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STATISTICAL INFERENCE PART V CONFIDENCE INTERVALS 1

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2 INTERVAL ESTIMATION Point estimation of : The inference is a guess of a single value as the value of . No accuracy associated with it. Interval estimation for : Specify an interval in which the unknown parameter, , is likely to lie. It contains measure of accuracy through variance.

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3 INTERVAL ESTIMATION An interval with random end points is called a random interval. E.g., is a random interval that contains the true value of with probability 0.95.

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Interpretation ( ) μ (unknown, but true value) 4 90% CI Expect 9 out of 10 intervals to cover the true μ

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5 INTERVAL ESTIMATION An interval (l(x 1,x 2,…,x n ), u(x 1,x 2,…,x n )) is called a 100 % confidence interval (CI) for if where 0< <1. The observed values l(x 1,x 2,…,x n ) is a lower confidence limit and u(x 1,x 2,…,x n ) is an upper confidence limit. The probability is called the confidence coefficient or the confidence level.

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6 INTERVAL ESTIMATION If Pr (l(x 1,x 2,…,x n ) )= , then l(x 1,x 2,…,x n ) is called a one-sided lower 100 % confidence limit for . If Pr( u(x 1,x 2,…,x n ))= , then u(x 1,x 2,…,x n ) is called a one-sided upper 100 % confidence limit for .

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7 METHODS OF FINDING PIVOTAL QUANTITIES PIVOTAL QUANTITY METHOD: If Q=q(x 1,x 2,…,x n ) is a r.v. that is a function of only X 1,…,X n and , and if its distribution does not depend on or any other unknown parameter (nuisance parameters), then Q is called a pivotal quantity. nuisance parameters: parameters that are not of direct interest

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8 PIVOTAL QUANTITY METHOD Theorem: Let X 1,X 2,…,X n be a r.s. from a distribution with pdf f(x; ) for and assume that an MLE (or ss) of exists: If is a location parameter, then Q= is a pivotal quantity. If is a scale parameter, then Q= / is a pivotal quantity. If 1 and 2 are location and scale parameters respectively, then are PQs for 1 and 2.

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Note Example: If 1 and 2 are location and scale parameters respectively, then is NOT a pivotal quantity for 1 because it is a function of 2 A pivotal quantity for 1 should be a function of only 1 and X’s, and its distribution should be free of 1 and 2. 9

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Example X 1,…,X n be a r.s. from Exp(θ). Then, is SS for θ, and θ is a scale parameter. S/θ is a pivotal quantity. So is 2S/θ, and using this might be more convenient since this has a distribution of χ²(2n) which has tabulated percentiles. 10

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11 CONSTRUCTION OF CI USING PIVOTAL QUANTITIES If Q is a PQ for a parameter and if percentiles of Q say q 1 and q 2 are available such that Pr{q 1 Q q 2 }= , Then for an observed sample x 1,x 2,…,x n ; a 100 % confidence region for is the set of that satisfy q 1 q(x 1,x 2,…,x n ; ) q 2.

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12 EXAMPLE Let X 1,X 2,…,X n be a r.s. of Exp( ), >0. Find a 100 % CI for . Interpret the result.

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13 EXAMPLE Let X 1,X 2,…,X n be a r.s. of N( , 2 ). Find a 100 % CI for and 2. Interpret the results.

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14 APPROXIMATE CI USING CLT Let X 1,X 2,…,X n be a r.s. By CLT, The approximate 100(1− )% random interval for μ: The approximate 100(1 − )% CI for μ: Non-normal random sample

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15 APPROXIMATE CI USING CLT Usually, is unknown. So, the approximate 100(1 )% CI for : When the sample size n ≥ 30, t /2,n-1 ~N(0,1). Non-normal random sample

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16 Lower confidence limit Upper confidence limit 1 - Confidence level Graphical Demonstration of the Confidence Interval for

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17 Inference About the Population Mean when is Unknown The Student t Distribution Standard Normal Student t 0

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18 Effect of the Degrees of Freedom on the t Density Function Student t with 10 DF 0 Student t with 2 DF Student t with 30 DF The “degrees of freedom” (a function of the sample size) determines how spread the distribution is compared to the normal distribution.

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19 Finding t-scores Under a t- Distribution (t-tables) t.100 t.05 t.025 t.01 t.005 Degrees of Freedom t 0.05, 10 = t 0

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20 EXAMPLE A new breakfast cereal is test-marked for 1 month at stores of a large supermarket chain. The result for a sample of 16 stores indicate average sales of $1200 with a sample standard deviation of $180. Set up 99% confidence interval estimate of the true average sales of this new breakfast cereal. Assume normality.

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21 ANSWER 99% CI for : ( , ) With 99% confidence, the limits and cover the true average sales of the new breakfast cereal.

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22 Checking the required conditions We need to check that the population is normally distributed, or at least not extremely nonnormal. Look at the sample histograms, Q-Q plots … There are statistical methods to test for normality

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