A random sample of size 81 whose variance is 26.25 and mean is 32. Construct a 98% confidence interval

Answer 1

To construct a 98% confidence interval for the population mean, we will use the formula:

CI = x̄ ± Z * (σ/√n)

Where:
CI = Confidence Interval
x̄ = Sample mean
Z = Z-score for the confidence level
σ = Population standard deviation
n = Sample size

Given:
Sample size (n) = 81
Sample variance = 26.25
Sample mean (x̄) = 32
Confidence level = 98%

First, we need to calculate the standard deviation (σ) from the variance:
σ = √26.25
σ = 5.124

Next, we need to find the Z-score for a 98% confidence level. The Z-score for a 98% confidence level is 2.33. (You can look this up in a Z-table or use a calculator)

Now, plug in the values into the formula:
CI = 32 ± 2.33 * (5.124/√81)
CI = 32 ± 2.33 * (5.124/9)
CI = 32 ± 2.33 * 0.569
CI = 32 ± 1.325

Therefore, the 98% confidence interval for the population mean is (30.675, 33.325).