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