A random sample of size 81 whose variant is 26.25 and mean is 32 construct a 98% confidence interval
To construct a 98% confidence interval, we can use the formula:
Confidence Interval = mean ± (Z * (standard deviation / √n))
Where:
- Z is the z-score corresponding to the confidence level (98% confidence level corresponds to a z-score of 2.33)
- standard deviation is the square root of the variance (sqrt(26.25) = 5.124)
- n is the sample size (81)
- mean is 32
Plugging in the values:
Confidence Interval = 32 ± (2.33 * (5.124 / √81))
Confidence Interval = 32 ± (2.33 * 0.569)
Confidence Interval = 32 ± 1.327
Confidence Interval = (30.673, 33.327)
Therefore, the 98% confidence interval for the mean of the population is (30.673, 33.327).