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# Statistics, Confidence Interval Estimation, Mean and Standard Deviation

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Confidence Interval Estimation
1. Compute a 95% confidence interval for the population mean, based on the sample 1, 2, 3, 4, 5, 6, and 30. Change the number from 30 to 10 and recalculate the confidence interval. Using the results, describe the effect of an outlier or extreme value on the confidence interval.

7.285714 mean
10.16061 std.dev
The 95% confidence since interval for true mean is
7.3 ± 2.447*10.16/sqrt(7)
7.3 ± 9.4 = (-2.1; 16.7)

4.428571 mean
2.992053 std.dev
The 95% confidence since interval for true mean is
4.4 ± 2.447*2.99/sqrt(7)
4.4 ± 2.8 = (1.6; 7.2)

As we see from the results, the extreme value make the confidence interval wider since due to this outlier the variation among data becomes larger comparing with the variation without that value.

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I am totally lost in this Statistics class, and I just want to make sure that I did this correctly and that this answer is somewhat close to the correct answer.

Confidence Interval Estimation
1. Compute a 95% confidence interval for the population mean, based ...

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