Chapter 7: Statistical Intervals Based on a Single Sample Dr Sharabati Purdue University March 23, 2014 Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 1 / 42 Chapter Overview Basic Properties of Confidence Intervals (CIs) Large-Sample CI for Population Mean and Proportion CI for mean µ CI for proportion p One-sided intervals Intervals Based on a Normal Population Distribution t distribution One sample t CI for mean µ CI for the Variance and Standard Deviation of a Normal Population Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 2 / 42 Point Estimate vs. Confidence Interval To estimate a parameter µ of a normal distribution. Given the observed value x1 , x2 , · · · , xn of a random sample X1 , X2 , · · · , Xn . Point Estimate Find an point estimate of µ using the sample mean x ¯= x1 +x2 +···+xn n For different observed values, we may have different estimates for µ. Which estimate is closer to the true value? No idea :( Instead, we may provide an interval of values of µ Include the true value of µ with a certain “level of confidence” Point estimate ± the error of estimation Narrow interval → precise estimate Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 3 / 42 ¯ Random interval centered at X Let X1 , X2 , · · · , Xn be a random sample from a normal distriution with µ and σ. ¯ is normally distributed with expected We know that regardless of sample size n, X √ value µ and standard deviation σ/ n. [see PROPOSITION on text p224] Z= ¯ −µ ¯ −µ¯ X X X = √σ σX¯ n 0.95 = P (−1.96 < Z < 1.96) −1.96 < =P ¯ −µ X √σ n ! < 1.96 ¯ − 1.96 · √σ < µ < X ¯ + 1.96 · √σ =P X n n ¯ − 1.96 · √σ , X ¯ + 1.96 · √σ A random interval X n n Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 4 / 42 Confidence Interval and Confidence Level Figure : Fifty different 95% CIs for µ (red ones do not include µ). CIs allow us to estimate population parameters using a range of values. A confidence level is a measure of the degree of the relaibility of the interval. Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 5 / 42 Definition of 95% Confidence Interval (CI) for µ After observing X1 = x1 , X2 = x2 , · · · , Xn = xn from a normal distribution with known σ, we compute the observed sample mean x ¯ and a fixed interval. The 95% CI for populaiton mean µ is σ σ x ¯ − 1.96 √ , x ¯ + 1.96 √ n n or σ x ¯ ± 1.96 √ n or with 95% confidence σ σ x ¯ − 1.96 √ < µ < x ¯ + 1.96 √ n n Example A normal population has unknown µ and σ = 2.0, if n = 31 and x ¯ = 80.0, what is the 95% CI? σ 2.0 x ¯ ± 1.96 √ = 80.0 ± 1.96 √ = 80.0 ± .7 = (79.3, 80.7) n 31 Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 6 / 42 Definition of 95% Confidence Interval (CI) for µ After observing X1 = x1 , X2 = x2 , · · · , Xn = xn from a normal distribution with known σ, we compute the observed sample mean x ¯ and a fixed interval. The 95% CI for populaiton mean µ is σ σ x ¯ − 1.96 √ , x ¯ + 1.96 √ n n or σ x ¯ ± 1.96 √ n or with 95% confidence σ σ x ¯ − 1.96 √ < µ < x ¯ + 1.96 √ n n Example A normal population has unknown µ and σ = 2.0, if n = 31 and x ¯ = 80.0, what is the 95% CI? σ 2.0 x ¯ ± 1.96 √ = 80.0 ± 1.96 √ = 80.0 ± .7 = (79.3, 80.7) n 31 Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 6 / 42 Interpreting a CI Given a 95% CI, is it correct to say that µ falls in the CI with probability 0.95? Why? ¯ − 1.96 · √σ < µ < X ¯ + 1.96 · √σ , No. Look at the probability P X n n ¯ with observed value x when substitute X ¯, no randomness left. For a particular sample, either µ is in the interval or µ is not. For instance, it is not correct to say P (79.3 ≤ µ ≤ 80.7) = 0.95 in the previous exmaple. This statement under P has no chance element for a given sample. Therefore, we talk about being confident (in the method). A precise way to interpret CI is: with 95% confidence, µ falls in the interval calculated. Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 7 / 42 Interpreting a CI - Continue Note that: It is the confidence interval itself that is random (i.e. it changes for each sample). µ is just a fixed number (the population mean). It is non-random, but usually unknown. The 95% Confidence Level The 95% in a 95% C.I. could be interpreted as P (the next sample will give a C.I. that contains µ) = 0.95. The 95% means that we used a method that captures the true mean 95% of the time. Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 8 / 42 Example We want to estimate the mean µ of a normal population with given ¯ σ = 2.0 and a random sample X1 , X2 , · · · , Xn . Let us use µ ˆ = X. ¯ ¯ What is the distribution of X−µ 1 What is the distribution of X? √σ ? n ¯ 2 Find c such that P −c < X−µ < c = 0.95? √σ n 3 Given the observed sample {2,3,1,6,5,7,10,4,9,8}. The estimate of µ is x ¯ = 5.5. What’s the 95% confidence interval for µ? Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 9 / 42 Notation zα/2 Let zα/2 denote the value such that P (Z > zα/2 ) = P (Z < −zα/2 ) = α/2, where Z ∼ N (0, 1). That is, the area between −zα/2 and zα/2 under the standard normal curve is 1 − α, which means P (−zα/2 < Z < zα/2 ) = 1 − α. Find the following zα/2 values: For α = 0.01, zα/2 = For α = 0.05, zα/2 = Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 10 / 42 Other Confidence Level Definition A 100(1 − α)% CI for the mean µ of a normal population when the value of σ is known σ σ x ¯ − zα/2 · √ , x ¯ + zα/2 · √ n n Several most commonly used zα/2 ’s: Confidence Level α/2 zα/2 Dr Sharabati (Purdue University) 90% 0.05 1.645 Confidence Intervals 95% 0.025 1.960 99% 0.005 2.576 Spring 2014 11 / 42 From 95% CI to 90% CI P −1.96 < i.e., P −z0.05/2 < ¯ −µ X √σ n ¯ −µ X √σ n ! < 1.96 = 0.95 ! < z0.05/2 = 0.95 The 95% CI of the mean µ is σ x ¯ ± z0.05/2 √ n Similarly, P −z0.10/2 < ¯ −µ X √σ n ! < z0.10/2 = 0.90 The 90% CI of the mean µ is σ x ¯ ± z0.10/2 √ n Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 12 / 42 Confidence Level, Sample Size and Precision Review A (1 − α)100% CI for the mean µ of a normal population when the value of σ is known σ σ x ¯ − zα/2 · √ , x ¯ + zα/2 · √ n n Want narrow CI with high confidence level. Normal population, with unknown mean µ, and standard deviation σ = 2.0. Sample of size 25 yields x ¯ = 1.0. 1 What’s the 100%, 95% and 90% CI for µ? 2 Suppose a sample of size 100 also yields x ¯ = 1.0, what’s the 90% CI? 3 Which of the above intervals is narrower? Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 13 / 42 Finding the Sample Size Want to estimate a normal population mean µ with σ = 25. What sample size is necessary to ensure that the 95% CI has a width of (at most) 10? Answer Width of CI=¯ x + zα/2 · √σn − x ¯ + zα/2 · √σn = 2zα/2 · √σn Here 10 = 2zα/2 √25n and α = 1 − 0.95 = 0.05 (z0.025 = 1.96) √ n = 2 · 1.96 25 10 n = 96.04, the sample size is at least 97. Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 14 / 42 Finding the Sample Size Want to estimate a normal population mean µ with σ = 25. What sample size is necessary to ensure that the 95% CI has a width of (at most) 10? Answer Width of CI=¯ x + zα/2 · √σn − x ¯ + zα/2 · √σn = 2zα/2 · √σn Here 10 = 2zα/2 √25n and α = 1 − 0.95 = 0.05 (z0.025 = 1.96) √ n = 2 · 1.96 25 10 n = 96.04, the sample size is at least 97. General Formula The sample size n necessary to ensure an interval width w for confidence level (1 − α)100% is σ 2 n = 2zα/2 · w [hint: we want w = 2zα/2 · √σn ] Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 14 / 42 Finding the Sample Size Want to estimate a normal population mean µ with σ = 25. What sample size is necessary to ensure that the 95% CI has a width of (at most) 10? Answer Width of CI=¯ x + zα/2 · √σn − x ¯ + zα/2 · √σn = 2zα/2 · √σn Here 10 = 2zα/2 √25n and α = 1 − 0.95 = 0.05 (z0.025 = 1.96) √ n = 2 · 1.96 25 10 n = 96.04, the sample size is at least 97. General Formula The sample size n necessary to ensure an interval width w for confidence level (1 − α)100% is σ 2 n = 2zα/2 · w [hint: we want w = 2zα/2 · √σn ] Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 14 / 42 Review: Confidence Level, Sample Size and Precision Recall that A (1 − α)100% CI for the mean µ of a normal population when σ is known σ σ x ¯ − zα/2 · √ , x ¯ + zα/2 · √ n n The width of the above CI is w = 2zα/2 · √σ . n Larger sample size n results in narrower CI, lower confidence level results in narrower CI. Confidence level (a measure of the degree of the relaibility of the interval) cannot be too high. 100% will result in (−∞, ∞). One strategy: specify desired confidence level and interval width then determine sample size. Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 15 / 42 Large-Sample CIs Large-Sample CI for Population Mean and Proportion CI for mean µ CI for proportion p One-sided intervals Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 16 / 42 Large-Sample CI for Population Mean Let X1 , X2 , · · · , Xn be a random sample from an unknown population having unknown mean µ and unknown standard deviation σ. How to find (1 − α)100% CI? ¯ is approximately normal distributed with mean µ, std dev By CLT, X ¯ −µ X Z= √σ n √σ . n ∼ N (0, 1) We find CI by: P −zα/2 < ¯ −µ X √σ n ! < zα/2 ≈1−α Then (1 − α)100% CI for µ is: ¯ − zα/2 √σ , X ¯ + zα/2 √σ X n n In the expression σ is still unknown. For large sample sizes, we may replace it with sample standard deviation S: ¯ − zα/2 √S , X ¯ + zα/2 √S X n n ¯ =x For a given sample x1 , x2 , · · · , xn , plug in X ¯ and S = s in the above expression. Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 17 / 42 Example Given a sample from an unknown distribution with mean µ and standard deviation σ, the sample size is 196, sample mean x ¯ = 2.0, and sample standard deviation s = 3.0. Find the 95% CI of the population mean. Answer s x ¯ ± zα/2 √ n 3.0 2.0 ± 1.96 √ 196 Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 18 / 42 Example Given a sample from an unknown distribution with mean µ and standard deviation σ, the sample size is 196, sample mean x ¯ = 2.0, and sample standard deviation s = 3.0. Find the 95% CI of the population mean. Answer s x ¯ ± zα/2 √ n 3.0 2.0 ± 1.96 √ 196 Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 18 / 42 Large Sample Confidence Interval for a Population Proportion p Suppose X ∼ Bin(n, p) (Binomial), the sampling distribution of the sample proportion pˆ = X/n is ! r p(1 − p) approx pˆ ∼ N µpˆ = p, σpˆ = n for large sample size n (np ≥ 10 and n(1 − p) ≥ 10). pˆ − p q ∼ N (0, 1) p(1−p) n To find (1 − α)100% CI, we get P −zα/2 < p Dr Sharabati (Purdue University) pˆ − p p(1 − p)/n ! < zα/2 Confidence Intervals ≈1−α Spring 2014 19 / 42 Finding the CI for a Population Proportion p Solving the inequality, we get the (1 − α)100% CI of p: r pˆ + lower confidence limit= 2 zα/2 2n 1+ pˆ + upper confidence limit= 2 zα/2 2n pˆ(1−ˆ p) n − zα/2 + 1+ 4n2 2 zα/2 n r + zα/2 2 zα/2 pˆ(1−ˆ p) n + 2 zα/2 4n2 2 zα/2 n When n is quite large, the above limits can be approximated by: r pˆ(1 − pˆ) lower