QAM – II (Academic Year 2014-15) Solution of Problem Set – 3 (Instructor: Gaurav Garg)

1) Calculated Chi Square = 7.286535, Critical Chi Square at 1% level and 6 d.f. = 16.812; Accept H0. The percentage distribution of the U.S. labor force by education attainment has not changed since 1992.

2) H0: Given data follows Poisson distribution. H0: Given data does not follow Poisson distribution.

�̂�=�̅�=1.94 X f0 P(X=x)=

𝑒−1.94

𝑥!(1.94)𝑥 fe=150*P(X=x) (𝑓0− 𝑓𝑒)

2

𝑓𝑒

0 21 0.1437 21.556 0.014 1 37 0.2788 41.818 0.555 2 44 0.2704 40.563 0.291 3 30 0.1749 26.231 0.542 4 14 0.0848 12.722 0.128 ≥5 4 0.0474 7.110 1.361 Total 150 1 150 2.891

χ𝑐2=Σ

(𝑓0− 𝑓𝑒)2

𝑓𝑒= 2.891

From the tables of χ2 distribution with 6-1-1= 4 d.f. we have the critical value of 9.49 at 5% level of significance. Since χ𝑐

2 < 9.49 we cannot reject the null hypothesis and conclude that the given data follows Poisson distribution.

3) Calculated Chi Square = 0.86482, Critical Chi Square at 2.5% level and 2 d.f. = 7.378; Accept H0. The credit card holders in this state are not different from all credit card holders with respect to making payments on their credit card bills.

4) Calculated Chi Square = 6.75, Critical Chi Square at 5% level and 4 d.f. = 9.488; Accept H0. The null hypothesis that the orders are evenly distributed over the days of the week is true.

5) After Pooling Calculated Chi Square = 10.96. Test of Goodness of fit is considered as a right tailed test. So p-value = P(χ2

(7) > 10.96) = 1- 0.859627 =0.140373. Fix level of significance at 5%. H0: Data follows Poisson, H1: Data does not follow Poisson. Since p-value > 0.05, we accept H0 at 5% level.

6) Calculated Chi Square = 10.11719, Critical Chi Square at 5% level and 5 d.f. = 11.0705; Accept H0. The observed distribution of X may be defined by the probability 𝑔(𝑥) = 𝑝 (1 − 𝑝)𝑥−1, where p =1/2.

7) Calculated Chi Square = 19.5833, Critical Chi Square at 5% level and 2 d.f. = 5.991465; Reject H0. Data is not consistent with the company’s claim.

8) Observed Frequencies

Employed Unemployed Total

Male 1480 5720 7200

Female 120 680 800

Total 1600 6400 8000

Calculated Chi Square = 13.8889, Critical Chi Square at 5% level and 1 d.f. = 3.8415; Reject H0. We conclude that distinction is made in appointment on the basis of sex.

9) Calculated Chi Square = 128.448, Critical Chi Square at 5% level and 2 d.f. = 5.991465;

Reject H0. Three media sources do not have the same promotional impact on customers.

10) Calculated Chi Square = 68.66277, Critical Chi Square at 5% level and 3 d.f. = 7.814728; Reject H0. The percentage distribution of the opinions of the users of the redesigned product is different from the percentage distribution of the users of this product before it was redesigned.

11)

(a)

Source of Variation Sum of Squares

D.F. Mean Sum of Squares

F Calculated

F Critical (α = 0.05)

Advertizing Media (M) 13172.02 1 13172.02 1.419351 4.01954

Marketing Strategy (S) 98838.63 2 49419.32 5.32518 3.16825

Interaction (MxS) 1609.633 2 804.8167 0.086723 3.16825

Error 501137 54 9280.309

Total 614757 59

(b) Null Hypothesis that Interaction effects are not significant is accepted 5% level of significance. Advertizing media and marketing strategy are independent at 5% level of significance.

(c) The average weekly sales due to the three marketing strategies are significantly different at 5% level of significance?

(d) The average weekly sales are not significantly different due to the two media sources at 5% level of significance?

12) Correction: Use 5 observations per cell.

Source of Variation SS DF MS VR

5% Critical F

Network 145 2 72.5 0.420968 3.27 Accept H0

News Time 160 2 80 0.464516 3.27 Accept H0

Interaction 240 4 60 0.348387 2.65 Accept H0

Error 6200 36 172.2222

Total 6745 44

13) (a)

ANOVA

Source of Variation SS df MS F P-value F crit

Between Groups 0.520833 1 0.520833 0.163433 0.694526 4.964603

Within Groups 31.86833 10 3.186833

Total 32.38917 11

(b) Two types of shock absorbers are not significantly different in mean strength at 5% level of significance level.

(c)

t-Test: Two-Sample Assuming Equal Variances

Manufacturer Competitor

Mean 10.71667 10.3

Variance 3.069667 3.304

Observations 6 6

Pooled Variance 3.186833

Hypothesized Mean Difference 0

df 10

t Stat 0.404268

t Critical two-tail (5%) 2.228139

Accept null hypothesis of no difference.

(d) F =0.163433, t =0.404268, t2=0.163433

14) (a)

Source Sum of Squares df Mean Square F Sig.

