Probability and Statistics Lecture 9: 1 and -Sample Estimation to accompany Probability and Statistics for Engineers and Scientists Fatih Cavdur
Introduction A statistic θ is said to be an unbiased estimator of the parameter θ if E θ = θ We can show that E X = μ and E S = σ
Introduction If we consider all possible unbiased estimators of θ, the one with the smallest variance is said to be the most efficient estimator of θ.
Introduction
1-Sample: Estimating the Mean If x is the mean of a random sample of size n from a population with known variance σ, a 100 1 α % confidence interval on μ: x z α/ σn < μ < x + z α/ σ n
1-Sample: Estimating the Mean
1-Sample: Estimating the Mean
1-Sample: Estimating the Mean
1-Sample: Estimating the Mean Example 9.: The average zinc concentration recovered from a sample of measurements taken in 36 different locations in a river is found to be.6 grams per milliliter. Find the 95% and 99% confidence intervals for the mean zinc concentration in the river. Assume that the population standard deviation is 0.3 grams per milliliter.
1-Sample: Estimating the Mean The 95% confidence interval is x z α/ σn < μ < x + z σ α/ n 0.3.6 z 0.05 36 < μ <.6 + z 0.3 0.05 36.6 1.96 0.3 0.3 < μ <.6 + 1.96 36 36.50 < μ <.70
1-Sample: Estimating the Mean The 99% confidence interval is x z α/ σn < μ < x + z σ α/ n 0.3.6 z 0.005 36 < μ <.6 + z 0.3 0.005 36.6.575 0.3 0.3 < μ <.6 +.575 36 36.47 < μ <.73
1-Sample: Estimating the Mean If x is the mean of a random sample of size n from a population with known variance σ, a 1-sided 100 1 α % confidence interval on μ: μ x + z α μ x z α σ n σ n
1-Sample: Estimating the Mean Example 9.4: In an experiment, 5 subjects are selected randomly and their reaction time to a particular stimulus is measured. Past experience suggests that the standard deviation of these types of stimuli is seconds and that the distribution of reaction times is approximately normal. The average time for the subjects is 6. seconds. Give an upper 95% bound for the mean reaction time.
1-Sample: Estimating the Mean μ x + z α σn = 6. + z 0.05 5 = 6. + 1.645 5 6.858
1-Sample: Estimating the Mean If x and s are the mean and standard deviation of a random sample of size n from a population with unknown variance σ, a 100 1 α % confidence interval on μ: x t α/;n 1 sn < μ < x + t α/;n 1 s n
1-Sample: Estimating the Mean
1-Sample: Estimating the Mean Example 9.: The contents of seven similar containers of sulfuric acid are 9.8, 10., 10.4, 9.8, 10.0, 10. and 9.6 liters. Find a 95% confidence interval for the mean contents of all such containers, assuming an approximately normal distribution.
1-Sample: Estimating the Mean We have x = 10.0, s = 0.83 and t 0.05;6 =.447. Hence, the 95% confidence interval is x t α/;n 1 sn < μ < x + t s α/;n 1 n 0.83 10.0 t 0.05;6 7 < μ < 10.0 + t 0.83 0.05;6 7 10.0.447 0.83 0.83 < μ < 10.0 +.447 7 7 9.74 < μ < 10.6
-Samples: Estimating the Difference of Means If x 1 and x are the means of independent random sample of size n 1 and n from a population with known variances σ 1 and σ, a 100 1 α % confidence interval for μ 1 μ : x 1 x z α/ σ 1 + σ < μ n 1 n 1 μ < x 1 x + z α/ σ 1 n 1 + σ n
-Samples: Estimating the Difference of Means Example 9.10: A study was conducted to compare two types of engines, A and B. Gas mileages, in miles per gallon, was measured. 50 experiments were conducted for engine type A and 75 experiments were conducted for engine type B. All conditions were held constant. The average gas mileages were 36 miles per gallon for type A and 4 miles per gallon for type B. Find a 96% confidence interval on μ B μ A. Assume that the population standard deviations are 6 and 8 for type A and B, respectively.
