A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean, x

Question

​, is found to be 107 and the sample standard​ deviation, s, is found to be 8
.
​(a) Construct a 99​% confidence interval about mu if the sample​ size, n, is 23.
Lower​ bound=....
Upper​ bound=....
.

Answer 1

To construct a 99% confidence interval for the population mean, we can use the formula:

Lower bound = x - (Z * (s / sqrt(n)))
Upper bound = x + (Z * (s / sqrt(n)))

Where:
- x is the sample mean (107)
- Z is the Z-score corresponding to a 99% confidence level (found in the Z-table or using a calculator, it is approximately 2.576)
- s is the sample standard deviation (8)
- n is the sample size (23)

Substituting the given values, we get:

Lower bound = 107 - (2.576 * (8 / sqrt(23)))
Upper bound = 107 + (2.576 * (8 / sqrt(23)))

Calculating the values:

Lower bound ≈ 107 - (2.576 * (8 / 4.7958)) ≈ 101.897
Upper bound ≈ 107 + (2.576 * (8 / 4.7958)) ≈ 112.103

Therefore, the 99% confidence interval for the population mean is approximately 101.897 to 112.103.