gm533 week 6 lecture april 2012

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Week 6 Checkpoint Help GM533 B Heard (Don’t copy or post without my permission, students may download a copy for personal use)

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Week 6 Charts for GM 533

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Page 1: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533

B Heard(Don’t copy or post without my permission, students may download a copy for personal use)

Page 2: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533•Always be on the lookout for typos…..•Remember I usually do these charts “on

the fly”▫Do not rely solely on me or these charts to

do well in this class!▫The live lectures are only part of what you

have available to you▫Good Luck!

Page 3: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM53313.8 THE REAL ESTATE SALES PRICE

CASE •A real estate agency collects data

concerning y = the sales price of a house (in thousands of dollars), and x = the home size (in hundreds of square feet). The data are given in the table below. The MINITAB output from fitting a least squares regression line to the data is on the next page.

Page 4: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533

a) By using the formulas illustrated in Example 13.2 (see page 497) and the data provided, verify that (within rounding) b0 = 48.02 and b1 = 5.700, as shown on the MINITAB output.

I put the data in Excel and did the math there. The formulas are provided in the text.

Page 5: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533

Size (x) Price (y)

23 180

11 98.1

20 173.1

17 136.5

15 141

21 165.9

24 193.5

13 127.8

19 163.5

25 172.5

Page 6: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533Size (x) Price (y) xy x^2

23 180 4140 52911 98.1 1079.1 12120 173.1 3462 40017 136.5 2320.5 28915 141 2115 22521 165.9 3483.9 44124 193.5 4644 57613 127.8 1661.4 16919 163.5 3106.5 36125 172.5 4312.5 625

Sum of x's Sum of y's sum of xy's sum of x^2's (sum of x's)^2188 1551.9 30324.9 3736 35344

n10

SSxy SSxx1149.18 201.6

b1 y bar x bar5.700298 155.19 18.8

b048.0244

y hat = b0 + b1*xy hat = 48.0244 plus 5.70029762 x

Page 7: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533b) Interpret the meanings of b0 and b1. Does the

interpretation of b0 make practical sense?

The b1 is 5.70 which basically is saying for every 100 square feet the average sales price increases that much

b0 is the y intercept when x is zero. In other words, it says that a house with zero square feet would cost about 48 thousand dollars. No, this doesn’t make sense (I will talk about this).

Page 8: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533c) Write the least squares prediction

equation.

y hat = 48.02 + 5.7xy hat = b0 + b1*x

y hat = 48.0244 plus 5.70029762 x

Page 9: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533d) Use the least squares line to obtain a

point estimate of the mean sales price of all houses having 2,000 square feet and a point prediction of the sales price of an individual house having 2,000 square feet.

Plug and chug (insert 20 for x remember the size was in 100’s of square feet)

y hat = 48.02 + 5.7(20) = 162.02 (in thousands)

So the predicted price would be $162,020

Page 10: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM53313.21 THE STARTING SALARY CASEThe MINITAB output of a simple linear

regression analysis of the data set for this case (see Exercise 13.4 on page 501) is given in Figure 13.11. Recall that a labeled MINITAB regression output is on page 509.

Page 11: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533

Page 12: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533

bo (Part a)b1

Page 13: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533

SSE (Part b)s^2

s

Page 14: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533

sb1 (Part c)t

t = b1/sb1 (show)

Page 15: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533

df (Part d)

Page 16: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533

Table fromhttp://www.statsoft.com/textbook/distribution-tables/#t(I just search on Internet, you have one in text)

(Part d continued)t.025 = 2.57 compared to 14.44 ?

Reject because it’s way out there in the rejection region Reject H0, there is strong evidence of something going on with x and y

Page 17: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533

Table fromhttp://www.statsoft.com/textbook/distribution-tables/#t(I just search on Internet, you have one in text)

(Part e)t.005 = 4.03 compared to 14.44 ?

Reject because it’s still way out there in the rejection region Reject H0, there is strong evidence of something going on with x and y (Very strong relationship)

Page 18: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533f) p value was .000 agrees with previous

two to reject at all alphas. Very very strong evidence of an x and y relationship

g) 95% CI Just use what you have found above

The interval is b1 +/- t.025 sb1h) 99% CI Just use what you have found

aboveThe interval is b1 +/- t.005 sb1

Page 19: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533

sbo (Part i)t

t = b0/sb0 (show)

bo

Page 20: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533j) p value was .000, reject at all alphas. Very very

strong evidence of an x and y relationshipk) Use the formulas and the data! (should give

you the same answer you got in part c for sb1 and in part i for sbo.

