Solutions to Statistics Problems (Stat364) - Regression and Chi-Square Tests, Assignments of Data Analysis & Statistical Methods

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Stat364 Solution to HW#

PROBLEM 1 (12.5 ) Do Graph → Scatterplot to get part (b):

Then make the graphics window current by clicking within it. Do Editor → Select Item → x scale. Then do Editor → Edit x Scale and specify appropriate ticks for x axis:

Do the same for y axis tick marks:

a.

b.

c. Plot (a) does not provide a good graphics since it’s too crowded. Plot (b) does suggest a nonlinear relationship.

PROBLEM 2 (12.12 )

x ¯ =

= 36. 9286 , ¯y =

= 24. 7143 , Sxy = 25825 −

517 × 346

= 13047. 71 ,

Sxx = 39095 −

= 20002. 93 , Syy = 17454 −

= 8902. 857.

a. β^ ˆ 1 =^13047.^71

20002. 93 = 0. 6523 , βˆ 0 = 24. 7143 − 0. 6523 × 36 .9286 = 0. 6258.

b. y ˆ = 0.6258 + 0. 6523 × 35 = 23. 4563 , ˆe = y − yˆ = 21 − 23 .4563 = − 2. 4563.

c. SSE = 17454 − 0. 6258 × 346 − 0. 6523 × 25825 = 391. 8257.

d. Note that SST = Syy = 8902. 857.

r^2 = 1 −

= 0. 9560.

e.

x ¯ =

517 − 103 − 142

=

= 22. 67 , y¯ =

346 − 75 − 90

=

= 15. 0833.

Sxy = (25825 − 103 × 75 − 142 × 90) −

272 × 181

= 5320 −

272 × 181

= 1217. 33

Sxx = (39095 − 1032 − 1422 ) −

= 8322 −

= 2156. 67

Syy = (17454 − 752 − 902 ) −

= 3729 −

= 998. 9167

βˆ 1 ′ =^1217.^33

2156. 67 = 0. 5644 , βˆ 0 ′ = 15. 0833 − 0. 5644 × 22 .67 = 2. 2884.

(r′)^2 = 1 −

= 0. 6879.

PROBLEM 3 (12.14 ) Do Stat → Regression → Regression. Fill in response and predictor vari- ables and click Results button and select most detailed output.

Regression Analysis: EfficiencyRatio versus Temperature

The regression equation is

EfficiencyRatio = - 15.2 + 0.0942 Temperature <--- part (a)

Predictor Coef SE Coef T P Constant -15.245 3.977 -3.83 0. Temperature 0.09424 0.02215 4.25 0.

S = 0.497245 R-Sq = 45.1% R-Sq(adj) = 42.6% (Note: Part (d): 45.1%)

Analysis of Variance

Source DF SS MS F P Regression 1 4.4757 4.4757 18.10 0. Residual Error 22 5.4396 0. Total 23 9.

Obs Temperature EfficiencyRatio Fit SE Fit Residual St Resid 1 170 0.840 0.775 0.234 0.065 0. 2 172 1.310 0.964 0.195 0.346 0. (Some Output Omitted) (the next four lines are for parts (b) and (c)) 16 182 0.900 1.906 0.116 -1.006 -2.08R 17 182 1.810 1.906 0.116 -0.096 -0. 18 182 1.940 1.906 0.116 0.034 0. 19 182 2.680 1.906 0.116 0.774 1. (Some Output Omitted) 23 186 1.870 2.283 0.176 -0.413 -0. 24 188 3.080 2.471 0.214 0.609 1.

R denotes an observation with a large standardized residual.

PROBLEM 4 (12.68 ) The following MINITAB output answered parts (a), (b), and (d):

Regression Analysis: Y versus X

The regression equation is Y = 81.2 - 0.133 X <--- part (a)

Predictor Coef SE Coef T P Constant 81.1731 0.9259 87.67 0. X -0.13326 0.03176 -4.20 0.006 <--- part (b)

S = 1.20642 R-Sq = 74.6% R-Sq(adj) = 70.3% (Note: Test for slope is significant and R-Sq is moderate and hence the result is good.)

Analysis of Variance

Source DF SS MS F P Regression 1 25.622 25.622 17.60 0. Residual Error 6 8.733 1. Total 7 34.

Predicted Values for New Observations

New Obs Fit SE Fit 95% CI 95% PI 1 77.842 0.427 (76.796, 78.888) (74.710, 80.973)

Values of Predictors for New Observations

New Obs X 1 25.

(Part (d): CI above. Yes!) Answer to part (c): Yes.

PROBLEM 5 (12.75 ) The following MINITAB output are used:

Regression Analysis: Y versus X

The regression equation is Y = - 0.0826 + 0.0446 X

Predictor Coef SE Coef T P Constant -0.08259 0.04909 -1.68 0. X 0.044649 0.002237 19.96 0.

S = 0.0611166 R-Sq = 98.3% R-Sq(adj) = 98.0%

Analysis of Variance

Source DF SS MS F P Regression 1 1.4879 1.4879 398.35 0. Residual Error 7 0.0261 0. Total 8 1.

Obs X Y Fit SE Fit Residual St Resid 1 4.0 0.1200 0.0960 0.0411 0.0240 0. 2 8.7 0.2800 0.3058 0.0324 -0.0258 -0.

3 12.7 0.5500 0.4844 0.0261 0.0656 1. 4 19.1 0.6800 0.7702 0.0205 -0.0902 -1. 5 21.4 0.8500 0.8729 0.0206 -0.0229 -0. 6 24.6 1.0200 1.0158 0.0229 0.0042 0. 7 28.9 1.1500 1.2077 0.0285 -0.0577 -1. 8 29.8 1.3400 1.2479 0.0300 0.0921 1. 9 30.5 1.2900 1.2792 0.0311 0.0108 0.

