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Answer: 11.2, β > 0
## Explanation The test statistic for the slope coefficient is calculated using the formula: $$t_{α,n−2} = \frac{β₁ - β₀}{se(β₁)}$$ Where: - $β₁$ = 0.2043 (estimated slope coefficient) - $β₀$ = 0 (hypothesized value under H₀) - $se(β₁)$ = 0.01828 (standard error of slope coefficient) - $n$ = 6 (number of observations) - Degrees of freedom = $n - 2 = 6 - 2 = 4$ **Test statistic calculation:** $$t = \frac{0.2043 - 0}{0.01828} = 11.1761 ≈ 11.2$$ **Critical value:** For a one-tailed test at 0.5% significance level with 4 degrees of freedom: $$t_{0.005,4} = 4.604$$ **Decision:** Since 11.2 > 4.604, we reject the null hypothesis H₀: β₁ = 0 and conclude that β₁ > 0. **Key points:** - This is a one-tailed test because the alternative hypothesis is Hₐ: β₁ > 0 - The test statistic (11.2) exceeds the critical value (4.604) - Therefore, we have sufficient evidence to conclude that the slope coefficient is significantly greater than zero
Author: Tanishq Prabhu
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Two variables have a linear relationship of the form:
Y = −4.6 + 0.2043X
The slope coefficient has a standard error equal to 0.01828. Carry out a test of H₀: β₁ = 0 vs Hₐ: β₁ > 0 at 0.5% significance, quoting the test statistic and the conclusion if the number of observations, n, is 6.
A
11.2, β > 0
B
11.2, β = 0
C
0.01828, β ≠ 0
D
11.2, β ≠ 0
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