A lawyer researched the average number of years served by 45 different justices on the Supreme Court. The average number of years served was 13.8 years with a standard deviation of 7.3 years. What is the 95% confidence interval for the average number of years served by all Supreme Court justices?
To calculate the 95% confidence interval for the average number of years served by all Supreme Court justices, we can use the formula:
Confidence Interval = X̄ ± Z * (σ/√n)
Where:
- X̄ is the sample mean (13.8 years)
- Z is the z-score corresponding to the desired confidence level (95% confidence level has a z-score of 1.96)
- σ is the population standard deviation (7.3 years)
- n is the sample size (45 justices)
Substituting the given values into the formula:
Confidence Interval = 13.8 ± 1.96 * (7.3/√45)
Calculating the standard error:
standard error = σ/√n = 7.3/√45 ≈ 1.088
Therefore, the 95% confidence interval is:
Confidence Interval ≈ 13.8 ± 1.96 * 1.088
Calculating the upper and lower limits:
Lower limit ≈ 13.8 - (1.96 * 1.088) ≈ 11.682
Upper limit ≈ 13.8 + (1.96 * 1.088) ≈ 15.918
Thus, the 95% confidence interval for the average number of years served by all Supreme Court justices is approximately 11.682 to 15.918 years.