Slope, Β1 = Ssxy / Ssxx

Problems: Chapter 12: 8. a) x -1 0 1 2 3 y -1 1 1 2.5 3.5 b) 4 3.5 3 2.5 2 1.5 1 0.5 0 -2 -1 -0.5 0 1 2 3 4 -1 -1.5

x y X2 x.y y2

-1 -1 1 1 1

0 1 0 0 1

1 1 1 1 1

2 2.5 4 5 6.25

3 3.5 9 10.5 12.25  x  5  y  7  x 2  15  xy  17.5  y 2  21.5

1 SSxy = ∑xy - (∑x)( ∑y)/n

= 17.5 – 5*7 / 5 = 10.5

2 2 SSxx = ∑x - (∑x) / n

= 15 - 52 / 5 = 10

Slope, β1 = SSxy / SSxx

= 10.5 / 10 = 1.05

Y-intercept, β0 = ybar - β1 xbar

= 7 / 5 - 1.05 (5 / 5) = 0.35

y = 0.35 + 1.05x c) Since the y-intercept is .35, and because the least square line will always pass through the point (xbar, ybar) = (1, 1.4), the least square line is like:

4 3.5 3 2.5 2 1.5 1 0.5 0 -2 -1 -0.5 0 1 2 3 4 -1 -1.5

It provides a good fit to the data points as it can be seen from the graph above. 20. a)

x -1 0 1 2 3 y -1 1 1 2.5 3.5

2 2 SSyy = ∑y - (∑y) / n = 21.5 - 7 2 / 5 = 11.7

2 β1=1.05 SSxy =10.5

SSE = SSyy - β1.SSxy = 11.7 - 1.05 * 10.5 = 0.675 b) s2 = SSE / (n - 2) = 0.675 / (5 - 2) = 0.225 s = √s2 = √0.225 = 0.474

29. a) H0: β1 = 0 Ha: β1 ≠ 0 n=5 α = 0.10 (5 - 2) = 3 df tα/2 = t0.05 = 2.353 (table-6)

We will reject the null hypothesis H0 if |t| > 2.353.

β1= 1.05 s=0.474342 SSxx = 0.474342 t = β1 / (s / √SSxx) = 1.05 / (0.474 / √10) = 6.999

Since the calculated t value falls in the rejection region, we will reject the null hypothesis and conclude that the slope β1 is not 0. b) The observed significance level of the test is P = 0.0060 c) 90% confidence interval = β1 + tα/2 (s/√SSxx) n=5 α = 0.10 (5 - 2) = 3 df tα/2 = t0.05 = 2.353

3 β1 + tα/2 (s / √SSxx) = 1.05 + 2.353 (0.474 / √10) = 1.05 + 0.35295

Therefore confidence interval is 0.69705 to 1.40295.

a) Positively correlated: As brand recognition increases, sales for that product increase. b) Negatively correlated: As the price of a product increases, the demand for it decreases.

2 r = (SSyy - SSE) / SSyy

SSyy = 11.7 SSE = 0.675. r2 = (11.7 - 0.675) / 11.7 = 0.9423 Approximately 94% of the total sum of squares of deviations of the y values about the mean can be attributed to the linear relationship between y and x.

SSxx = 10 s = 0.474342 Xbar = 1 Yhat = 0.35 + 1.05x = 0.35 + (1.05).1 = 1.4

a) 90% confidence interval: 2 2 yhat + (tα/2) s√(1/n + (x - xbar) / SSxx) = 1.4 + (2.353) 0.474√ (1/5 + (1 - 1) / 10) = 1.4 + 0.499 = (.901, 1.899)

b) 90% prediction interval: 2 2 yhat + (tα/2) s√(1 + 1/n + (x - xbar) / SSxx) = 1.4 + (2.353) 0.474√ (1 + 1/5 + (1 - 1) /10) = 1.4 + 1.223 = (0.177, 2.623)

4 c) The prediction interval is wider than the confidence interval.

Using E-Labs and Computational Tools: Regression Analysis with Diagnostic Tools: Regression models are often constructed based on certain conditions that must be verified for the model to fit the data well, and to be able to predict accurately. This site provides the necessary diagnostic tools for the verification process and taking the right remedies such as data transformation. I used the data in the table 12.1. The results are:

5 6 Scattered Diagram: This JavaScript construct the scatter-diagram of your data, and determines the possible outliers. I used the data in the table 12.1. The results are: x-outliers: 22; y-outliers: none

Testing the Population Correlation Coefficient: This Java applet tests a claimed on a population's correlation coefficient value based on a set of random paired-observations.

7 H0: The population's correlation is about the claimed value. Ha: The population's correlation is quite different from the claimed value. I used the data in example 12.8. The results are:

Website Review:

The review for the website “Ordinary Least Squares”:

The site has a tool for calculating the least squares. I used the data in example 12.8 and got the following results:

QuickFit: Results The results of a QuickFit performed 5 data pairs (x,y): ( -1.00 , -1.00 ); ( 0.000E+00, 1.00 ); ( 1.00 , 1.00 ); ( 2.00 , 2.50 ); ( 3.00 , 3.50 ); y = a + bx where: a= 0.350 (Sa = 0.24 ) b= 1.05 (Sb = 8.86E-02 ) degrees of freedom = 3 r = 0.971 (p = 0.006)

8 It also has another tool that plots a graph which is very useful.