Question

Use the following linear regression equation to answer the questions.

x₁ = 1.9 + 3.4x₂ – 8.2x₃ + 1.5x₄

Suppose that n = 15 data points were used to construct the given regression equation and that the standard error for the coefficient of x₂ is 0.342. Construct a 99%confidence interval for the coefficient of x₂. (Use 2 decimal places.)

Lower Limit:

Upper Limit:

Using the information of part (e) and level of significance 10%, test the claim that the coefficient of x₂ is different from zero. (Use 2 decimal places.)

t:

t critical+/-

Answer 1

t-critical value at df = n - p - 1 = 15 - 3 - 1 = 11 and alpha = 0.01 is 3.11

The 99% confidence interval is calculated by the following formula:

Coefficient +- t*SE

Coefficient = 3.4
SE = 0.342
t* = 3.11

Lower limit: 3.4 - 0.342*3.11 = 2.34

Upper limit: 3.4 + 0.342*3.11 = 4.46

Let's find the t-value:

t = b₂/SE = 3.4/0.342 = 9.94

t-critical value at df = 11 and alpha = 0.1 is 1.80

As t>1.80, we can reject the null hypothesis and conclude that the coefficient is different than zero.