# F n j 1 x j x 2 f t n 2 if 0 17 if β 0 and then the

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\$ \$ \$ F \$ \$ ' \$ \$ ' n j ' 1 ( X j & ¯ X ) 2 \$ F - t n & 2 if \$ ' 0. (17) If β > 0 and then the t-value of converges in probability to + 4 , and if β < 0 and n 6 4 ˆ β then the t-value of converges in probability to !4 . Moreover, if the sample size n is n 6 4 ˆ β large then by Proposition 7 we may use the standard normal distribution instead of the t distribution to find critical values of the test. Similarly, \$ t \$ " ( ' t & value of \$ " ) ' def . \$ " \$ F \$ " - t n & 2 if " ' 0. (18) However, the hypothesis α = 0 is often of no interest. In the ice cream example, ' n j ' 1 ( X j & ¯ X ) 2 ' 18 Y ' n j ' 1 ( X j & ¯ X ) 2 ' 18 . 4.24264, ' n j ' 1 ( Y j & ¯ Y ) 2 ' ' n j ' 1 Y 2 j & n . ¯ Y 2 ' 1020 & 8×11 2 ' 52 and by (13), ˆ σ 2 ' 1 n & 2 j n j ' 1 ˆ U 2 j ' 1 n & 2 j n j ' 1 ( Y j & ¯ Y ) 2 & ˆ β 2 1 n & 2 j n j ' 1 ( X j & ¯ X ) 2 ' 52 & (1.5) 2 .18 8 & 2 ' 11.5 6 . 1.916667 Y ˆ σ . 1.384437 Hence, \$ t \$ \$ ' \$ \$ ' n j ' 1 ( X j & ¯ X ) 2 \$ F ' 1.5×4.24264 1.384437 . 4.597 (19) Assuming that the conditions of Proposition 6 hold, the null hypothesis can be tested H 0 : β ' 0

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11 against the alternative hypothesis using the two-sided t-test at say the 5% H 1 : β 0 significance level, as follows. Under the null hypothesis, (19) is a random drawing from the t distribution with n ! 2 = 6 degrees of freedom. Look up in the table of the t distribution the value such that for This value is Then accept the null t ( T - t 6 , P [| T | > t ( ] ' 0.05. t ( ' 2.447. hypothesis if and reject the null hypothesis in favor of the & t ( ' & 2.447 # ˆ t ˆ β # 2.447 ' t ( , alternative hypothesis if Thus, in the ice cream example we reject the null | ˆ t β | > t ( ' 2.447. hypothesis because H 0 : β ' 0 | ˆ t β | ' 4.597 > 2.447 ' t ( . This test is illustrated in Figure 2 below. The curved line in Figure 2 is the density of the t distribution with 6 degrees of freedom. The grey areas are each 0.025, so that the total grey area is 0.05. Figure 2 Two-sided t-test of against the alternative hypothesis H 0 : β ' 0 H 1 : β 0. The null hypothesis can be tested against the alternative hypothesis H 0 : β ' 0 at the 5% significance level by the right-sided t-test. Now look up in the table of the t H 1 : β > 0 distribution the value such that for This value corresponds to the t ( T - t 6 , P [ T > t ( ] ' 0.05. critical value of the two-sided t-test at the 10% significance level: Then accept the t ( ' 1.943. null hypothesis if and reject the null hypothesis in favor of the alternative ˆ t ˆ β # t ( ' 1.943, hypothesis if Thus, in the ice cream case we reject the null hypothesis ˆ t β > t ( ' 1.943.
12 in favor of the alternative hypothesis H 0 : β ' 0 H 1 : β > 0. This right-sided t-test is illustrated in Figure 3 below. Again, the curved line in Figure 3 is the density of the t distribution with 6 degrees of freedom, and the grey area is 0.05. Figure 3 Right-sided t-test of against the alternative hypothesis H 0 : β ' 0 H 1 : β > 0. If the sample size n is large, so that then an alternative way of ˆ t ˆ β - N (0,1) if β ' 0, testing the null hypothesis β = 0 against the alternative hypothesis β 0 is to use the (two-sided) p-value: \$ p \$ \$ ( ' p & value of \$ \$ ) ' def . P [| U | > | \$ t \$ \$ |], where U - N (0,1). (20) For example, if we reject the null hypothesis β = 0 in favor of the alternative ˆ p ˆ β < 0.05 hypothesis β 0 at the 5% significance level, and if we accept the null hypothesis β ˆ p ˆ β \$ 0.05 = 0. The p-value for is defined and used similarly. ˆ α

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