Question

1. Use the paired dataset given in the below table to conduct a simple regression analysis as follows:

1. (10 pts) Using these 4 paired data points, fill up the table calculating all of the terms such as (x², x*y) used in the formulas of this question. Fill up the other column titles on your own and make the calculations. If needed add coloumns. (the table should be written in your answer paper)

	xi	yi	xi2	xi*yi
1	2	32
2	1	27
3	5	93
4	8	101
Total
Avr.

2. (15 pts.) Using the values in the table, find the intercept and slope of the estimated regression line.
3. (15 pts.) Check the homogenous variance assumption of the error of your estimated regression model using a graphic.
4. (15 pts.) Conduct hypothesis testing to conclude whether or not there is a significant linear relationship between variable x and variable y.

5. (15 pts.) Sample three scores from a standard normal distribution, square each score, and sum the squares. What is the probability that the sum of these two squares will be 8 or higher? Explain how you find the result.

(Hint: use a special type of distribution we learned that represents the described case to find the probability)

Answer 1

a.

	xi	yi	xi2	xi*yi
1	2	32	4	64
2	1	27	1	27
3	5	93	25	465
4	8	101	64	808
Total	16	253	94	1364
Avr.	4	63.25	23.5	341

For linear regression, y = a+bx

a = intercept

and b = slope

then, , and

then, = 1408/120 = 11.73

and a = 63.25 - 11.73*4 = 16.33

therefore, intercept = 16.33 and slope = 11.73

Equation: y = 16.33 + 11.73*x

b. To test this we will plot residuals with predicted y

	xi	yi	xi2	xi*yi	yi hat	ei = yi - yihat
1	2	32	4	64	39.79	-7.79
2	1	27	1	27	28.06	-1.06
3	5	93	25	465	74.98	18.02
4	8	101	64	808	110.17	-9.17
Total	16	253	94	1364	253	0
Avr.	4	63.25	23.5	341	63.25	0

Then, plot:

The pattern is random or we can say residuals do not form any specific pattern.

Hence, variances are constant and not variable , we have homoscedasticity and no heteroscedasticity.

c.

Running regression in R:

Regression Statistics
Multiple R	0.947475
R Square	0.897709	explains 89.77 % variability
Adjusted R Square	0.846563
Standard Error	15.33976
Observations	4
ANOVA
	df	SS	MS	F	Significance F
Regression	1	4130.133	4130.133	17.55201	0.052525	not less than 0.05, not significant at 5%
Residual	2	470.6167	235.3083
Total	3	4600.75
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	16.31667	13.57663	1.20182	0.352433	-42.0989	74.73219
X Variable 1	11.73333	2.800645	4.189511	0.052525	-0.31687	23.78354

The model is not significant at 5% , however it can be significant at alpha values greater than output p-value 0.0525

a. For standard normal distribution, z ~ N(0,1) i.e, mean = 0 and variance = 1

Then, let 3 scores be , X1 = 0, X2 = 1, X3 = -1 as Z ranges from -3 to 3 for 99.73% times

then distribution of sum of these, i.e,W = X1²+X2²+X3² ~ chi square with 3 degrees of freedom.

then, P(W greater than or equal to 8) = 0.05 (using calculator)

Please rate my answer and comment for doubt, it took alot of effort