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A recent survey of 9 social networking sites has a mean of 13.01 million visitors for a specific month. The standard deviation was 4.3 million. Find the 80% confidence interval of the true mean number of visitors in that month. Assume the variable is normally distributed. Margin of error: Interval Calculation E = 2.0021 13.01 - 2.0021 < p < 13.01 +2.0021 11.0079 << 15.0121 11.01 << 15.01 Write the confidence statement in the box below:

A recent survey of 9 social networking sites has a mean of 13.01 million visitors for a specific month. The standard deviation was 4.3 million. Find the 80% confidence interval of the true mean number of visitors in that month. Assume the variable is normally distributed. Margin of error: Interval Calculation E = 2.0021 13.01 - 2.0021 < p < 13.01 +2.0021 11.0079 << 15.0121 11.01 << 15.01 Write the confidence statement in the box below:

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.3: Measures Of Spread
Problem 1GP
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A recent survey of 9 social networking sites has a mean of 13.01 million visitors for a specific month. The standard deviation was 4.3 million. Find the 80% confidence interval of the true mean number of visitors in that month. Assume the variable is normally distributed. Margin of error: Interval Calculation 13.01 - E = 2.0021 2.0021 << 13.01 +2.0021 11.0079 << 15.0121 11.01 << 15.01 Write the confidence statement in the box below: Accessibility: Investigate Q Search T Estimate of a Mean Confidence Level 0.8 Mean 13.01 s 4.3 N 9 Result S Mean T Estimate of a Mean S SE N df Sample Lower Limit Upper Limit Interval 13.01 4.3 1.4333 9 8 11.0079 15.0121 13.01 2.0021 Focus BB
Transcribed Image Text:A recent survey of 9 social networking sites has a mean of 13.01 million visitors for a specific month. The standard deviation was 4.3 million. Find the 80% confidence interval of the true mean number of visitors in that month. Assume the variable is normally distributed. Margin of error: Interval Calculation 13.01 - E = 2.0021 2.0021 << 13.01 +2.0021 11.0079 << 15.0121 11.01 << 15.01 Write the confidence statement in the box below: Accessibility: Investigate Q Search T Estimate of a Mean Confidence Level 0.8 Mean 13.01 s 4.3 N 9 Result S Mean T Estimate of a Mean S SE N df Sample Lower Limit Upper Limit Interval 13.01 4.3 1.4333 9 8 11.0079 15.0121 13.01 2.0021 Focus BB
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