The true mean number of siblings my Facebook friends have is X=2.9 My estimated sample mean was X=3 The margin of error: T=(s/√n)=1.96(1.5/√40) = 5 Siblings Confidence Interval: √ 2.9+5= (2.4,3.4) Conclusion I am 95% confident that 40 of my Facebook friends have between 2.4 and 3.4 siblings. HO:μ= 3 H1 : μ/3 Test Statistic: 1-x-μo s √n = Hypothesis Test 2.9-3 1.5/√40 = -42 Critical Value: At the .05 level of significance, my critical values are + 1.96 P-Value*: P(z-- 42)+P(-42)=3372+.3372-6744 Decision Because my test statistics is -42 is less than 1.96 and because my P-Value of 6744 is not significant at p= .05

The true mean number of siblings my Facebook friends have is X=2.9 My estimated sample mean was X=3 The margin of error: T=(s/√n)=1.96(1.5/√40) = 5 Siblings Confidence Interval: √ 2.9+5= (2.4,3.4) Conclusion I am 95% confident that 40 of my Facebook friends have between 2.4 and 3.4 siblings. HO:μ= 3 H1 : μ/3 Test Statistic: 1-x-μo s √n = Hypothesis Test 2.9-3 1.5/√40 = -42 Critical Value: At the .05 level of significance, my critical values are + 1.96 P-Value*: P(z-- 42)+P(-42)=3372+.3372-6744 Decision Because my test statistics is -42 is less than 1.96 and because my P-Value of 6744 is not significant at p= .05

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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