A sample of size =n92 is drawn from a normal population whose standard deviation is =σ5.1. The sample mean is =x35.33.Part: 0 / 20 of 2 Parts CompletePart 1 of 2(a) Construct a 99.9% confidence interval for μ. Round the answer to at least two decimal places.A 99.9% confidence interval for the mean is <<μ.
Question
A sample of size =n92 is drawn from a normal population whose standard deviation is =σ5.1. The sample mean is =x35.33.Part: 0 / 20 of 2 Parts CompletePart 1 of 2(a) Construct a 99.9% confidence interval for μ. Round the answer to at least two decimal places.A 99.9% confidence interval for the mean is <<μ.
Solution
To construct a 99.9% confidence interval for the population mean (μ), we will use the formula for the confidence interval:
CI = x̄ ± Z * (σ/√n)
where: x̄ is the sample mean, Z is the Z-score corresponding to the desired confidence level, σ is the population standard deviation, and n is the sample size.
Given: x̄ = 35.33, σ = 5.1, and n = 92.
The Z-score for a 99.9% confidence level is approximately 3.29 (you can find this value in a standard Z-table or using a calculator).
Substituting these values into the formula, we get:
CI = 35.33 ± 3.29 * (5.1/√92)
Calculating the value inside the parenthesis:
σ/√n = 5.1/√92 ≈ 0.531
Substituting this back into the formula:
CI = 35.33 ± 3.29 * 0.531
Calculating the multiplication:
3.29 * 0.531 ≈ 1.75
So, the confidence interval is:
CI = 35.33 ± 1.75
Therefore, the 99.9% confidence interval for μ is (33.58, 37.08).
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