A study is conducted in a company that employs 800 engineers. A random sample of 50 engineers reveals that the average sample age is 34.3 years. Historically, the population standard deviation of the age of the company’s engineers is approximately 8 years. Construct a 98% confidence interval to estimate the average age of all the engineers in this company.
"n=800 \\\\\n\n\\bar{x}=34.3 \\\\\n\n\\sigma= 8"
Two-sided confidence interval:
"CI = (\\bar{x} - \\frac{Z_c \\times \\sigma}{\\sqrt{n}}, \\bar{x} + \\frac{Z_c \\times \\sigma}{\\sqrt{n}})"
For 98 % confidence "\u03b1= 1 -0.98 = 0.02"
"\\frac{\u03b1}{2} = \\frac{0.02}{2} = 0.01 \\\\\n\nZ_c = Z_{0.01} = 2.33 \\\\\n\nCI = (34.3 - \\frac{2.33 \\times 8}{\\sqrt{800}}, 34.3 + \\frac{2.33 \\times 8}{\\sqrt{800}}) \\\\\n\n= (34.3 -2.64, 34.3 +2.64) \\\\\n\n= (31.66, 36.94)"
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