Using the t test 2 10972 7640 12151 2768 se 680 1046 3556 0941 SALES PRICE

Using the t test 2 10972 7640 12151 2768 se 680 1046

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Using the t-test: 2 109.72 7.640 12.151 2.768 (se) (6.80) (1.046) (3.556) (0.941) SALES PRICE ADVERT ADVERT The t-value for testing H 0 : β 2 = 0 against H 1 : β 2 ≠ 0 is t = 7.640/1.045939 = 7.30444 . Its square is t = (7.30444)2 = 53.355, identical to the F-value. When testing a single “equality” null hypothesis against a “not equal to” alternative hypothesis, either t or F-test can be used, the test outcomes are identical. The square of a t random variable with df degree of freedom is an F random variable with 1 degree of freedom in the numerator and df freedom in the denominator, t df 2 = F (1, df) . We can summarize the elements in an F-test as follows: The null hypothesis H 0 consists of one or more equality restrictions on the model parameters β k ; the null hypothesis may not include any “greater than” or “less than” options. The alternative hypothesis states that one or more of the equalities in the null hypothesis is not true . The alternative hypothesis may not include any “greater than” or “less than” options. The test statistic is the F-statistic in (24). If the null hypothesis is true, F has the F-distribution with J numerator degrees of freedom and N - K denominator degrees of freedom When testing a single equality null hypothesis, it is perfectly correct to use either the t- or F-test procedure: they are equivalent. There is another important relationship between F and R 2 , where K is the number of parameters, N is the sample size.   2 2 2 2 1 1 ( ) . 1 ( / ) 1 . 1 1 / ( 1) (1 ) / ( ) SST SSE K SSR K F SSE N K SST SSR N K SSR SST N K SSR SST K R N K R K R K R N K 3.8.1.4 More general F-tests So far we have discussed the F-test where the null hypotheses are equal to zero. The F-test can also be used for much more general hypotheses. Any number of conjectures (≤ K) involving linear
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86 hypotheses with equal signs can be tested. Deriving the restricted model implied by the null can be tricky, but the same general principles hold. Example: We found out that the optimal amount for Andy to spend on advertising is such that 3 4 0 β 1 ADVERT (3.31) Big Andy has been spending $1,900 (ADVERT 0 ) per month on advertising, and he wants to know if this amount could be optimal. Consider the issue of this testing: 0 3 4 1 3 4 H : β 2 β 1.9 1 H : β +2 β 1.9 1 Since ADVERT 0 = $1,900 per month, then: 0 3 4 1 3 4 H : β 3.8β 1 H : β +3.8β 1 How to find a restricted model? Note that when H 0 is true, then β 3 = 1 3.8β 4 so that: 2 1 2 4 4 β β 1 3.8β β SALES PRICE ADVERT ADVERT e Rearrange it in a convenient form for estimation: 2 1 2 4 2 2 3 β β β 3.8 reconstruct a model by creating new variables dependent variable: = explanatory variables: , and 3.8 .
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