STA 6166 – Exam 2 – Fall 2006
PRINT Name _________________
A researcher is comparing the breaking strenghts of two types of aluminum tubes. She
samples 12 tubes of each type and measures pressure necessary to bend the tubes. Labelling
1 as the true mean for all tubes of type 1, and 2 as the true mean for all tubes of type 2,
conduct the test of H0: 1-2 = 0 vs HA: 1-2 ≠ 0 at the  = 0.05 significance level, based
on the following output (assume population variances are equal):
Independent Samples Test
Levene's Test for
Equality of Variances
F
y
Equal variances
as sumed
Equal variances
not ass umed
1.534
Sig.
.229
t-test for Equality of Means
t
df
Sig. (2-tailed)
Mean
Difference
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower
Upper
8.574
22
.000
8.87316
1.03485
6.72702
11.01930
8.574
21.174
.000
8.87316
1.03485
6.72216
11.02416
Test Statistic tobs = 8.574
Reject H0 if the test statistic falls in the range(s)
< -2.0739
or > 2.0739
P-value 2P(t≥ 8.574)  0
Conclude (Circle One)
Do Not Conclude 1-2 ≠ 0
1-2 <0
1-2 > 0
A study is conducted to compare 4 formulations of a new drug in terms of the availability of
the drug in the bloodstream over time. Ten healthy subjects are selected and each subject
receives each drug in random order in a randomized block design. The researcher conducts
the appropriate F-test for testing for formulation differences. If the test is conducted at the
=0.05 significance level, he will conclude formulation differences exist if the F-statistic
falls in what range? Fobs ≥ F.05,3,27 = 2.960
A study is conducted to compare golf ball distances among 4 brands of golf balls. A
mechanical driver is set up and hits 12 balls of each brand, and the distance travelled (meters)
is measured. The F-test determines that there are differences among the brands. A follow-up
comparison based on Tukey’s method yields the following table:
Multiple Comparisons
Dependent Variable: y
Tukey HSD
(I) group
1
2
3
4
(J) group
2
3
4
1
3
4
1
2
4
1
2
3
Mean
Difference
(I-J)
St d. E rror
-21.40796*
2.51529
-.00121
2.51529
-12.43549*
2.51529
21.40796*
2.51529
21.40675*
2.51529
8.97247*
2.51529
.00121
2.51529
-21.40675*
2.51529
-12.43428*
2.51529
12.43549*
2.51529
-8. 97247*
2.51529
12.43428*
2.51529
Sig.
.000
1.000
.000
.000
.000
.005
1.000
.000
.000
.000
.005
.000
95% Confidenc e Int erval
Lower Bound Upper Bound
-28.1238
-14.6921
-6. 7171
6.7146
-19.1513
-5. 7196
14.6921
28.1238
14.6909
28.1226
2.2566
15.6883
-6. 7146
6.7171
-28.1226
-14.6909
-19.1501
-5. 7184
5.7196
19.1513
-15.6883
-2. 2566
5.7184
19.1501
*. The mean difference is significant at the .05 level.
Clearly state what can be said of all pairs of brands at the 0.05 experimentwise error rate.
Brands:
1 vs 2:

1 vs 3:

1 vs 4:

2 vs 3:

2 vs 4:

3 vs 4:

