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Inference in the Simple Regression Model Hill et al Chapter 5.

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Interval Estimation Provide an estimate of the range in which an unknown parameter is likely to lie.

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Standardising the sampling distribution of the least squares estimator Using the estimate of the variance:

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Critical values from the t distribution

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Interpretation Note that b 2 and se(b 2 ) are both random. The interval estimate is therefore also random. It is an estimate. There is no guarantee that it will contain the true value of the parameter of interest. If we construct 95% intervals from repeated samples, 95% of them will contain the true parameter value.

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Example b 2 =.1283

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Hypothesis testing What is the probability that the sample estimate would be obtained given an assumption about the true value of the parameter. Null Hypothesis (H 0 ), a belief which we maintain until we have sufficient evidence to convince us otherwise, 2 = c. Alternative Hypothesis (H 1 ) – 2 c – 2 >c – 2 < c

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The test statistic Random variable Has a known distribution if the null is true. Has a different distribution if the alternative is true. Consider: –H 0 : 2 = c –H 1 : 2 c

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The rejection region Rejection rule for a two-tailed test: If the value of the test statistic falls in the rejection region, either tail of the t-distribution, then we reject the null hypothesis and accept the alternative. If the value of the test statistic falls between the critical values t c and t c, in the non-rejection region, then we do not reject the null hypothesis.

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Example H 0 : 2 =.10. H 1 : =.05. Degrees of freedom: (T 2) = 38 Critical value t c is Since t=.93 < t c =2.024 we do not reject the null hypothesis

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Errors Type I –Reject a null when it is true. –The level of significance is the probability of a type I error. Type II –A false null is not rejected. –Increased chance of type II error with: reduced significance level. smaller sample.

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p-values The lowest level of significance at which the null hypothesis would not be rejected. Rejection rule for a two-tailed test: When the p-value of a hypothesis test is smaller than the chosen value of, then the test procedure leads to rejection of the null hypothesis.

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Tests of significance H 0 : 2 = 0. H 1 : 2 0 Since t=4.20 > t c =2.024 we reject the null hypothesis and accept the alternative P=2* =

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The least squares predictor Given a value x 0, what is the prediction of the explanatory variable: e 0 is unobservable, replace with its expectation E(e 0 )=0

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The forecast error The predictor is unbiassed.

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Variance of the forecast error The forecast error is estimated by substituting the estimate of Confidence interval:

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