Financial Risk Manager Part 1

Financial Risk Manager Part 1

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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.

TTanishq



Explanation:

Explanation

The test statistic for the slope coefficient is calculated using the formula:

tα,n2=β1β0se(β1)t_{α,n−2} = \frac{β₁ - β₀}{se(β₁)}

Where:

  • β1β₁ = 0.2043 (estimated slope coefficient)
  • β0β₀ = 0 (hypothesized value under H₀)
  • se(β1)se(β₁) = 0.01828 (standard error of slope coefficient)
  • nn = 6 (number of observations)
  • Degrees of freedom = n2=62=4n - 2 = 6 - 2 = 4

Test statistic calculation: t=0.204300.01828=11.176111.2t = \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: t0.005,4=4.604t_{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
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