Lecture 12: Small Sample Intervals Based on a Normal Population Distribution

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1 Lecture 12: Small Sample Intervals Based on a Normal Population MSU-STT-351-Sum-17B (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 1 / 24

2 In this lecture, we will discuss (i) Student s t-distribution; (ii) Confidence intervals based on small sample size n; (iii) Prediction intervals. We know from CLT that (a) If the population is N(µ, σ) and σ known, then for any n (and hence for small n as well) the sampling distribution of X is approximately normally distributed and n(x µ) Z = N(0, 1). (b) If n is large, then for any population Z = σ n(x µ) S N(0, 1). (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 2 / 24

3 (c) However, when n is small and σ is unknown, if we replace σ by s in the formula (a) above, the distribution of the above statistic can be far from standard normal N(0, 1). So, we require a new probability distribution which performs well for small samples. Student s t-distribution Let X 1,..., X n be a random sample from a normal distribution, where n is small. The statistic T = X µ S/ n is said to follow a t-distribution with (n 1) degrees of freedom (df). It is denoted by t n 1. (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 3 / 24

4 Properties of the t-distribution (i) Each t ν density is symmetric about origin. (ii) Each t ν curve is more spread out/heavy tails than the standard normal. (iii) As ν, the spread of t ν decreases and approaches standard normal. 0.4 ν = 1 ν = 3 ν = (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 4 / 24

5 t-critical value t α,ν : t ν curve Shaded area = α 0 t α,ν (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 5 / 24

6 Definition: A t-critical value, denoted by t α,ν, is the point on the t ν curve with probability α to the right. That is, t-critical value t α,ν satisfies P(T ν > t α,ν ) = α, where T ν is the t-random variable with ν df. Critical value: The critical values for selected ν and α are given in Table A.5. Let X and S be the sample mean and sample SD from a normal population with mean µ. Then (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 6 / 24

7 Example 1: Find the t-critical value for (a) Central area=.95 for df = 10 (b) Upper tail area=.01, df = 25 (c) Lower tail area=.025, df = 5. Solution: (a) Since standard t-distribution is symmetric about zero, we have from Table A.5 (see column corresponding to 0.025) the critical value is t 0.025,10 = (b) An upper tail area of.01 is t 0.01,25 = (see Table A.8 and column corresponding to 25). (c) A lower tail area of.025 corresponds to the negative value of the one corresponding to the upper tail area of.025. But t 0.025,5 = (see column 3 of table A.5). Hence, t 0.975,5 = (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 7 / 24

8 One Sample t Confidence Interval for µ We have a random sample from normal population N(µ, σ 2 ), both µ and σ 2 are unknown and n is small. Then (i) The general two-sided interval for µ is X ± t α/2,n 1 S n. (ii) An upper one-sided confidence bound for µ is provided by X + t α,n 1 S n. (iii) A lower one-sided confidence bound for µ is provided by X t α,n 1 S n. (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 8 / 24

9 Example (Ex. 33): An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range: 418, 412, 421, 422, 425, 427, 431, 434, 437, 439, 446, 447, 448, 453, 454, 463, 465 (a) Construct a boxplot of the data and comment on its features. (b) Does the given sample observations come from a normal population? (c) Calculate a two-sided 95% confidence interval for µ, the true average degree of polymerization. Does the interval suggest that 440 is a possible value for µ? (d) What about 450? (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 9 / 24

10 Solution: (a) The lower and upper quartiles are 428 and 448; the IQR = = 23. To check for outliers, we calculate the values (IQR) = (23) = 390.5; (IQR) = (23) = Since the maximum and minimum are 465 and 418, there are no outliers in the data. The median is 437 and a boxplot of the data appears below: (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 10 / 24

11 (b) A quantile plot can be used to check for normality. The quantile plot below shows a fairly strong linear pattern, supporting the assumption that this data came from a normal population N o r m a l q u a n t i le 1 (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 11 / 24

12 (c) Since n is small (n = 17), we use an interval based on the t-distribution. Here, x = and s = Also, for 16 df, the 95% critical value is (from Table A.5). Therefore, ± = [430.5, 446.1] is the required CI for µ. This interval suggests that 440 (which is inside the interval) is a plausible value for the mean polymerization. (d) The value of 450, however, is not plausible, since it lies outside the interval. (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 12 / 24

13 Prediction Intervals Often, we want to predict the value of a rv to be observed at some future time. A prediction interval is a confidence interval for a future observation. Prediction Interval for a Single Observation X n+1 Note 95% confidence interval means that we are 95% confident that the population mean µ is between the interval limits. However, if we observe one element X n+1, its value can be far from the mean. Prediction interval is a CI for a future observation X n+1 of X, based on the sample X 1,..., X n. (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 13 / 24

