stat 322

Stat 322 – Review II Solutions

updated 3/5, 9:15pm

1) (a) What are the characteristics of a data set that involves ANOVA as the inference procedures?

Comparing several means or one categorical and one quantitative variable.

(b) Repeat (a) for regression.

Deciding if there is a relationship between two quantitative variables.

factor = explanatory variable = predictor; levels = the outcomes of the factor; treatment = factor level combination

response variable = the one we are trying to explain.

2) What is the relationship between S_xx and s_x²?

S_xx = S(x_i-)²

s_x² = Sxx/(n-1)

3) problem 3, p. 420. State the hypotheses in symbols and in words. Provide a complete ANOVA table.

H₀: m₁=m₂=m₃ (the mean output is the same for each brand)

H_a: at least one brand has a different mean output.

source	df	SS	MS	F
brand	I-1=2	591.2	591.2/2=295.6	F=295.6/237.4=1.30
error	23-2=21	4773.3	4773.3/21=227.3
total	24-1=23

Compare F=1.30 to F(.05, 2, 21)=3.47

Since our observed test statistic is smaller than the table value, our p-value < .05 and we fail to reject H₀ at the 5% level. In fact, F<F(.1,2,21) = 2.57 indicating that p-value > .10.

4) (a) problem 9 (p. 421) in Minitab. Remember to check and discuss the technical conditions for the validity of your procedure.

Analysis of Variance for thiamin

Source	DF	SS	MS	F	P
grain	3	8.983	2.994	3.96	0.023
Error	20	15.137	0.757
Total	23	24.120

Since the p-value < .05, we would reject the null hypothesis, and conclude there is a difference in the mean thiamin content at least two of the grains. Histogram of residuals looks bizarre, but normal probability plot is fine. Variances appear equal (1.04 vs. .669).

(b) Produce and discuss a visual display of these data.

Visually appears that barley and oats are above the other two.

(b) If the data are statistically significant, perform Tukey’s multiple comparison to see which means are different at the 5% (overall) level. Produce an underscoring graph to show the significant and insignificant differences.

Individual confidence level = 98.89%

type = Barley subtracted from:

type Lower Center Upper

Maize -2.5064 -1.1000 0.3064

Oats -1.0231 0.3833 1.7898

Wheat -2.2898 -0.8833 0.5231

type = Maize subtracted from:

type Lower Center Upper

Oats 0.0769 1.4833 2.8898

Wheat -1.1898 0.2167 1.6231

type = Oats subtracted from:

type Lower Center Upper

Wheat -2.6731 -1.2667 0.1398

This indicates that only maize and oats differ.

(c) What is the individual error rate used by Minitab. How does this compare to a Bonferroni correction? Which procedure is more “conservative”?

Individual error rate = 0.0111

A bonferroni correction would have used a/6 = .0083 making this procedure more conservative (more likely to fail to reject).

(d) Explain the consequences of making the individual error rate smaller.

Making the individual error rate smaller makes the confidence intervals larger and we are less likely to declare significant differences.

5) problem 3, p. 454

General Linear Model: C1 versus C2, C3

Factor Type Levels Values

C2 fixed 4 1(200), 2(400), 3(700), 4(1100)

C3 fixed 4 1(190), 2(250), 3(300), 4(400)

Analysis of Variance for C1, using Adjusted SS for Tests

Source DF Seq SS Adj SS Adj MS F P

C2 3 324082 324082 108027 105.31 0.000

C3 3 39934 39934 13311 12.98 0.001

Error 9 9232 9232 1026

Total 15 373248

With p-value ≈ 0 (F = 105.31), we reject the null hypothesis of no gas-rate effect and conclude that at least one gas rate does have a different mean heat transfer coefficient.

With a p-value = .001 < .01, we reject the null hypothesis of no liquid-rate effect and conclude that at least one liquid rate has a different mean heat transfer coefficient..

S = 32.0279 R-Sq = 97.53% R-Sq(adj) = 95.88%

Unusual Observations for C1

Obs C1 Fit SE Fit Residual St Resid

16 733.000 683.438 21.184 49.563 2.06 R

R denotes an observation with a large standardized residual.

