A recent survey of 9 social networking sites has a mean of 13.01 millionvisitors for a specific month. The standard deviation was 4.3 million.Find the 80% confidence interval of the true mean number of visitors inthat month. Assume the variable is normally distributed.Margin of error:Interval Calculation13.01-E = 2.00212.0021 << 13.01 +2.002111.0079 << 15.012111.01 << 15.01Write the confidence statement in the box below:Accessibility: InvestigateQ SearchT Estimate of a MeanConfidence Level 0.8Mean 13.01s 4.3N 9ResultSMeanT Estimate of a MeanSSENdfSampleLower LimitUpper LimitInterval13.014.31.43339811.007915.012113.01 2.0021Focus BB

We have given thatsample size (n) = 9Sample mean (x̅) = 13.01 million Standard deviations (s) = 4.3…

Answered: A recent survey of 9 social networking…

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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