Question

In a random sample of 12 senior-level civil engineers, the mean annual earnings were $$133,326 and the standard deviation was 36$$,729. Assume the annual earnings are normally distributed and construct a 95% confidence interval for the population mean annual earnings for senior-level civil engineers. Interpret the results. (Adapted from Salary.com)

Accepted Answer

Step 1: Identify the given values. From the problem, the sample mean ($$ \bar{x} ) is$$ $$133,326, the sample standard deviation ( s$$ $$) is 36$$,729, the sample size ($$ n ) is 12$$, and the confidence level is 95%. Step 2: Determine the critical value ($$ t^* ) $$for a 95% confidence interval. Since the sample size is small (n \u003c 30) and the population standard deviation is unknown, use the t-distribution. The degrees of freedom (df) are $$ n - 1 = 12 - 1 = 11 . $$Look up the critical value $$ t^*  in a t-$$distribution table or use statistical software for a two-tailed test with 95% confidence and 11 degrees of freedom. Step 3: Calculate the standard error of the mean (SE). The formula for the standard error is $$ SE = \frac{s}{\sqrt{n}} $$, where $$ s  is $$the sample standard deviation and $$ n  is $$the sample size. Step 4: Compute the margin of error (ME). The formula for the margin of error is $$ ME = t^* \cdot SE $$, where $$ t^*  is $$the critical value from Step 2 and $$ SE  is $$the standard error from Step 3. Step 5: Construct the confidence interval. The formula for the confidence interval is $$ \bar{x} \pm ME $$, where $$ \bar{x}  is $$the sample mean and $$ ME  is $$the margin of error. Interpret the results by stating that you are 95% confident the true population mean annual earnings for senior-level civil engineers fall within this interval.

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Key Concepts

Confidence Interval

Normal Distribution

Standard Deviation