Question

# Use the following linear regression equation to answer the questions. x1 = 1.7 + 3.6x2 –...

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.

 constant x2 coefficient x3 coefficient x4 coefficient

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.)

 lower limit upper limit

(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.)

 t t critical ±

#### Homework Answers

Answer #1
 constant 1.7 x2 coefficient 3.6 x3 coefficient -7.7 x4 coefficient 2.4

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

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