Tukey Simultaneous Tests

Response Variable C1

All Pairwise Comparisons among Levels of C2

C2 = 1(200) subtracted from:

Difference SE of Adjusted

C2 of Means Difference T-Value P-Value

2(400) 93.50 22.65 4.129 0.0113

3(700) 209.25 22.65 9.240 0.0000

4(1100) 381.50 22.65 16.845 0.0000

C2 = 2(400) subtracted from:

Difference SE of Adjusted

C2 of Means Difference T-Value P-Value

3(700) 115.8 22.65 5.111 0.0029

4(1100) 288.0 22.65 12.717 0.0000

C2 = 3(700) subtracted from:

Difference SE of Adjusted

C2 of Means Difference T-Value P-Value

4(1100) 172.3 22.65 7.606 0.0002

This indicates that, at the 1% level, the mean heat transfer coefficient is the same for 200 and 400 but differ for the other levels (so we would underscore 200 and 400 and that’s it).

Tukey Simultaneous Tests

Response Variable C1

All Pairwise Comparisons among Levels of C3

C3 = 1(190) subtracted from:

Difference SE of Adjusted

C3 of Means Difference T-Value P-Value

2(250) 45.50 22.65 2.009 0.2535

3(300) 82.50 22.65 3.643 0.0229

4(400) 136.25 22.65 6.016 0.0009

C3 = 2(250) subtracted from:

Difference SE of Adjusted

C3 of Means Difference T-Value P-Value

3(300) 37.00 22.65 1.634 0.4085

4(400) 90.75 22.65 4.007 0.0134

C3 = 3(300) subtracted from:

Difference SE of Adjusted

C3 of Means Difference T-Value P-Value

4(400) 53.75 22.65 2.373 0.1522

This indicates that only 190 and 400 differ (we could just underscore 250, 300, and 400).

6) problem 14, p. 456

E(_i.- _..) = E(_i.) - E(_..)

E(_i.) = 1/J E(S_j m + a_i + b_j) =1/J( Jm + Ja_i + 0 ) = m + a_i

E(..)=1/IJE(S_iS_j m + a_i + b_j) = 1/IJ(IJ m + 0 + 0) = m

E(_i.) - E(_..) = m + a_i - m = a_i

7) problem 19, p. 464. Be very explicit about the hypothesis statements and check of technical conditions.

Analysis of Variance

Source DF SS MS F P

coal 2 1.00241 0.501206 29.49 0.000

NaOH 2 0.12431 0.062156 3.66 0.069

Interaction 4 0.01456 0.003639 0.21 0.924

Error 9 0.15295 0.016994

Total 17 1.29423

(a) With p-value = .924 we fail to reject H₀: no interaction between coal type and NaOH concentration in favor of H_a: there is an interaction.

Thus we can examine the main effects. With p-value =.000 we do reject H₀: m₁=m₂=m₃ in favor of H_a: the mean acidity does differ for at least one coal type. With p-value = .069 we fail to reject H₀: b₁=b₂=b₃=0 (at the .01 level) and conclude that the NaOH concentration doesn't make a difference.

Examining residuals plots: There is some concern about the technical conditions, though that is mainly due to two outliers.

(b) Could use the two-way ANOVA for (a) but here need to use General Linear Model

Tukey Simultaneous Tests

Response Variable acidity

All Pairwise Comparisons among Levels of Coal

coal = Maddingley subtracted from:

Difference SE of Adjusted

coal of Means Difference T-Value P-Value

morwell 0.2117 0.07526 2.812 0.0485

yallourn 0.5717 0.07526 7.595 0.0001

coal = morwell subtracted from:

Difference SE of Adjusted

C2 of Means Difference T-Value P-Value

yallourn 0.3600 0.07526 4.783 0.0026

If we use a=.01, then we have that coal type 1 and 2 differ, and coal types 2 and 3 differ.

8) problem 19, p. 518

(a) The regression equation is NOx-hat = - 45.6 + 1.71 burner

(b) = -45.6 + 1.71(225) = 339.15

(d) No, the value 500 is too far outside the range of x values used to construct the regression line.

9) problem 33, p. 528

(a) CI for b₁: .10748 + (t_25,.025=2.06) (.0128) = (.081, .134). We are 95% confident that the true average change in strength associated with a 1 Gpa increase in modulus of elasticity is between .081 MPa and .134 MPa.

(b) Since .01 is contained in the interval we know we would fail to reject a two-sided alternative at the 5% level. But since they said "at most" we really want a one-sided alternative.

H₀: b₁=.1 (at most .1)

H_a: b > .1 (more than .1)

t=(.10748-.1)/.0128 = .58, table A.5 gives p-value > .10, thus we fail to reject H₀, there is not enough evidence to contradict the prior belief.

