An Aggie Does Not Lie, Cheat, Or Steal Or Tolerate Those Who Do s1

“An Aggie does not lie, cheat, or steal or tolerate those who do.”

Answer Key for Homework 10 (Due date: April 27, 2005, Wednesday in class)

 If the  is not given numerically in any question use 0.05.

1. (10 pt.) Exercise 9.49 (Hint: Use the following MINITAB output to answer. Make sure to write the hypothesis, indicate the test statistics, make a decision specifying the P-value or indicating the critical value, then write your conclusion) Test and CI for Two Proportions X: number returned Sample X n Sample p 1 (Plain) 104 207 0.502415 2 (Skydiver) 109 213 0.511737

Estimate for p(1) - p(2): -0.00932163 95% CI for p(1) - p(2): (-0.104954, 0.0863110) Test for p(1) - p(2) = 0 (vs not = 0): Z = -0.19 P-Value = 0.848

H 0 : p1  p2  0 versus H a : p1  p2  0 Test statistics is z=-0.19 The P-value=P(z<-0.19)=0.848/2

Since the P-value > =0.10, fail to reject H0. The return rate is not lower for the Plain cover

2. Exercise 9.72 (Hint: Use the given output to answer each part, make sure to write the hypothesis, indicate the test statistics, make a decision specifying the P-value or indicating the critical value, then write your conclusion) Notice that this really is a dependent data (THANKS TO STEVEN POINTING THIS OUT) but the MINITAB outputs are given assuming independent samples (MY MISTAKE) and you are expected to answer based on the given info. (a) (10 pt.) Answer the question given in the exercise (b) (8 pt) Test for equal true variances.

Two-sample T for Commutator vs Pinion n Mean StDev SE Mean (1) Commutator 17 259.9 31.3 7.6 (2) Pinion 17 264.1 27.4 6.6

Difference = mu Commutator - mu Pinion Estimate for difference: -4.2

Assuming unequal variances, 95% CI for difference: (-24.7, 16.4) T-Test of difference = 0 (vs not =): T-Value = -0.41 P-Value = 0.682 DF = 31

Assuming equal variances, 95% CI for difference: (-24.7, 16.4) T-Test of difference = 0 (vs not =): T-Value = -0.41 P-Value = 0.682 DF = 32 Both use Pooled StDev = 29.4

F-Test (normal distribution) Test Statistic: 1.302 P-Value : 0.603

Levene's Test (any continuous distribution) Test Statistic: 0.127 P-Value : 0.724

(a) 95% confidence interval for 1-2 is (-24.7, 16.4) assuming unequal true variances

95% confidence interval for 1-2 is (-24.7, 16.4) assuming true variances

Testing H 0 : 1   2  0 versus H a : 1  2  0 . 0 does fall between the limits of the confidence interval. Population mean penetration does not differ for two types of bearings

2 2 2 2 (b) H 0 : 1   2 versus H a : 1   2 The test statistics is F=1.302 and the P-value=0.603. Since the P-value is very large, fail to reject H0. There are no differences between the true variances.

3. Exercise 10.6 (Hint: Use the given output to answer the question) “An Aggie does not lie, cheat, or steal or tolerate those who do.”

(a) (10 pt.) Fill in the blanks on the ANOVA table (b) (8 pt.) You are answering if there are differences between the true means of four types of iron formations. Make sure to write the hypothesis, indicate the test statistics, make a decision specifying the P-value or indicating the critical value, then write your conclusion. (c) (4 pt.) If you reject the null hypothesis, make sure to look at the multiple comparison test and identify the different means) (d) (4 pt.) Is the constant variance assumption satisfied for ANOVA. Make sure to write the hypothesis, indicate the test statistics, make a decision specifying the P-value or indicating the critical value, then write your conclusion.