14 The natural predictor is X with the error (X X n+1 ). Note E(X X n+1 ) = µ µ = 0 Hence, V(X X n+1 ) = V(X) + V(X n+1 ) = σ 2 ( 1 n + 1). X X n+1 N(0, 1). σ n But for a small n, X X n+1 t n 1. S n (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 14 / 24

15 So, ( P t α/2,n 1 < X X n+1 S n < t α/2,n 1 ) = 1 α. Hence, ( P X t α/2,n 1 S n < X n+1 < X + t α/2,n 1 S ) = 1 α. n Thus, we obtain a two-sided prediction interval for X n+1 as 1 X ± t α/2,n 1 S n + 1. The upper and lower prediction one-sided intervals can be obtained using t α,n 1 instead of t α/2,n 1. (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 15 / 24

16 Example (Ex 36): The n = 26 observations on escape exercise of oil workers yield x = and s = (a) Calculate an upper confidence bound for population mean escape time using a confidence level of 95%. (b) Calculate an upper 95% prediction interval for the escape time X n+1 of a single additional worker, and compare with part (a)? Solution: Given, n = 26, x = , s = (a) A 95% upper confidence bound for µ is ( S n ) x + (t.05,25 ) = (1.708)( ) = = That is, with a confidence level of 95%, the value of µ, is less than (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 16 / 24

17 (b) A 95% upper prediction bound for X n+1 is ( x + (t.05,25 ) s ) n ( = (1.708) ) 26 = = Hence, X n+1 will be less than 0, The prediction bound is larger than the confidence bound computed in part (a). This is because there is more variation in predicting a single worker s escape time than there is in estimating the populations mean escape time. (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 17 / 24

18 Confidence intervals for Variance σ 2 (unknown) Let X 1,..., X n be a random sample from N(µ, σ 2 ), where both µ and σ 2 ) are unknown. Then (n 1)S 2 (Xi X) 2 = χ 2 σ 2 σ 2 n 1. Let the critical value χ 2 α,ν be defined by P(χ2 ν > χ 2 α,ν) = α. Then 21 α/2,n 1 (n ) 1)S2 P (χ < < χ 2 σ 2 α/2,n 1 = 1 α. That is, ( (n 1)S 2 P < σ 2 (n ) 1)S2 < = χ 2 χ 2 1 α α/2,n 1 1 α/2,n 1 (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 18 / 24

19 Thus, 100(1 α)% CI for σ 2 is (n 1)S2, χ 2 α/2,n 1 (n 1)S 2. χ 2 1 α/2,n 1 The critical values of χ 2 distribution are shown in the following figures. χ 2 ν pdf Shaded area = α χ 2 α,ν (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 19 / 24

20 χ 2 ν pdf χ 2 1 α/2,ν χ 2 α/2,ν (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 20 / 24

21 Example (Ex 43): Determine the following: (a) The 95th percentile of χ (b) The 5th percentile of χ (c) P( χ ). Solution: (a) The 95th percentile of χ 2 10 is nothing but χ2 0.05,10 = (see Table A.7). (b) The 5th percentile of χ 2 10 is χ2.95,10 = (c) Since = χ 2.975,22 and = χ2.025,22, we have P(χ 2.975, 22 χ2 22 χ2.025, 22 ) = P(χ2 22 χ2.975,22 ) P(χ2 22 χ2.025,22 ) = = (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 21 / 24

22 Example (Ex 46): The results of a Wagner turbidity test performed on 15 samples of standard Ottawa testing sand were; 26.7, 25.8, 24.0, 24.9, 26.4, 25.9, 24.4, 21.7, 21.4, 25.9, 27.3, 26.9, 27.3, 24.8, (a) Is it possible that this sample comes from a normal population distribution? (b) Calculate an upper 95% confidence bound for the standard deviation σ. Solution: (a) Using a normal probability plot, we ascertain that this sample comes from a normal population distribution. (b) With S = 1.579, n = 15 and χ 2.95,14 = the 95% upper confidence bound for σ is (n 1)S 2 14(1.579) 2 χ 2.95,14 = = (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 22 / 24

23 Example (Ex 44): The amount of lateral expansion was determined for a sample of 9 pulsed power gas metal arc welds used in LNG ship containment tanks. The resulting sample SD was s = Assuming normality, find 95% CI for σ 2 and σ. Solution: For n 1 = 8, So, the 95% interval for σ 2 is χ 2.025,8 = , χ2.975,8 = ( 8(7.90) , 8(7.90) ) = (3.60, 28.98) The 95% interval for σ is ( 3.60, 28.98) = (1.90, 5.38). (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 23 / 24

24 Home work: Section 7.3: 31, 37, 39(c) Section 7.4: 42(a, b, c), 44, 45 (P. Vellaisamy: MSU-STT-351-Sum-17B) Probability & Statistics for Engineers 24 / 24

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