For the following estimate confidence interval for population mean with confidence level 90%. Sample size n=9, sample mean=80, sample standard deviation s=25

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The sample size is n = 9, the sample mean is `barx = 80` and the sample standard deviation is s = 25.

The standard error of the mean is `s/sqrt n` = `25/sqrt 9`

The critical value as a t-score is determined in the following way:

First, we find `alpha` that is defined as 1 - (confidence level/100). Here, `alpha = 1 - 90/100 = 0.1`

The critical probability is `1 - alpha/2` = `1 - 0.1/2 = 0.95`

The degrees of freedom is n - 1 = 8

The critical value is the t-score with 8 degrees of freedom and cumulative probability equal to 0.95. Using a t-distribution calculator, the critical value for these parameters is 1.86

The margin of error is defined as critical value*standard error = `1.86*(25/sqrt 9) = 15.5`

The confidence interval for the population mean with confidence level 90% is `80+-15.5`

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