Answer:
Step-by-step explanation:
Hello!
The given data corresponds to the variables
Y: Annual Maintenance Expense ($100s)
X: Weekly Usage (hours)
n= 10
∑X= 253; ∑X²= 7347; 
= ∑X/n= 253/10= 25.3 Hours
∑Y= 346.50; ∑Y²= 13010.75; 
= ∑Y/n= 346.50/10= 34.65 $100s
∑XY= 9668.5
a)
To estimate the slope and y-intercept you have to apply the following formulas:
^Y= a + bX
^Y= 10.53 + 0.95X
b)
H₀: β = 0
H₁: β ≠ 0
α:0.05
F= 47.62
p-value: 0.0001
To decide using the p-value you have to compare it against the level of significance:
If p-value ≤ α, reject the null hypothesis.
If p-value > α, do not reject the null hypothesis.
The decision is to reject the null hypothesis.
At a 5% significance level you can conclude that the average annual maintenance expense of the computer wheel alignment and balancing machine is modified when the weekly usage increases one hour.
b= 0.95 $100s/hours is the variation of the estimated average annual maintenance expense of the computer wheel alignment and balancing machine is modified when the weekly usage increases one hour.
a= 10.53 $ 100s is the value of the average annual maintenance expense of the computer wheel alignment and balancing machine when the weekly usage is zero.
c)
The value that determines the % of the variability of the dependent variable that is explained by the response variable is the coefficient of determination. You can calculate it manually using the formula:
![R^2 = \frac{b^2[sumX^2-\frac{(sumX)^2}{n} ]}{[sumY^2-\frac{(sumY)^2}{n} ]} = \frac{0.95^2[7347-\frac{(253)^2}{10} ]}{[13010.75-\frac{(346.50)^2}{10} ]} = 0.86](https://tex.z-dn.net/?f=R%5E2%20%3D%20%5Cfrac%7Bb%5E2%5BsumX%5E2-%5Cfrac%7B%28sumX%29%5E2%7D%7Bn%7D%20%5D%7D%7B%5BsumY%5E2-%5Cfrac%7B%28sumY%29%5E2%7D%7Bn%7D%20%5D%7D%20%3D%20%5Cfrac%7B0.95%5E2%5B7347-%5Cfrac%7B%28253%29%5E2%7D%7B10%7D%20%5D%7D%7B%5B13010.75-%5Cfrac%7B%28346.50%29%5E2%7D%7B10%7D%20%5D%7D%20%3D%200.86)
This means that 86% of the variability of the annual maintenance expense of the computer wheel alignment and balancing machine is explained by the weekly usage under the estimated model ^Y= 10.53 + 0.95X
d)
Without usage, you'd expect the annual maintenance expense to be $1053
If used 100 hours weekly the expected maintenance expense will be 10.53+0.95*100= 105.53 $100s⇒ $10553
If used 1000 hours weekly the expected maintenance expense will be $96053
It is recommendable to purchase the contract only if the weekly usage of the computer is greater than 100 hours weekly.