10) problem 52, p. 537

The regression equation is y = 6.45 + 10.6 x

Predictor	Coef	SE Coef	T	P
Constant	6.449	2.795	2.31	0.054
x	10.6026	0.9985	10.62	0.000

S = 2.546 R-Sq = 94.2% R-Sq(adj) = 93.3%

Analysis of Variance

Source	DF	SS	MS	F	P
Regression	1	730.69	730.69	112.76	0.000
Residual	7	45.36	6.48
Total	8	776.06

(a) With a p-value of .000 (F = 112.76), we would strongly reject H₀: b₁ =0 indicating that the model does specify a useful relationship.

(b) Estimate for b₁: 10.6026 + (t_7,.025=2.365)(.9985) = (8.24, 12.96)

(c), (d) Estimate for (x=3):

New Obs Fit SE Fit 95.0% CI 95.0% PI

1 38.256 0.911 ( 36.100, 40.413) ( 31.858, 44.655)

(e) =2.67. Since 2.5 is closer to this will shrink the intervals.

(f) No, the value of 6.0 is not in the range of observed x values, therefore predicting at that point is meaningless.

11) problem 71, p. 550

(a) R-square from SAS output: .5073

(b) correlation coefficient = sqrt(.5073) = .7122

(d) = .787218 + .00757(50) = 1.166

+ (t_15,.025=2.131)(s=.20308)sqrt(1/17+(=42.31-50)²/16(26.38²))

1.166 + 2.131 (.20308) sqrt(1/17 + .0053)

1.166 + 2.131(.20308)(.253) = 1.166 + .1096 = (1.006, 1.276)

(e) = .787218 + .00757(30) = 1.014

observed = .80

residual = -.214

12) problem 67, p. 627

(a) For a one-minute increase in the 1-mile walk time, we would expect the VO2max to

decrease by .0996, while keeping the other predictor variables fixed.

(b) We would expect male to have an increase of .6566 in VO2max over females, while

keeping the other predictor variables fixed.

(d) R²= 1-SSE/SST = 1-30.1033/102.3922 = .706 or 70.6% of the observed variations in

VO2max can be attributed to the model relationship.

(e) H₀: b₁ = b₂ = b₃ = b₄ = 0

H_a: at least one b_i ≠ 0

F = (.706/4) / (1-.706)/15 = 9.005. This is greater than F_{.05, 4, 15} = 8.25 so we reject H₀ and conclude the model specifies a useful relationship between VO2max and at least one of the other predictors.

13) problem 70, p. 628, interpret the coefficient of x₃ as well.

(a) For the model excluding the interaction term, R² = 1- 5.18/8.55 = .394, or 39.4% of the observed variation in lift/drag ratio can be explained by the model without the interaction accounted for. However, including the interaction term increases the amount of variation in lift/drag ratio that can be explained by the model to R² = 1-3.07/8.55 = .641 or 64.1%.

(b) Without interaction, we are testing : H₀: b₁ = b₂ = 0 vs. Ha: either b₁ or b₂ ≠ 0.

The test statistic is F = R²/k/ (1-R²)/(n-k-1).

F = 1.95 < F_{.05, 2, 26} = 5.14, so fail to reject, model does not appear to be useful.

With the interaction term, we are testing H₀: b₁ = b₂ = b₃ = 0 vs. Ha: at least one b_i ≠ 0

F = (.641/3) / ((1-.641)/5 = 2.98, so again fail to reject.

Even with the interaction term, there is not enough of a significant relationship between lift/drag ratio and the two predictor variables to make the model useful (a bit of a surprise!)

14) problem 75, p. 629

(a) H₀: b₁ = b₂ = 0

H_a: at least one b_i ≠ 0

Analysis of Variance

Source DF SS MS F P

Regression 2 237.52 118.76 30.81 0.000

Residual Error 7 26.98 3.85

Total 9 264.50

Reject null hypothesis (F = 30.31, p-value < .001) and conclude quadratic model is useful.

(b)

Predictor Coef SE Coef T P

Constant 41.7422 0.8522 48.98 0.000

log added Mn 6.581 1.002 6.57 0.000

log added Mn sq -2.3621 0.3073 -7.69 0.000

The quadratic predictor is significant (t = -7.69, p-value < .001) and should be retained in the model.

(c)

New

Obs Fit SE Fit 90% CI 90% PI

1 45.961 1.031 (44.007, 47.915) (41.760, 50.162)

Values of Predictors for New Observations

log log

New added added

Obs Mn Mn sq

1 1.00 1.00

We are 90% confident that the expected height for wheat treated with 10 mM of Mn is between 44.007 cm and 47.915 cm.

45.961 + 1.895(1.031) where t_{.05, 7} = 1.895