confidence limit=ˆ p − zα/2 n r pˆ(1 − pˆ) upper confidence limit=ˆ p + zα/2 n Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 20 / 42 The 100(1 − α)% CI for Population Proportion p The 100(1 − α)% CI for the population proportion p for a sample of size n, with sample proportion pˆ, is given by: r pˆ + 2 zα/2 2n ± zα/2 1+ pˆ(1−ˆ p) n + 2 zα/2 4n2 2 zα/2 n Although this formula was based on a large-sample distribution of pˆ, it works quite well even when the sample size n is reasonably small. When n is quite large, the CI reduces to r pˆ(1 − pˆ) pˆ ± zα/2 n Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 21 / 42 Example Among 10000 cats in Indiana state, 20% are found to be long-hairs. What is the 99% CI for the proportion p of long-hairs in Indiana? Hint r pˆ − zα/2 Dr Sharabati (Purdue University) pˆ(1 − pˆ) , pˆ + zα/2 n Confidence Intervals r pˆ(1 − pˆ) n ! Spring 2014 22 / 42 Example Among 10000 cats in Indiana state, 20% are found to be long-hairs. What is the 99% CI for the proportion p of long-hairs in Indiana? Hint r pˆ − zα/2 Dr Sharabati (Purdue University) pˆ(1 − pˆ) , pˆ + zα/2 n Confidence Intervals r pˆ(1 − pˆ) n ! Spring 2014 22 / 42 One-sided Confidence Intervals (Confidence Bounds) The 100(1 − α)% upper confidence bound for µ is s µ<x ¯ + zα √ n The 100(1 − α)% lower confidence bound for µ is s µ>x ¯ − zα √ n Confidence Level α zα Dr Sharabati (Purdue University) 90% 0.10 1.28 Confidence Intervals 95% 0.05 1.645 99% 0.01 2.33 Spring 2014 23 / 42 CI for Normal Population with Unknown µ and Unknown σ t-distribution and properties One sample t CI for normal mean µ with unknown µ and unknown σ Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 24 / 42 CI For Normal Mean µ with Unknown µ and Unknown σ Let X1 , X2 , · · · , Xn be a random sample from a normal distriution with unknown µ and σ. To find a CI for µ, We need to introduce a new distribution. Theorem ¯ is the mean of a random sample of size n, and S is the sample When X standard deviation from a normal distribution with mean µ, then the rv: T = ¯ −µ X √S n has a t distribution with degree of freedom(df) n − 1, denoted by tn−1 . Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 25 / 42 Properties of t Distribution The density curve of a t with df ν: Bell-shaped More spread out than the standard normal (z) curve − heavier tails t curve becomes less spread out when ν increases t distribution becomes standard normal distribution when df ν → ∞ (because as n increases, s → σ) Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 26 / 42 Properties of t Distribution Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 27 / 42 tα,ν Notation and t Table The notation tα,ν is the value on the measurement axis for which the area under the t curve with df=ν to the right of tα,ν is α; tα,ν is called t critical value. The tα,ν values are tabulated. Exercises 1 Find t 0.1,10 , t0.001,30 , t0.05,120 . 2 Determine t critical value that will capture the desired t curve when Central area= 0.95, df=16 Lower-tail area= 0.1, df=16 3 For a rv T which follows a t distribution with df=n − 1, what is P −tα/2,n−1 < T < tα/2,n−1 ? Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 28 / 42 The One-sample t Confidence Interval 2 Given a random sample X1 , X2 , ..., Xn from an unknown normal distri. N (µ, σ ), we ¯ ¯ have T = X−µ ∼ tn−1 . For a given α, P −tα/2,n−1 < X−µ < −tα/2,n−1 = 1 − α. S S √ √ n n Proposition Let x ¯ and s be the sample mean and sample standard deviation computed from a random sample from a normal population. The (1 − α)100% CI for µ is s s ¯ + tα/2,n−1 · √ x ¯ − tα/2,n−1 · √ , x n n If the sample size is large, the critical value can be taken from the standard normal table. The t CIs are robust to small or even moderate deviations from normality unless n is quite small. Compare to the (1 − α)100% CI for normal mean µ when σ is known, x ¯ − zα/2 √σn , x ¯ + zα/2 √σn . Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 29 / 42 Example A random sample of size 10 from a normal population yields: 2,3,1,6,5,7,10,4,9,8. Find the 95% CI for the normal mean µ. Answer s s ¯ + tα/2,n−1 · √ x ¯ − tα/2,n−1 · √ , x n n α = 0.05, t0.05/2,10−1 = t0.025,9 = 2.262, x ¯= s s= 2 + 3 + ··· + 8 = 5.5 10 22 + 32 + · · · + 82 − 10 − 1 (2+3+···+8)2 10 = 3.0277 so the 95% CI is: 3.0277 3.0277 (5.5 − 2.262 · √ , 5.5 + 2.262 · √ ) = (3.3343, 7.6657) 10 10 Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 30 / 42 Example A random sample of size 10 from a normal population yields: 2,3,1,6,5,7,10,4,9,8. Find the 95% CI for the normal mean µ. Answer s s ¯ + tα/2,n−1 · √ x ¯ − tα/2,n−1 · √ , x n n α = 0.05, t0.05/2,10−1 = t0.025,9 = 2.262, x ¯= s s= 2 + 3 + ··· + 8 = 5.5 10 22 + 32 + · · · + 82 − 10 − 1 (2+3+···+8)2 10 = 3.0277 so the 95% CI is: 3.0277 3.0277 (5.5 − 2.262 · √ , 5.5 + 2.262 · √ ) = (3.3343, 7.6657) 10 10 Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 30 / 42 More Intervals on Normal Populations Prediction interval (PI) for a single observation CI for variance σ 2 and standard deviation σ Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 31 / 42 Prediction Interval for a Single Future Value ColorredIn many applications, we wish to predict a single value of a variable using a given sample. Suppose we have a random sample from a normal population X1 , · · · , Xn . Now we wish to predict the the value of Xn+1 , a single future observation. We need a new rv: T = ¯ − Xn+1 X q S 1 + n1 It can be shown that: T ∼ tn−1 . By solving the below inequality for Xn+1 ¯ − Xn+1 X P −tα/2,n−1 < q < tα/2,n−1 = 1 − α S 1 + n1 we can find the (1 − α)100% PI. The prediction level is (1 − α). Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 32 / 42 Prediction Interval for a Single Obs Proposition A PI for a single observation to be selected from a normal population distribution is r 1 x ¯ ± tα/2,n−1 · s 1 + n The prediction level is (1 − α)100%. Interpretation Similar to the CI: we are (1 − α)100% sure that a future obs will fall in the PI Or, when the PI is calculated sample after sample, in the long run, 95% will include the true value of the future obs. Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 33 / 42 Example A random sample of size 10 from a normal population yields: x ¯ = 21.90, s = 4.134, given t0.025,9 = 2.262, find: 95% CI for the normal mean µ, how do you interpret the CI? 4.134 4.134 21.90 − 2.262 √ , 21.90 + 2.262 √ 10 10 ⇒ (18.94, 24.86) 95% PI for a single future value, how do you interpret the PI? ! r r 1 1 21.90 − 2.262 × 4.134 1 + , 21.90 + 2.262 × 4.134 1 + 10 10 ⇒ (12.09, 31.71) Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 34 / 42 CIs for Variance and Standard Deviation of a Normal Given a sample from a normal population, want to estimate the CI of the variance σ 2 or std dev σ. Need a new distribution: χ2 . Theorem Let X1 , X2 , · · · , Xn be a random sample from a normal distribution with parameters µ and σ 2 . Then the rv: P ¯ (n − 1)S 2 (X − Xi )2 = 2 σ σ2 has a χ2 distribution with (n − 1) df. Denoted by χ2n−1 . Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 35 / 42 χ2 Distribution Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 36 / 42 χ2 Distributions and χ2α,ν Notation χ2 curves A group of positive distributions: x ≥ 0. Not symmetric. Each has a positive skew with long upper tail. different df ν has different density shape, and density curve becomes more symmetric when ν increases χ2α,ν Notation χ2α,ν is called a χ2 critical value, denote the number on the measurement axis such that α of the area under the χ2 curve with ν degrees of freedom lies to the right of χ2α,ν . χ2α,ν values are tabulated. Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 37 / 42 Examples 1 Find χ20.1,12 , χ20.05,15 , χ20.95,15 . 2 Determine the following 2 The 95% percentile of the χ dist2with df2 ν = 15. 2 P 10.52 < χ < 40.65 , where χ is a χ rv with df ν = 25. 3 For a random sample of size n from a normal population with variance σ 2 . What is: (n − 1)S 2 2 2 P χ1−α/2,n−1 < < χα/2,n−1 ? σ2 Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 38 / 42 CI for σ 2 and σ (n − 1)S 2 2 χ21−α/2,n−1 < < χ =1−α α/2,n−1 σ2 Solve the inequality for σ 2 , we get: P (n − 1)S 2 (n − 1)S 2 < σ2 < 2 2 χα/2,n−1 χ1−α/2,n−1 Proposition A (1 − α)100% CI for variance σ 2 of a normal population is: ! (n − 1)s2 (n − 1)s2 , χ2α/2,n−1 χ21−α/2,n−1 A (1 − α)100% CI for std dev σ is then: ! s s (n − 1)s2 (n − 1)s2 , χ2α/2,n−1 χ21−α/2,n−1 Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 39 / 42 Example A random sample from a normal population of size 10 yields 2,3,1,6,5,7,10,4,9,8. Find the 95% CI for the normal variance σ 2 and std dev σ. Answer (n − 1)s2 (n − 1)s2 , χ2α/2,n−1 χ21−α/2,n−1 ! Because 95% CI ⇒ α = 0.05, sample size n = 10 ⇒df= 9, we have χ20.975,9 = 2.700, χ20.025,9 = 19.022. Also from the sample, we can compute s2 = (2+3+···+8)2 10 22 +32 +···+82 − 10−1 = 9.167 95% CI for σ 2 : 9 × 9.167 9 × 9.167 , 19.022 2.700 ⇒ (4.337, 30.557) 95% CI for σ is then: √ Dr Sharabati (Purdue University) √ 4.337, 30.557 ⇒ (2.083, 5.528) Confidence Intervals Spring 2014 40 / 42 Example A random sample from a normal population of size 10 yields 2,3,1,6,5,7,10,4,9,8. Find the 95% CI for the normal variance σ 2 and std dev σ. Answer (n − 1)s2 (n − 1)s2 , χ2α/2,n−1 χ21−α/2,n−1 ! Because 95% CI ⇒ α = 0.05, sample size n = 10 ⇒df= 9, we have χ20.975,9 = 2.700, χ20.025,9 = 19.022. Also from the sample, we can compute s2 = (2+3+···+8)2 10 22 +32 +···+82 − 10−1 = 9.167 95% CI for σ 2 : 9 × 9.167 9 × 9.167 , 19.022 2.700 ⇒ (4.337, 30.557) 95% CI for σ is then: √ Dr Sharabati (Purdue University) √ 4.337, 30.557 ⇒ (2.083, 5.528) Confidence Intervals Spring 2014 40 / 42 Summary CI for Normal Population A (1 − α)100% CI for the mean µ when σ is known, x ¯ ± zα/2 · √σn when σ is unknown, x ¯ ± tα/2,n−1 · √s n A (1 − α)100% CI for the variance 2 (n−1)s2 , (n−1)s χ2α/2,n−1 χ21−α/2,n−1 Large-Sample CI for Population Mean and Proportion A (1 − α)100% CI for mean µ: x ¯ ± zα/2 √sn A (1 − α)100% CI for proportion p: approx. pˆ ± zα/2 q pˆ(1−ˆ p) n Notation: Let x ¯ and s be the sample mean and sample standard deviation computed from a random sample. Confidence level is (1 − α). Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 41 / 42 Summary CI for Normal Population A (1 − α)100% CI for the mean µ when σ is known, x ¯ ± zα/2 · √σn when σ is unknown, x ¯ ± tα/2,n−1 · √s n A (1 − α)100% CI for the variance 2 (n−1)s2 , (n−1)s χ2α/2,n−1 χ21−α/2,n−1 Large-Sample CI for Population Mean and Proportion A (1 − α)100% CI for mean µ: x ¯ ± zα/2 √sn A (1 − α)100% CI for proportion p: approx. pˆ ± zα/2 q pˆ(1−ˆ p) n Notation: Let x ¯ and s be the sample mean and sample standard deviation computed from a random sample. Confidence level is (1 − α). Dr Sharabati (Purdue University) Confidence Intervals Spring 2014 42 / 42
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