Use the following linear regression equation to answer the questions. x 1 = 1.7
ID: 3350899 • Letter: U
Question
Use the following linear regression equation to answer the questions.
x1 = 1.7 + 3.6x2 – 7.7x3 + 2.4x4
Which number is the constant term? List the coefficients with their corresponding explanatory variables.
If x2 = 1, x3 = 8, and x4 = 9, what is the predicted value for x1? (Use 1 decimal place.)
Suppose x3 and x4 were held at fixed but arbitrary values and x2 increased by 1 unit. What would be the corresponding change in x1?
Suppose x2 increased by 2 units. What would be the expected change in x1?
Suppose x2 decreased by 4 units. What would be the expected change in x1?
(e) Suppose that n = 11 data points were used to construct the given regression equation and that the standard error for the coefficient of x2 is 0.363. Construct a 99% confidence interval for the coefficient of x2. (Use 2 decimal places.)
(f) Using the information of part (e) and level of significance 5%, test the claim that the coefficient of x2 is different from zero. (Use 2 decimal places.)
Explanation / Answer
x1 = 1.7 + 3.6x2 – 7.7x3 + 2.4x4
= 1.7 + 3.6 - 7.7 *8 + 2.4 *9
= -34.7
x2 increased by 1 unit. What would be the corresponding change in x1?
increase by 3.6
Suppose x2 increased by 2 units. What would be the expected change in x1?
increase by 3.6*2 = 7.2
Suppose x2 decreased by 4 units. What would be the expected change in x1?
decrease by 3.6 *4 = 14.4
e)
df = n-k -1 = 11 - 3-1 = 7
for 99 % confidence interval
t = 3.5
(3.6 - 3.5* 0.363 , 3.6 + 3.5 * 0.363)
=(2.3295 , 4.8705)
f) t = (3.6 - 0)/0.363 = 9.917
critical value = 2.365
9.917 > 2.365
we reject the null
constant 1.7 x2 coefficient 3.6 x3 coefficient -7.7 x4 coefficient 2.4