Predicted Values for New Observations

New Obs Fit SE Fit 95% CI 95% PI 1 0.8104 0.0204 (0.7622, 0.8586) (0.6580, 0.9627)

Values of Predictors for New Observations

New Obs X 1 20.

a.

b. Y = − 0 .0826 + 0. 0446 X

c. 98 .3%. d. The fitted response value is 0.7702 with residual of − 0. 0902. e. Yes.

f. A 95% CI for the slope is

1. 044649 ±t. 025 , 7 × 0 .002237 = 0. 044649 ± 2. 365 × 0 .002237 = 0. 044649 ± 0 .005290 = (. 039359 , .049939).

g. The predicted mean response when x = 20 is 0.8104 with a 95% CI of (0.7622, 0.8586).

PROBLEM 6 (14.1 )

a. χ^2 = 12. 25 ≥ 9 .488 = χ^2. 05 , 4 , reject H 0. b. χ^2 = 8. 54 6 ≥ 11 .344 = χ^2. 01 , 3 , do not reject H 0. c. χ^2 = 4. 36 6 ≥ 4 .605 = χ^2. 10 , 2 , do not reject H 0. d. χ^2 = 10. 20 6 ≥ 15 .085 = χ^2. 01 , 5 , do not reject H 0.

PROBLEM 7 (14.4 ) H 0 : pi = 18 for all i = 1, 2 ,... , 8 versus all alternatives. Note that Ei = 120 × 18 = 15 for all i. Hence,

χ^2 =

[(12 − 15)^2 + (16 − 15)^2 + (17 − 15)^2 + (15 − 15)^2

+(13 − 15)^2 + (20 − 15)^2 + (17 − 15)^2 + (10 − 15)^2 ] = 4. 8 6 ≥ 12 .017 = χ^2. 10 , 7.

Do not reject H 0. The data does support the hypothesis.

PROBLEM 8 (14.25 ) The MINITAB output (edited) is used to answer the question. It is a test of homogeneity.

Chi-Square Test: L, LL, Y+YL, O, Others

Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts

L LL Y+YL O Others Total Long 409 11 22 7 277 726 449.66 7.32 17.58 8.79 242. 3.677 1.846 1.113 0.364 4.

Short 512 4 14 11 220 761 471.34 7.68 18.42 9.21 254. 3.508 1.761 1.062 0.347 4.

Total 921 15 36 18 497 1487

Chi-Sq = 23.179, DF = 4, P-Value = 0.

With the small P -value, the null hypothesis is rejected. The most noticeable chi-square contri- butions are for Mark Types of L and Others.

PROBLEM 9 (14.28 ) The MINITAB output (edited) is used to answer the question. It is a test of independence.

Chi-Square Test: NoResponse, WildRunning, ClonicSeizure, TonicSeizure

Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts

NoResponse WildRunning ClonicSeizure TonicSeizure Total Thienylalanine 21 7 24 44 96 24.06 9.99 21.56 40. 0.388 0.893 0.276 0.

Solvent 15 14 20 54 103 25.81 10.71 23.13 43. 4.528 1.008 0.424 2.

Sham 23 10 23 48 104 26.06 10.82 23.36 43. 0.360 0.062 0.005 0.

Unhandled 47 13 28 32 120 30.07 12.48 26.95 50. 9.531 0.021 0.041 6.

Total 106 44 95 178 423

Chi-Sq = 27.664, DF = 9, P-Value = 0.

With small P -value, the null hypothesis is rejected. The most noticeable chi-square contri- butions are maked above. There is a noticeable higher observed count of ‘No Response’ under ‘Unhandled’ treatment than would be expected if response were independent of the treatment.

PROBLEM 10 (14.31 ) The MINITAB output (edited) is used to answer the question. It is a test of independence.

Chi-Square Test: 0-<10, 10-<20, >=

Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts

0-<10 10-<20 >=20 Total

Subcompact 6 27 19 52 10.19 26.21 15. 1.724 0.024 0.

Compact 8 36 17 61 11.96 30.74 18. 1.309 0.899 0.

Midsize 21 45 33 99 19.40 49.90 29. 0.131 0.480 0.

Full-size 14 18 6 38 7.45 19.15 11. 5.764 0.069 2.

Total 49 126 75 250

Chi-Sq = 14.158, DF = 6, P-Value = 0.

Since the P -value is less than 0.05, the null hypothesis is rejected. The most noticeable chi- square contribution is marked above. The higher than expected observed count for Full-size cars that commute less than 10 miles contributes the most in the chi-square statistic that leads to the rejection of null hypothesis.

PROBLEM 11 (10.6 ) Data entry and analysis in MINITAB are described in the Solution Key to HW#4. The ANOVA table obtained from the computer output is given below:

Source DF SS MS F P formulation 3 509.1 169.7 10.85 0. Error 36 563.1 15. Total 39 1072. At 0.01 significance level, the complete null hypothesis (of equal treatment means) is rejected since P -value = 0. 000 < 0. 01.

The mean total Fe appears to differ for the four types of iron formation.

PROBLEM 12 (10.16 )

a. The variances may differ since σ ˆmax σ ˆmax

=

≈ 2. 14 > 2.

b. Suppose a level α = 0. 01 is desired. The complete null hypothesis (of equal treatment means) is rejected since P -value = 0. 000 < α. The means appear to differ.

c. The means, reported by using underscoring method, are

plate length 4 6 8 10 12 mean μ 4 μ 6 μ 8 μ 10 μ 12 estimate 333.21 368.06 375.13 407.36 437.