SALES_TECHNIQUE 23.333 3 7.778 6.512 .026 Reject H0 at 5%

STORE_SIZE 136.167 2 68.083 57.000 .000

Error 7.167 6 1.194

Corrected Total 166.667 11

(b) Reject H0 of equality at 5%.

(c)

Difference Std. Error Sig.

95% Confidence Interval

Lower Bound Upper Bound

A & D -2.6667 .89235 .118 -6.0377 .7044

Accept H0 of No Difference. 15)

Source Sum of Squares df Mean Square F Sig. At 5%

INTERACTION 16.680 6 2.780 2.393 .093 NO DIFFERENCE

CAR 5.333 2 2.667 2.296 .143 NO DIFFERENCE

GASOLINE_ADD 85.280 3 28.427 24.471 .000 DIFFERENT

Error 13.940 12 1.162

Total 121.233 23

16) a) b) 3.1884

17) a) b) 67.39

18) The regression equation is Y=8.103 + 7.602 X1 + 3.111 X2

Predictor Coef SE Coef T Constant 8.103 2.667 3.038245 X1 7.602 2.105 3.611401 X2 3.111 0.613 5.075041 S=3.335 R-sq = 92.3% R-sq (adj) = 91.02% Analysis of Variance Table SOURCE DF SS MS F Regression 2 1612 806 71.9219 Residual Error 12 134.4789 11.2066 Total 14 1746.4789

(S is square root of MSE) a) See table. b) Critical Value F at (2, 12) d.f, 5% level of significance=3.8853, Reject H0. Model is valid.

Critical Value t at 12 d.f, 5% level of significance=2.17881, Both β1 and β2 are significant.

c) See table. d) See table.

19) The admissions officer for Clearwater College developed the following estimated regression equation relating final college GPA to the student’s SAT mathematics score and high-school GPA.

Ŷ= - 1.41+ .0235x1 + .00486x2, where x1 = high-school grade point average, x2 = SAT mathematics score, y = Final college grade point average. A portion of the Minitab computer output follows:

The regression equation is Y= -1.41 + .0235 X1 + .00486 X2

Predictor Coef SE Coef T Constant -1.4053 0.4848 -2.89872 X1 0.023467 0.008666 2.707939 X2 0.00486 0.001077 4.512535 S=0.1298 R-sq = 93.73% R-sq (adj) = 91.94% Analysis of Variance Table SOURCE DF SS MS F Regression 2 1.76209 0.881045 52.30528 Residual Error 7 0.11791 0.016844 Total 9 1.88 0.208889

a) See table. b) Critical Value F at (2, 7) d.f, 5% level of significance=4.7374, Reject H0. Model is

valid. c) R-sq (adj) is good enough. H0 in F-test is rejected. d) Critical Value t at 7 d.f, 5% level of significance=2.3646. H0:β1 = 0 and H: β2 = 0 are

rejected in favor of two sided alternatives.

20) The personnel director for Electronics Associates developed the following estimated regression equation relating an employee’s score on a njob satisfaction test to length of service and wage rate.

Ŷ= - 14.4- 8.69x1 + 13.5x2, where x1 = length of service (year), x2 = wage rate (dollars), y = job satisfaction test score (higher scores indicate greater job satisfaction). A portion of the Minitab computer output follows:

The regression equation is Y= 14.4 – 8.69 X1 +13.52 X2

Predictor Coef SE Coef T Constant 14.448 8.191 1.76 X1 -8.69 1.555 -5.58842 X2 13.517 2.085 6.482974 S=3.733 R-sq =90.12% R-sq (adj)= 86.16% Analysis of Variance Table SOURCE DF SS MS F Regression 2 648.83 324.4150 22.7916 Residual Error 5 71.17 14.2340 Total 7 720.00

a) See table. b) Critical Value F at (2, 5) d.f, 5% level of significance=5.7861, Reject H0. A significant

relationship is present. c) Yes. Adj R-sqis good enough. Linear Relationship is present. d) Critical Value t at 5 d.f, 5% level of significance=2.57058. H0:β1 = 0 and H: β2 = 0 are

rejected in favor of two sided alternatives.

21) a) Score = Ŷ= 50.6095 + 1.5621 RecRes. The H0 of F test rejected at the .05 level of

significance. Linear model is valid. b) R-Sq = 0.43, Adj R-Sq = 0.40. Model is not good enough. c) Score = Ŷ= 33.4848 + 1.8998 RecRes + 2.6108 Afford. The H0 of F test rejected at

the .05 level of significance. Linear model is valid. R-Sq = 0.81, Adj R-Sq = 0.78. Model is improved and good enough.

22) a) Speed = 97.570 + 0.069 Price + 0.000 Curb_Weight + 0.059 Horsepower – 2.484

zero_to_60. b) Using the F test, we determine that the linear model is valid. c) At 10% level of significance, Curb_Weight is insignificant, all other regressors are

significant. At 5% level of significance, Curb_Weight and Price are insignificant d) Speed = 95.384 + 0.073 Price + 0.057 Horsepower – 2.474 zero_to_60.

Notice that the values of coefficients do not change much.

