A sample of 230 observations is selected from a normal population with a population standard deviation of 26. The sample mean is 18. (Use t Distribution Table & z Distribution Table.)Required:a. Determine the standard error of the mean. (Round your answer to 3 decimal places.)c. Determine the 99% confidence interval for the population mean. (Round your answers to 3 decimal places.)
Question
A sample of 230 observations is selected from a normal population with a population standard deviation of 26. The sample mean is 18. (Use t Distribution Table & z Distribution Table.)Required:a. Determine the standard error of the mean. (Round your answer to 3 decimal places.)c. Determine the 99% confidence interval for the population mean. (Round your answers to 3 decimal places.)
Solution
a. The standard error of the mean is calculated by dividing the population standard deviation by the square root of the sample size.
In this case, the population standard deviation (σ) is 26 and the sample size (n) is 230.
So, the standard error (SE) is calculated as follows:
SE = σ / √n SE = 26 / √230 SE = 26 / 15.16575 SE = 1.714 (rounded to three decimal places)
c. The 99% confidence interval for the population mean is calculated using the formula:
CI = x̄ ± (Z*SE)
Where: x̄ is the sample mean Z is the Z-score from the Z distribution table that corresponds to the desired confidence level SE is the standard error
For a 99% confidence level, the Z-score is approximately 2.576.
So, the confidence interval is calculated as follows:
CI = 18 ± (2.576*1.714) CI = 18 ± 4.415 CI = [13.585, 22.415] (rounded to three decimal places)
So, we can be 99% confident that the population mean is between 13.585 and 22.415.
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