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 certain counselor wants to compare mean IQ scores for two different social groups. A random sample of 11 IQ scores from group 1 showed a mean of 120 and a standard deviation of 16, while an independently chosen random sample of 10 IQ scores from group 2 showed a mean of 113 and a standard deviation of 14. Assuming that the populations of IQ scores are normally distributed for each of the groups and that the variances of these populations are equal, construct a 90% confidence interval for the difference u,-, between the mean u, of IQ scores of group 1 and the mean u, of IQ scores of group 2. Then find the lower limit and upper limit of the 90% confidence interval. Carry your intermediate computations to at least three decimal places. Round your responses to at least two decimal places. (If necessary, consult a list of formulas.) Lower limit: Upper limit: Explanation Check 02021 McGraw H Education. All Rights Reserved Terms of UseI PivcyI Accessibty MacBook Air ese 888 %23 2$ & 3 8 Q E…
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The newest model of a car battery from a popular brand is supposed to have a lifetime of 60 months, but the lifetime varies slightly from battery to battery. It is known that the population of all lifetimes (in months) of this model of car battery is approximately normally distributed. You are a product reviewer who wants to estimate the standard deviation for this population with a random sample of 22 car batteries. Follow the steps below to construct a 99% confidence interval for the population standard deviation of all lifetimes of this model of car battery. (If necessary, consult a list of formulas.) (a)Click on "Take Sample" to see the results from the random sample. Take Sample Number of carbatteries Sample mean Sample standard deviation Sample variance To find the confidence interval for the population standard deviation, first find the confidence interval for the population variance. Enter the values…
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The breaking strengths of cables produced by a certain manufacturer have historically had a mean of 1825 pounds and a standard deviation of 70 pounds. The company believes that, due to an improvement in the manufacturing process, the mean breaking strength, μ, of the cables is now greater than 1825 pounds. To see if this is the case, 17 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1844 pounds. Assume that the population is normally distributed. Can we support, at the 0.05 level of significance, the claim that the population mean breaking strength of the newly-manufactured cables is greater than 1825 pounds? Assume that the population standard deviation has not changed. Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places, and round your responses as specified below. (If necessary, consult a list of formulas.) (a) State the null hypothesis H0…

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