8.72 Given M = 88, M, = 83, t = 2.042, t = 2.040, and a two-tailed testwith a = .05, (a) decide whether the null hypothesis was rejected ornot, (b) tell whether the difference between sample means is astatistically significant one or not, and (c) make a statement aboutthe direction of the difference between the population means.

Given M=88, M2=83, tcv= tcritical =2.042 , t=2.04=test statistic Two tailed hypothesis Level of…

Answered: 8.72 Given M = 88, M2 = 83, ty = 2.042,…

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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Question 4 The average student loan debt is reported to be 525,235. A student belives that the student loan debt is higher in her area. She takes a random sample of 100 college students in her area and determines the mean to be $27,524 and the standard devition to be $6000. Is there sufficient evidence to support the student claim at a 5% significance level? a) Determine the null and alternative hypotheses. Ho: P H.: µSelect an answer (Put in the correct symbol and value) b) Determine the test statistic. Round to two decimals. t%3D c) Find the p-value, Round to 4 decimals. P-value = d) Make a decision. O Reject the null hypothesis O Fail to reject the null hypothesis e) Write the conclusion. O There is sufficient evidence to support the claim that student loan debt is higher than 525,235, O There is not sufficient evidence to support the claim that student loan debt is higher than S25,235. Submit Question
Before the semester began, Professor Keithley predicted that 15% of hisphilosophy students would receive an A, 30% a B, 40% a C, 10% a D, and 5%an F. At the end of the semester, 14 of Professor Keithley’s philosophy studentsearned an A, 19 a B, 12 a C, 5 a D, and 2 an F.Use the Chi-Square test and a 0.05 level of significance to determine if ProfessorKeithley’s predicted percentages were accurate. (State your null and alternativehypothesis, find the critical values and test statistics, make a decision, and writea conclusion based on your results).
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6.24 Sleep deprivation, CA vs. OR, Part II. Exercise 6.22 provides data on sleep deprivation rates of Californians and Oregonians. The proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. (a) Conduct a hypothesis test to determine if these data provide strong evidence the rate of sleep deprivation is different for the two states. (Reminder: Check conditions) (b) It is possible the conclusion of the test in part (a) is incorrect. If this is the case, what type of error was made?
Question 83 Parts A and B
Question 3 A fruit farmer has a large number of almond trees, all of the same variety and the same age. One year he wishes to estimate the mean yield of his trees. He collects all the almonds from 8 trees and records the following weights (kg); 36 53 78 67 92 77 59 66 a) Use these data to estimate the mean and standard deviation of the yields of all the farmer's trees. b) Construct a 95% confidence interval for the mean yield. c) What statistical assumption is required for your procedure to be valid? d) How might you select a sample of eight trees from those growing in a large field. Criteria [2.1, 2.2, 2.3]
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Wk Qu. 13-11 A simulation returns... A simulation returns the following statistics: Mean 100 Standard Deviation: 5 Mode: 96 Minimum:75 Maximum: 135 Standard Error: 8. What is the most likely value? Most likely value A 68% confidence interval for the true mean would start at a value of 92 and end at a value of 108. A 68% confidence interval for the true mean would start at a value of and end at a value 108
ch.8 proportions question4 Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 143 with 65% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.C.I. =

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