Given the linear regression equation:

y = 1.6 + 3.5x1 - 7.9x2 +2.0x3

a. Which variable is the response variable? How many explanatory variables are there?
b. If x1 = 2, x2 = 1 and x3 = 5, what is the predicted value for y?
c. Supposed the n = 12 data points were used to construct the given regression equation above, and that the standard error for the coefficient x1 is 0.419. Construct a 90% confidence interval for the coefficient of x1.
d. Using the information from part c and 5% level of significance, test the claim that the coefficient of x1 is different from 0. What does your conclusion mean in relation to x1 predicting y?

a. y is the response variable. There are 3 explanatory variables, namely x1, x2 and x3.

b. y = 1.6 + 3.5 x 2 - 7.9 x 1 + 2 x 5 = 10.7

c. 90% confidence interval for the coefficient of x1 = (2.721, 4.279)

d. The claim is true. x1 has predictability towards y.

Step-by-step explanation:

c. Use the following four-step approach to construct a confidence interval.

- Identify a sample statistic. From the regression equation, we see that the coefficient of x1 is 3.5

- Select a confidence level. We are working with a 90% confidence level.

- Find the margin of error

•Find standard error for the coefficient of x1. The standard error is given as 0.419

•Find critical value. The critical value is a t score with degrees of freedom equal to n - k (n= number of data points, k = number of parameters). To find the critical value, we take these steps.

oCompute alpha (α):

α = 1 - (confidence level / 100)

α = 1 - 90/100 = 0.1

oFind the critical probability (p*):

p* = 1 - α/2 = 1 - 0.1/2 = 0.95

oFind the degrees of freedom (df):

df = n - k = 12 - 4 = 8.

oThe critical value is the t statistic having 8 degrees of freedom and a cumulative probability equal to 0.95. From the t Distribution Calculator, we find that the critical value is 1.86.

•Compute margin of error (ME):

ME = critical value * standard error

ME = 1.86 * 0.419 = 0.779

Specify the confidence interval. The range of the confidence interval is defined by the sample statistic + margin of error. And the uncertainty is denoted by the confidence level.

Therefore, the 90% confidence interval for the coefficient of x1 is 3.5 ± 0.779, which is 2.721 to 4.279 .

d. 5% level of significance is conversely translates to a 95% level of confidence. Using the same approach, we identify the 95% confidence interval for the coefficient of x1 is 3.5 ± 0.966, which is 2.534 to 4.466 .

0 falls out of the confidence interval, therefore the claim that the coefficient of x1 is different from 0. It means x1 has predictability towards y.

its 69.420

step-by-step explanation:

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol