HP 48gII hp 48gII_user's manual_English_E_HDPMSG48E67_V2.pdf - Page 626

Confidence intervals for the slope (Β) and intercept (A): • First, we obtain t n-2,α/2 = t3,0.025 = 3.18244630528 (See chapter 17 for a program to solve for tν,a): • Next, we calculate the terms (t n-2,α/2)⋅se/√Sxx = 3.182...⋅(0.1826.../2.5)1/2 = 0.8602... (t n-2,α/2)⋅se⋅[(1/n)+x2/Sxx]1/2 = 3.1824...⋅√0.1826...⋅[(1/5)+32/2.5] 1/2 = 2.65 • Finally, for the slope B, the 95% confidence interval is (-0.86-0.860242, -0.86+0.860242) = (-1.72, -0.00024217) For the intercept A, the 95% confidence interval is (3.24-2.6514, 3.24+2.6514) = (0.58855,5.8914). Example 2 -- Suppose that the y-data used in Example 1 represent the elongation (in hundredths of an inch) of a metal wire when subjected to a force x (in tens of pounds). The physical phenomenon is such that we expect the intercept, A, to be zero. To check if that should be the case, we test the null hypothesis, H0: Α = 0, against the alternative hypothesis, H1: Α ≠ 0, at the level of significance α = 0.05. The test statistic is t0 = (a-0)/[(1/n)+x2/Sxx]1/2 = (-0.86)/ [(1/5)+32/2.5] ½ = 0.44117. The critical value of t, for ν = n - 2 = 3, and α/2 = 0.025, can be calculated using the numerical solver for the equation α = UTPT(γ,t) developed in Chapter 17. In this program, γ represents the degrees of freedom (n-2), and α represents the probability of exceeding a certain value of t, i.e., Pr[ t>tα] = 1 - α. For the present example, the value of the level of significance is α = 0.05, g = 3, and tn-2,α/2 = t3,0.025. Also, for γ = 3 and α = 0.025, tn-2,α/2 = t3,0.025 = 3.18244630528. Because t0 > - tn-2,α/2, we cannot reject the null hypothesis, H0: Α = 0, against the alternative hypothesis, H1: Α ≠ 0, at the level of significance α = 0.05. This result suggests that taking A = 0 for this linear regression should be acceptable. After all, the value we found for a, was -0.86, which is relatively close to zero. Page 18-55