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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Steve believes that his wife's cell phone battery does not last as long as his cell phone battery. On eight different occasions, he measured the length of time his cell phone battery lasted and calculated that the mean was 19.7 hours with a standard deviation of 9.4 hours. He measured the length of time his wife's cell phone battery lasted on twelve different occasions and calculated a mean of 18.9 hours with a standard deviation of 5.3 hours. Assume that the population variances are the same. Let Population 1 be the battery life of Steve's cell phone and Population 2 be the battery life of his wife's cell phone. Step 1 of 2: Construct a 90 % confidence interval for the true difference in mean battery life between Steve's cell phone and his wife's. Round the endpoints of the interval to one decimal place, if necessary.

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