- 2068 Chaitra · 5 marks
Fiducial limits for mean, sample of 900
Short answer
Large sample, use z-interval with s: SE=0.087 cm. 95% limits: 3.4 ± 1.96*0.087 = (3.23, 3.57) cm. 98% limits: 3.4 ± 2.326*0.087 = (3.20, 3.60) cm.
The sample of 900 members has a mean of 3.4 cms and standard deviation 2.61 cms. If the population is normal and its mean is unknown, find 95% and 98% fiducial limits of true mean.
Fiducial (confidence) limits for a population mean are x̄ ± z·s/√n. The sample is large (n = 900), so the normal distribution is used and s replaces the unknown σ.
n = 900, x̄ = 3.4 cm, s = 2.61 cm
SE = s/√n = 2.61/√900 = 2.61/30 = 0.0870
95% limits
z = 1.960
margin = 1.960 × 0.0870 = 0.1705
limits = 3.4 ± 0.1705 = (3.229, 3.571)
98% limits
z = 2.326
margin = 2.326 × 0.0870 = 0.2024
limits = 3.4 ± 0.2024 = (3.198, 3.602)
Answer: 95% fiducial limits: 3.23 cm to 3.57 cm; 98% fiducial limits: 3.20 cm to 3.60 cm. The higher the confidence level, the wider the interval.
✓ Answer checked by Claude Sonnet 5.5
Related questions
- CI: sample of 16, mean 52, SD 4 · Probability and Statistics, chapter 6
- Standard error; population 12,19,13,16 · Probability and Statistics, chapter 4
- Standard normal; normal approx of Poisson · Probability and Statistics, chapter 3 · asked 3×
- Standard normal: definition, properties, importance · Probability and Statistics, chapter 3 · asked 2×
- Normal characteristics; relation to standard normal · Probability and Statistics, chapter 3
- Standard normal definition and importance · Probability and Statistics, chapter 3
- Gumbel 500-year flood, 95% limits, 30 years · Engineering Hydrology, chapter 6
- FM numerical: deviation, index, 98% bandwidth, power · Communication Systems, chapter 3 · asked 2×
- Standard deviation and mean of array · Computer Programming, chapter 7
- HR trickling filter, 2.5 MLD; change if standard · Sanitary Engineering, chapter 7