Where c is a constant and is an important value in

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Where c is a constant and is an important value in the context of specific regression model Step-by-step procedure for testing hypothesis 1. Determine the null and alternative hypothesis 2. Specify the test statistic and its distribution if the null hypothesis is true 3. Select α and determine the rejection region 4. Calculate the sample value of the test statistic 5. State your conclusion Testing the Significance of a linear relationship o We can test the significance of a linear relationship between X and Y by testing: H 0 : β 2 = 0 H 1 : β 2 ≠ 0 The distribution of the slope coefficient estimator, b 2 , and the test statistic o Based on assumptions SR1-SR6 we can show that b 2 N ( β 2 , [ σ 2 ( x i x ) 2 ] ) o A standardised normal random variable is obtained from b 2 by subtracting its mean and dividing by its standard deviation Z = b 2 β 2 σ 2 ( x i x ) 2 N ( 0,1 ) o The standardized random variable Z is normally distributed with mean 0 and variance 1 o Usually σ 2 is unknown and replaced by an estimate of σ 2 , ^ σ 2 The Test Statistic o Replacing σ 2 with ^ σ 2 creates a random variable t: t = b 2 β 2 ^ σ 2 ( x i x ) 2 t ( n 2 ) Where β 2 = H 0 The ratio has a t-distribution with (n-2) degrees of freedom y = β 1 + β 2 x + e (so β 1 & β 2 means n-2 degrees of freedom) Hypothesis Tests o The test statistic for testing the significance of β 2 H 0 : β k = 0 t = b k β k se ( b k ) for k = 1,2 Revisiting the Food Expenditure Example – Testing Significance of the Relationship
o Earlier we estimated a linear relationship b/w food expenditure and income in form y = β 1 + β 2 x + e where y = food expenditure, and x = income o Estimated regression model is given by ^ y = 83.42 + 10.21 x o We want to test the claim that income has a significantly positive influence on food expenditure In terms of our regression model this can be tested by testing H 0 : β 2 = 0 H 1 : β 2 > 0 The test statistic to perform this hypothesis test is t = b 2 0 se ( b k ) for k = 1,2 If the null hypothesis is true N = 40 DF = N-2 = 40 – 2 = 38 Let us select α = 0.05 Critical Value: t (1-α,N-2) = t (0.95,38) = 1.686 Rejection Region: Reject the null hypothesis H 0 if t ≥ 1.686 Using the food example data, b2 = 10.21; se(b2) = 2.09. The value of the test statistic is t = b 2 0 se ( b k ) = 10.21 2.09 = 4.88 Since t = 4.88 > 1.686 we reject the null H 0 : β 2 = 0 hypothesis and accept the alternative H 1 : β 2 > 0. That is we reject the null hypothesis that there is no relationship b/w income and food expenditure at the 5% level of significance Therefore there is enough evidence to conclude that there is statistically significant positive linear relationship b/w household income and food expenditure p-Value o Reject the null hypothesis when p-value is less than or equal to, level of significance α I.e. if p-value ≤ α, then reject H 0 I.e. if p-value > α, then do not reject H 0 Testing for the significance: p-value for a right-tail test o Null hypothesis is H 0 : β 2 = 0 o

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