-Samples: Estimating the Difference of Means We have, x 1 x z α/ 4 36 z 0.0 σ 1 + σ < μ n 1 n 1 μ < x 1 x + z α/ 64 75 + 36 50 < μ B μ A < 4 36 + z 0.0 3.43 < μ B μ A < 8.57 σ 1 n 1 + σ n 64 75 + 36 50
-Samples: Estimating the Difference of Means If x 1 and x are the means of independent random sample of size n 1 and n from a population with unknown variances σ 1 = σ, a 100 1 α % confidence interval for μ 1 μ : x 1 x ± t α/;n1 +n s p 1 n 1 + 1 n where s p is the pooled estimate of the population standard deviation s p = n 1 1 s 1 + n 1 s n 1 + n
-Samples: Estimating the Difference of Means Example 9.11: Assume that we have two samples with n 1 = 1, x 1 = 3.11, s 1 = 0.771 and n = 10, x =.04, s = 0.448. Find a 99% confidence interval for the difference between the population means assuming that the populations are approximately normally distributed with equal variances.
-Samples: Estimating the Difference of Means We have s p = = n 1 1 s 1 + n 1 s n 1 + n 11 0.771 + 9 0.448 1 + 10 = 0.646
-Samples: Estimating the Difference of Means and x 1 x ± t α/;n1 +n s p 3.11.04 ± t 0.05;0 0.646 1 n 1 + 1 n 1 1 + 1 10 0.59 < μ 1 μ < 1.55
-Samples: Estimating the Difference of Means If x 1 and x are the means of independent random sample of size n 1 and n from an approximately population with unknown variances σ 1 and σ, a 100 1 α % confidence interval for μ 1 μ : x 1 x ± t s 1 α/;v + s n 1 n where v is the degrees of freedom of the t distribution and defined as v = s 1 n + s 1 n s 1 n 1 n 1 1 + s n 1 n
-Samples: Estimating the Difference of Means We have, v = s 1 n + s 1 n s 1 n 1 n 1 1 + s 1 n 1 n 1 = 3.07 15 3.07 15 14 + 0.80 1 + 0.80 1 11 = 16.3 x 1 x ± t s 1 α/;v + s n 1 n 3.84 1.49 ± t 0.05;16 3.07 15 + 0.80 1 0.6 < μ 1 μ < 4.1
-Samples: Estimating the Difference of Means If d and s d are the mean and standard deviation of the normally distributed differences of n random pairs of measurements, a 100 1 α % confidence interval for μ 1 μ : d t α/;n 1 s d n < μ 1 μ < d + t α/;n 1 s d n
-Samples: Estimating the Difference of Means Example 9.13: Assume that we have n = 0, d = 0.87 and s d =.9773. We also assume that the distribution of the differences is approximately normal. Find a 95% confidence interval for the difference. We have s d d t α/;n 1 n < μ s d 1 μ < d + t α/;n 1 n.9773 0.8700 t 0.05;19 0 < μ.9773 1 μ < 0.8700 + t 0.05;19 0.634 < μ 1 μ < 0.534
1-Sample: Estimating a Proportion If p is the proportion of successes in a random sample of size n and q = 1 p, an approximate 100 1 α % confidence interval for p: or p z α/ p q n < p < p + z p q α/ n p + z α/ n 1 + z α/ n ± z α/ 1 + z α/ n p q n + z α/ 4n
-Samples: Estimating the Difference of Proportions If p 1 and p are the proportions of successes in random samples of sizes n 1 and n, respectively, and q 1 = 1 p 1 and q = 1 p, an approximate 100 1 α % confidence interval for p 1 p : p 1 p ± z α/ p 1 q 1 n 1 + p q n
1-Sample: Estimating the Variance If s is the variance of a random sample of size n from a normal population, a 100 1 α % confidence interval for σ : n 1 s < σ < χ α/;n 1 n 1 s χ 1 α/;n 1
1-Sample: Estimating the Variance
-Samples: Estimating the Ratio of Variances If s 1 and s are the variances of independent samples of sizes n 1 and n from normal populations, respectively, a 100 1 α % confidence interval for σ 1 /σ : s 1 s 1 f α/ n 1 1, n 1 < σ 1 σ < s 1 s f α/ n 1 1, n 1
-Samples: Estimating the Ratio of Variances
End of Lecture Thank you! Questions?