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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

Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015
Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015
1st Edition
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
Publisher:HOUGHTON MIFFLIN HARCOURT
Chapter11: Data Analysis And Displays
Section: Chapter Questions
Problem 5CR
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I do not know if I reject or accept null hypothesis of this guess is reasonable.. please help
Insert ainter Fy C = Design Draw H0 μ= 3 H1 : μ/3 Test Statistic: t=x-μo s √n = Layout Times New Roma 12 A A Aa A E E T V BIU ab x₂ x² A - A. Font Type here to search References 2.9-3 Quantitative Data Portion Saved V Mailings Confidence Interval 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. G 1.5/√40-42 Hypothesis Test t . Review View Critical Value: At the .05 level of significance, my critical values are + 1.96 P-Value*: P(Z - 42)+P(z>-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 ericut Paragraph He O
Transcribed Image Text:Insert ainter Fy C = Design Draw H0 μ= 3 H1 : μ/3 Test Statistic: t=x-μo s √n = Layout Times New Roma 12 A A Aa A E E T V BIU ab x₂ x² A - A. Font Type here to search References 2.9-3 Quantitative Data Portion Saved V Mailings Confidence Interval 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. G 1.5/√40-42 Hypothesis Test t . Review View Critical Value: At the .05 level of significance, my critical values are + 1.96 P-Value*: P(Z - 42)+P(z>-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 ericut Paragraph He O
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