N.S.D.
A study is conducted to compare whether incidence of muscle aches differs among athletes
exposed to 5 types of pain medication. A total of 500 people who are members of a large
fitness center are randomly assigned to one of the medications. After a lengthy workout, each
is given a survey to determine presence/absence of muscle pain. For the 5 groups: 25, 38, 32,
40, and 35 are classified as having muscle pain, respectively. The following output gives the
results for the Pearson chi-square statistic for testing (=0.05):
H0: True incidence rate of muscle pain doesn’t differ among medications
HA: Incidence rates are not all equal
Chi-Square Tests
Pearson Chi-Square
Value
6.150a
df
4
Asymp. Sig.
(2-sided)
.188
Test Statistic 2obs = 6.150
Reject H0 if the test statistic falls in the range(s)
≥ 9.488
P-value .188
Conclude (Circle One):
Medication effects not all equal
No differences in effects
Give the expected number of incidences of muscle pain for each medication under H0:
Overall incidence rate: (25+38+32+40+35)/500 = 0.34
Expected incidences per med = .34(100) = 34
In a 2-Factor ANOVA, measuring the effects of 2 factors (A and B) on a response (y), there
are 3 levels each for factors A and B, and 4 replications per treatment combination. Give the
values of of the F-statistic for the AB interaction for which we will conclude the effects of
Factor A levels depend on Factor B levels and vice versa (=0.05): FAB ≥ F.05,4,27 = 2.728
A simple linear regression model is fit, relating plant growth over 1 year (y) to amount of
fertilizer provided (x). Twenty five plants are selected, 5 each assigned to each of the
fertilizer levels (12, 15, 18, 21, 24). The results of the model fit are given below:
Coefficientsa
Unstandardized
Coefficients
Model
1
(Constant)
B
8.624
Std. Error
1.810
t
4.764
Sig.
.000
.527
.098
5.386
.000
x
a. Dependent Variable: y
Can we conclude that there is an association between fertilizer and plant growth at the 0.05
significance level? Why (be very specific).
Yes, P-value for test of H0: 1 = 0 versus HA: 1 ≠ 0 is .000 (tobs = 5.386)
Give the estimated mean growth among plants receiving 20 units of fertilizer.
^
Y  8.624  0.527(20)  19.164
The estimated standard error of the estimated mean at 20 units is 2.1
Give a 95% CI for the mean at 20 units of fertilizer.
t.025,23 = 2.0687
CI: 19.164 ± 2.0687(0.46) ≡ (18.212 , 20.116)
1 (20  18) 2

 0.46
25
450
A multiple regression model is fit, relating salary (Y) to the following predictor variables:
experience (X1, in years), accounts in charge of (X2) and gender (X3=1 if female, 0 if male).
The following ANOVA table and output gives the results for fitting the model. Conduct all
tests at the 0.05 significance level:
Y = 0 + 1X1 + 2X2 + 3X3 + 
ANOVA
df
3
21
24
Regression
Residual
Total
Intercept
experience
accounts
gender
Coefficients
39.58
3.61
-0.28
-3.92
SS
2470.4
224.7
2695.1
Standard
Error
1.89
0.36
0.36
1.48
MS
823.5
10.7
t Stat
21.00
10.04
-0.79
-2.65
F
76.9
P-value
.0000
P-value
0.0000
0.0000
0.4389
0.0149
Test whether salary is associated with any of the predictor variables:
H0: HA: Not all i = 0 (i=1,2,3)
Test Statistic Fobs = 76.9
Reject H0 if the test statistic falls in the range(s) ≥ 3.072
P-value .0000
Conclude (Circle One) Reject H0
Set-up the predicted value (all numbers, no symbols) for a male employee with 4 years of
experience and 2 accounts.
^
Y  39.58  3.61(4)  0.28(2)  3.92(0)  53.46
The following tables give the results for the full model, as well as a reduced model,
containing only expereience.
Test H0: 2 = 3 = 0 vs HA: 2 and/or 3 ≠ 0
Complete Model: Y = 0 + 1X1 + 2X2 + 3X3 + 
ANOVA
df
3
21
24
Regression
Residual
Total
SS
2470.4
224.7
2695.1
MS
823.5
10.7
F
76.9
Reduced Model: Y = 0 + 1X1 + 

Regression
Residual
Total
df
1
23
24
SS
2394.9
300.2
2695.1
MS
2394.9
13.1
F
183.5
P-value
0.0000
Test Statistic:
Fobs 
(2470.4  2394.9) / 2
 3.53
224.7 / 21
Rejection Region: Fobs ≥ 3.467
Conclude (Circle one):
Reject H0
Fail to Reject H0
P-value
.0000