23) a) CityMPG = 24.121 – 2.102 Displacement. All H0 under t test and F test are rejected

at 5% level of significance. Model is valid. R-Sq and Adj R-Sq are very small. b) CityMPG = 26.383 – 2.439 Displacement -1.201 Drive4. All H0 under t test and F

test are rejected at 5% level of significance. Model is valid. R-Sq and Adj R-Sq are improved a bit.

c) Dummy variable added in part (b) is significant at α = .05. d) CityMPG = 33.333 – 4.149 Displacement -1.237 Drive4 +2.161 EightCyl. R-Sq and

Adj R-Sq are improved. e) All H0 under t test and F test are rejected at 5% level of significance. Model is valid.

24) Today’s marketplace offers a wide choice to buyers of sport utility vehicles (SUVs) and pickup trucks. An important factor too many buyers is the resale value of the vehicle. The following table shows the resale value (%) after two years and the suggested retail price for 10 SUVs, 10 small pickup trucks, and 10 large pickup trucks (Kiplinger’s New Cars and Trucks 2000 Buyer’s Guide).

a) Resale_Value = 0.002Suggested_Price. All H0 under t test and F test are rejected at 5% level of significance. Model is valid. R-Sq and Adj R-Sq are close to 1.

b) Yes. All H0 under t test and F test are rejected at 5% level of significance. Model is valid. R-Sq and Adj R-Sq are close to 1.

c) Resale_Value = 0.002 Suggested_Price + 2.516 Type. [Type = 0, 1 and 2 for Sport Utility, Small Pickup and Full-size Pickup respectively.]

d) Adj R-Sq is not improved. F test suggests that model is valid. Using t test, we find that Type is not significant.

25) A 10-year study conducted by the American Heart Association collected data on how age, blood pressure, and smoking relate to the risk of strokes. Risk is interpreted as the probability (times 100) that the patient will have a stroke over the next 10-year period. A linear model is fitted to predict the Risk of strokes (Risk) using age (Age) and blood pressure (BP). The output of SPSS is as follows:

Model Summary

Model R R Square

Adjusted R

Square

Std. Error of

the Estimate

1 .898a .806 0.783 6.90826

a. Predictors: (Constant), BP, Age

ANOVAb

Model

Sum of

Squares df Mean Square F

1 Regression 3377.9060 2 1688.9529 35.3144

Residual 813.0443 17 47.8261

Total 4190.9500 19 220.5763

a. Predictors: (Constant), BP, Age

b. Dependent Variable: Risk

Coefficientsa

Model

Unstandardized Coefficients

t B Std. Error

1 (Constant) -110.942 16.470 -6.7360

Age 1.315 .173 7.6012

BP .296 .051 5.8039

a. Dependent Variable: Risk

(a) See Table.

(b) At 5% level of significance, Model is valid (Using F test) Critical Value of F at (2,17) d.f. and

5% level of significance is 3.5915.

(c) At 5% level of significance, and 17 d.f., t = 2.10982. All regressors are significant.

(d) Added variable is significant at 5% level of significance. Adj R-Sq is improved a bit. Model is

improved.

26) a)

Model F Statistic Critical Value

I 15.64103 4.964603 Reject H0

II 20.58104 4.964603 Reject H0

III 9.740506 4.256495 Reject H0

b) Test the significance of individual regression coefficients in all three models.

Model Coefficients T Statistic Critical Value

I β0 β1

3.399639

3.952562

2.2281 2.2281

Significant Significant

II β0 β2

0.587002

4.537125

2.2281 2.2281

Insignificant Significant

III β0 β1 β2

0.77078

0.558036

1.455102

2.2622 2.2622 2.2622

Insignificant Insignificant Insignificant

c) Model II is the best. It has highest Adjusted R-Sq. It could be improved by removing

intercept.

d) Normality is required for the validity of F and t tests.

27) Predicted Y = 3.239 X2 (No intercept model will be better as intercept comes to be insignificant.) Inclusion of both regressors creates multicollinearity. Regressor X2 is included as it has higher correlation with Y.

28) Predicted Y = 1.708 X1 +0.016 X2 Adj R-Sq is close to 1. MSE is small. F-test suggests model is valid. t test suggests both regressors are significant. No eigen value is close to zero. Only VIFs are more than 5. If we use stepwise method for variable elimination, it eliminates X2. But Adj R-Sq is also reduced a bit. MSE is also increased. This means that the multicollinearity is not a serious problem here. And a no intercept model taking both regressors is a better model.

29) When we fit a linear regression model taking only X1 as regressor Intercept model has very small R-Sq and Adj R-Sq. In F test and t-test, we conclude that X1 is not significant. No-intercept model looks much better. R-Sq and Adj R-Sq are much higher. MSE is reduced. In F-test and t-test, we conclude that X1 is significant. So this model is much better. When we take X1 and X2 both in the model, R-Sq and Adj R-Sq are increased. MSE is reduced. In F-test and t-test, we conclude that X1 and X2 are significant. So a no-intercept model taking X1 and X2 both is the best. Predicted Y = 0.019 X1 +17.456 X2 Adj R-Sq=0.903