Analysis of Variance for Fe Source DF SS MS F P formations ? 509.1 ? ? 0.000 Error ? 563.1 ? Total 39 1072.3

Level N Mean StDev 1 10 26.080 3.391 2 10 24.690 4.425 3 10 29.950 2.854 4 10 33.840 4.831 Pooled StDev = 3.955

Tukey's pairwise comparisons Family error rate = 0.0100 Individual error rate = 0.00194 Critical value = 4.73

Intervals for (column level mean) - (row level mean) 1 2 3 2 -4.526 7.306

3 -9.786 -11.176 2.046 0.656

4 -13.676 -15.066 -9.806 -1.844 -3.234 2.026

Bartlett's Test (normal distribution) Test Statistic: 2.884 P-Value : 0.410

Levene's Test (any continuous distribution) Test Statistic: 0.444 P-Value : 0.723

(a) Analysis of Variance for Fe Source DF SS MS F P formations 3 509.1 169.7 10.85 0.000 Error 36 563.1 15.6 Total 39 1072.3

(b) H 0 : 1   2  3   4 versus H a : at least one i   j , i,j=1,2,3,4 Test statistics: F=10.85 and the P-value=0

Since the P-value < =0.05 then reject H0. There are differences between the true means of four types of iron formations

(c) According to multiple comparison test,

1  2 , 1  3 , 1  4 , 2  3 , 2  4 , 3  4 Formations 1 and 2 differ from formation 4.

(d) H 0 :constant variance versus H a :nonconstant variance The test statistics is F=2.884 and the P-value=0.410. Since the P-value is very large, fail to reject H0. “An Aggie does not lie, cheat, or steal or tolerate those who do.”

There are no differences between the true variances. Constant variance assumption is satisfied.

4. (16 pt.) Exercise 12.32 (Hint: Use the given output to answer the question) The model has the t-ratio=22.64 with the P-value=0.000 to test the null hypothesis of slope being zero (no linear relationship) which leads us to reject the hypothesis of no linear relationship because the P-value will be less than any . There is a useful linear relationship between rainfall and runoff.

Assume =0.05. The 95% confidence interval for the slope is estimated slope  critical value (standard error for the

estimated slope)=0.8267  2.16(0.03652) = (0.7478 , 0.9056) where the critical value is t/2;n-1= t0.025;15-2 = 2.16, the estimated slope and the standard error for the slope can be found on the output.

5. (25 pt.) Exercise 12.62 (Hint: Use the given graph and output to answer the question) Additional to parts (a) and (b), answer (c) Does a scatter plot of this data support the use of simple linear regression? Answer it looking at the graph. (d) Does it appear that there is a useful linear relationship between these two variables? To answer, make sure to write the hypothesis, indicate the test statistics, make a decision specifying the P-value or indicating the critical value, then write your conclusion.

Pearson correlation of x and y = 0.449 P-Value = 0.107

The regression equation is y = 0.100 + 1.89 x

Predictor Coef SE Coef T P Constant 0.1005 0.2667 0.38 0.713 x 1.894 1.088 1.74 0.107

S = 0.1906 R-Sq = 20.2% R-Sq(adj) = 13.5%

Analysis of Variance Source DF SS MS F P Regression 1 0.11012 0.11012 3.03 0.107 Residual Error 12 0.43581 0.03632 Total 13 0.54592

Unusual Observations Obs x y Fit SE Fit Residual St Resid 14 0.370 0.7500 0.8013 0.1496 -0.0513 -0.43 X

X denotes an observation whose X value gives it large influence.

(a) H 0 :   0 versus H 0 :   0 0.449 15  2 The test statistics, t  =1.74 with n-2=14-2=12 degrees of freedom. 1 0.4492

H0 is rejected if |t| > t α/2;n-2= t 0.025;12=2.179. Since 1.74 < 2.179, fail to reject H0. Conclude that the population correlation is zero.

OR the p-value is given on the output as 0.10 which leads to the same conclusion.

(b) That is the definition of R2. In simple linear regression, R2=r2=0.4492=0.2016=20.16%. OR again it is given on the output.

(c) No it does not.

(d) H 0 : 1  0 (no useful linear relationship) versus H a : 1  0 (useful linear relationship) The test statistics is F=3.03 if you look at the Analysis of variance table t=1.74 if you look at the table with predictors Since P-value=0.107 > =0.05 in both tables, fail to reject H0 There is no useful linear relationship “An Aggie does not lie, cheat, or steal or tolerate those who do.”

0.6 y 0.5

0.2 0.18 0.28 0.38 x