Page 21: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM53313.30 THE FUEL CONSUMPTION CASEThe following partial MINITAB regression

output for the fuel consumption data relates to predicting the city’s fuel consumption (in MMcf of natural gas) in a week that has an average hourly temperature of 40°F.

Page 22: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533

Page 23: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533

Part a

Page 24: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533

Part b

Page 25: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533c) Remembering that s = .6542; SSxx =

1,404.355; n = 8, hand calculate the distance value when x0 = 40. Remembering that the distance value equals , use s and from the computer output to calculate (within rounding) the distance value using this formula. Note that, because MINITAB rounds sy, the first hand calculation is the more accurate calculation of the distance value.

Page 26: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533Distance Value (dv) = 1/8 + (40-43.98)2 /

1404.355 = 0.1363

AndDistance Value (dv) = (0.241 / 0.6542)2 =

0.1357

Page 27: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533d) Remembering that for the fuel

consumption data b0 = 15.84 and b1 = -.1279, calculate (within rounding) the confidence interval of part a and the prediction interval of part b.

CI: 15.84 - 0.1279 (40) ± 2.447*0.6542*√(0.1363) = [10.13299, 11.31501]

For PI, just substitute 1.1363 for 0.1363

Page 28: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533e) Remember you are predicting for one

day, so use prediction interval. Since 9.01 < 9.595 and 12.43 > 11.847 thecity cannot be ____ % confident it won’t pay

a fine. (Fill in the blank)

Page 29: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533

THE FRESH DETERGENT CASE In Exercises 13.50 through 13.55, we give MINITAB and Excel outputs of simple linear

regression analyses of the data sets related to six previously discussed case studies. Using the appropriate computer output,

a Use the explained variation and the unexplained variation as given on the computer output to calculate (within rounding) the F (model) statistic.

b Utilize the F (model) statistic and the appropriate critical value to test H0 : β1 = 0 versus Hα : α1 ≠ 0 by setting a equal to .05. What do you conclude about the regression relationship between y and x?

c Utilize the F (model) statistic and the appropriate critical value to test H0 : β1 = 0 versus Hα : β1 ≠ 0 by setting a equal to .01. What do you conclude about the regression relationship between y and x?

d Find the p -value related to F (model) on the computer output and report its value. Using the p -value, test the significance of the regression model at the .10, .05, .01, and .001 levels of significance. What do you conclude?

e Show that the F (model) statistic is (within rounding) the square of the t statistic for testing H0 : β1 = 0 versus Hα : b1 ≠ 0. Also, show that the F.05 critical value is the square of the t025 critical value.

Note that in the lower right hand corner of each output we give (in parentheses) the number of observations, n, used to perform the regression analysis and the t statistic for testing H0 : β1 = 0 versus Hα : β1 ≠ 0.

Page 30: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533

Part a

Page 31: Gm533 week 6 lecture april 2012

Used a table at http://www.statsoft.com/textbook/distribution-tables/#f (I was lazy)

Page 32: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533F.05 =4.196, reject H0 (df1 (top) = 1, df2

(left) = 28). Looks like there is STRONG evidence of a significant relationship between x and y.

Page 33: Gm533 week 6 lecture april 2012

Used a table at http://www.statsoft.com/textbook/distribution-tables/#f (I was lazy)

c) F.01 =7.636, reject H0 (df1 (top) = 1, df2 (left) = 28). Looks like there is STRONG (Very because of .01) evidence of a significant relationship between x and y.

Page 34: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533d) p-value is ______ (smaller than all levels

of significance), reject H0 . Again, pick your “ly” ending word that means there is definitely strong evidence of a significant relationship between x and y.

Page 35: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533

Part eSquare this number, you should see that it gives you a result within rounding error.Then get your table value for t.025 and verify(t.025)2 = 4.19 = F.05

Page 36: Gm533 week 6 lecture april 2012

Week 6 Checkpoint Help GM533I will post these in the Stat Cave atwww.facebook.com/statcave