Estimation part I
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This document discusses different types of estimation and confidence intervals. It provides examples of calculating confidence intervals for a population mean (μ) in different situations: when the population standard deviation (σ) is known or unknown, and when the sample size is small (n < 30) or large (n ≥ 30). Confidence intervals are constructed using the point estimate, standard deviation or error, sample size, and critical value from the t-distribution or z-distribution based on the confidence level. The document also gives an example of determining the necessary sample size needed to estimate a population mean within a given level of precision.
Estimation part I
- 2. Estimation:- The entire process of using an estimator to produce an estimate of the parameter is known as Estimation. Or estimation is a process to estimate unknown value of population parameter with the help of sample. Estimator:- Any statistic used to estimate a parameter is an estimator. Estimate:- Any specific value of a statistic is an estimate. An estimate of a population parameter may be expressed in two ways:
- 3. Point Estimation:- A single number used to estimate a population parameter with the help of a sample is called a point estimate. The process of estimating a single number is known as point estimation. Interval Estimation:- A range of values used to estimate a population parameter with the help of a sample is called an interval estimate, and the process of estimating with a range of values is known as interval estimation.
- 4. Confidence Interval:- The interval estimates based on specific confidence levels are known as confidence intervals, and the upper and lower limits of the intervals are known as confidence limits. The Level of Confidence:- The degree of probability associated with an interval estimate is known as the confidence level or confidence coefficient.
- 5. CONFIDENCE INTERVAL FOR : A (1-α)100% confidence interval for population mean when σ is known and size of sample either small (n < 30) or large (n ≥ 30) given as 𝑿 ± 𝒁∝ 𝟐 𝝈 𝒏
- 6. Example-1: We wish to estimate the mean serum indirect bilirubin level of 4-day-old infants. The mean for a sample of 16 infants was found to be 5.98 mg/dl. Assuming bilirubin levels in 4-day-old infants are approximately normally distributed with a standard deviation of 3.5 mg/dl find the 95% confidence interval for mean.
- 7. Solution: n = 16 𝑥 = 5.98 𝜎 = 3.5 = 1 – 0.95 = 0.05 𝑍∝ 2 = 𝑍0.05 2 = 𝑍0.025 = 1.96 (From z- table) 𝑥 ± 𝑍∝ 2 𝜎 𝑛 5.98 ±1.96 3.5 16 5.98 ± 1.715 4.265 7.695
- 8. CONFIDENCE INTERVAL FOR : A (1-α)100% confidence interval for population mean when σ is unknown and size of sample large (n ≥ 30) given as 𝑿 ± 𝒁∝ 𝟐 𝒔 𝒏
- 9. Example-2: In an automotive safety test conducted by the North Carolina Highway Safety Research Center, the average tire pressure in a sample of 62 tires was found to be 24 pounds per square inch, and the standard deviation was 2.1 pounds per square inch. Construct a 95 percent confidence interval for the population mean. Also interpret the result.
- 10. Solution: n = 62 𝑥 = 24 S = 2.1 = 1 – 0.95 = 0.05 𝑍∝ 2 = 𝑍0.05 2 = 𝑍0.025 = 1.96(From z-table) 𝑥 ± 𝑍∝ 2 𝑆 𝑛 24 ±1.96 2.1 62 24 ± 0.523 23.48 24.52 per square inch The population tires will have pressure more than 23.48 but less than 24.52 per square inch.
- 11. CONFIDENCE INTERVAL FOR : A (1-α)100% confidence interval for population mean when σ is unknown and size of sample small (n < 30) given as 𝑿 ± 𝒕∝ 𝟐,(𝒏−𝟏) 𝒔 𝒏 Where 𝒔 = 𝟏 𝒏−𝟏 𝒙 𝟐 − 𝒙 𝟐 𝒏 or 𝒔 = 𝒏 𝒙 𝟐− 𝒙 𝟐 𝒏(𝒏−𝟏)
- 12. Example-3: Seven homemaker were randomly sampled, and it was determined that the distance they walked in their house work had an average of 39.2 miles per week and a sample standard deviation of 3.2 miles per week. Construct a 95% confidence interval for the population mean.
- 13. Solution: n = 7 𝑥 = 39.2 S = 3.2 = 1 – 0.95 = 0.05 𝑡∝ 2,(𝑛−1) = 𝑡0.05 2 (7−1) = 𝑡0.025,6 = 2.447 (From t- table) 𝑥 ± 𝑡∝ 2,(𝑛−1) 𝑆 𝑛 39.2±2.447 3.2 7 39.2 ± 2.9596 36.240 42.160miles
- 14. Example-4: The contents of 7 similar containers of sulfuric acid are 9.8, 10.2, 10.4, 9.8, 10.0, 10.2 and 9.6 liters. Find a 95% confidence interval for the mean content of all such containers, assuming an approximate normal distribution for container contents. Solution:
- 15. X X2 9.8 96.04 10.2 104.04 10.4 108.16 9.8 96.04 10.0 100 10.2 104.04 9.6 92.16 x =70 x2 = 700.48 Where n = 7 x =70 x2 = 700.48 𝑥= Σ𝑥 𝑛 = 70 7 = 10 𝑆 = 𝑛Σ𝑥2− Σ𝑥 2 𝑛 𝑛−1 S = 7 700.48 − 70 2 7 7−1 S = 0.283 = 1 – 0.95 = 0.05
- 16. 𝑡∝ 2,(𝑛−1) = 𝑡 0.05 2 (7−1) = 𝑡0.025,6 = 2.447 (From t- table) 𝑥 ± 𝑡∝ 2 𝑛 − 1 𝑆 𝑛 10 ± 2.447 0.283 7 10 ± 0.2617 9.74 10.26
- 17. Example-5: The mean and S.D for the quality grade – point averages are 2.6 and 0.3. How large a sample is required if we want to be 95% confidence that our estimate of is within 0.05?
- 18. Solution: = 1 – 0.95 = 0.05 𝑍∝ 2 = 𝑍0.05 2 = 𝑍0.025 = 1.96 = 0.3 and e = 0.05 𝑛 = 𝑍∝ 2 𝛿 𝑒 2 𝑛 = 1.96 0.3 0.05 2 n = 138.3 n 139