SMAM 314
Exam 3n
Name_________________
1. Mark the following statements True T or False F. (6 points -2 each)
_T__A. A paired t test is appropriate for comparison of scores on a standardized English
test before and after a course in remedial English.
_F__B. A 95 % confidence interval for the difference of two means is (.042, .501). You
would fail to reject the null hypothesis H 0 µ1 = µ 2 in favor of the alternative hypothesis
H1 µ1 ! µ 2 at α=.05.
_T__C. A process is out of control if at a particular point in time the reading is more than
3 standard deviations from the historical average.
2. Pick the correct choice. (4 points-2 each)
__d__A. You are doing a t test on independent random samples. Each of the two samples
has 10 observations. It is assumed that both populations have the same standard
deviation. The correct number of degrees of freedom is a. 10 b.9 c.20 d. 18 e. 19.
__e_B..A correlation coefficient of - 0.92 means that a. there is a weak positive linear
association between two variables b. there is a strong positive linear association between
two variables c. the regression line has a positive slope d. there is a weak negative linear
association between two variables e. there is a strong negative linear association
between the two variables.
3. A. Spurrier and Thombs (1990) gave data for 29 consecutive measurements on the
inner diameter of a landing gear triunion. The target for the diameters is 98.00.The
historical standard deviations is 4. At each interval of time they take the mean of four
observations. What is the upper and lower control limits? (8 points)
UCL = 98 + 3(4 / 4) = 104
LCL = 98 ! 3(4 / 4) = 92
B. Consider n observations of subgroups with four observations each. x = 19.01, R = 0.4 .
What is the center line and the upper and lower control limits for
(1) An Xbar chart (6 points)
CL=19.01
UCL=19.01+(0.729)(.4) =19.3016
LCL=19.01-(0.729)(.4) = 18.8083
(2)A range chart? (6 points)
CL=0.4
UCL=2.282(.4)=.9128
LCL =0
4. The data below resulted from an experiment in which weld diameter x and shear
strength y in pounds were determined for five different spots on steel
x 200.1 210.1 220.1 230.1 240.0
y 813.7 785.3 960.4 1118.0 1076.2
A. Use your calculator to fit the regression line. Regression equation =
__y=8.599x-941.7_______________ (5 points)
B. What percentage of the variation is accounted for by the regression? (1 point)
81.9%
C. What is the correlation coefficient? (1 point)
R= .904
D. Predict y when x = 225. (5 points)
Y =8.599(225) -941.7 =993.075
E. The standard error of the slope is 2.337. A confidence interval for the slope would be
b ± t !/2,n"2se(b)
(1)Find a 98% confidence interval for b.(5 points)
8.599 ± 4.541(2.337)
8.599 ± 10.6123
(!2.0223,19.2113)
(2) Based only on the confidence interval would you reject H0 slope=0 in favor of H1
slope≠0 at α=.02. Explain you answer. (3 points)
You would fail to reject H0. The confidence interval contains zero.
5. A large automobile manufacturing company is trying to decide whether to purchase
brand A or brand B tires for its new models. To help arrive at a decision an experiment is
conducted using 12 of each brand. The tires are run until they wear out. The summary
statistics are
Brand A: x1 = 37,900,s1 = 5100 kilometers
Brand B : x 2 = 39,800,s 2 = 5900 kilometers
A. Test the hypothesis at α=.05 that there is no difference in the two brands of tires.
Assume the populations are approximately normal with equal variances. (15 points)
H 0 µ1 = µ 2
H 1 µ1 ! µ 2
Assumptions
Independent random samples.
Equal but unknown variances
Normal populations
Region of rejection
T>2.074 T< –2.074
Calculation
x1 ! x 2
T=
sp
s 2p =
1 1
+
n1 n 2
(n1 ! 1)s12 + (n 2 ! 1)s 22
n1 + n 2 ! 2
11(5100)2 + 11(5900)2
= 30410000
22
s p = 5514.5
s 2p =
T=
37900 ! 39800
5514.5
1 1
+
12 12
=
!1900
= !.844
2251.3
Reject or do not reject H0+
Cannot reject H0
Conclusion
There is insufficient evidence to show that the mileage for the two kinds of tires is
significantly different.
B.(1) Find a 95% confidence interval for the difference between the mean wearout
mileage of the tires
x1 ! x 2 ± t "/ 2,n +n !2s p
1
2
1 1
(7 points)
+
n1 n 2
!1900 ± 2.074(5514.5)
1 1
+
12 12
!1900 ± 4669.1
(!6569.1,2769.1)
(2) Is this confidence interval consistent with the results of the test of
hypothesis.?Explain. (3 points)
Yes it is because it contains zero.
C. Consider the Minitab output below
Is the assumption of equal variances justified. Explain your answer given specific
reference to the above computer output. (3 points)
The assumption would be justified because the p value of the F statistic is .75 greater than
.05.
6. A taxi company manger is trying to decide whether the use of radial tires instead of
regular belted tires improves fuel economy. Twelve cars were equipped with radian tires
and driven over a prescribed test course . Without changing drivers the same cars were
equipped with regular belted tires and driven once again over the test course. The
gasoline consumption in kilometers per liter was recorded as follows,
Row
1
2
3
4
5
6
7
8
9
10
11
12
Radial
Tires
4.2
4.7
6.6
7.0
6.7
4.5
5.7
6.0
7.4
4.9
6.1
5.2
Belted
Tires
4.1
4.9
6.2
6.9
6.8
4.4
5.7
5.8
6.9
4.7
6.0
4.9
Difference
0.1
-0.2
0.4
0.1
-0.1
0.1
0.0
0.2
0.5
0.2
0.1
0.3
Consider the normal probability plot below.
A. Based on this normal probability plot is it reasonable to assume that the differences
between the mileages for radial and belted tires is normally distributed? Explain. (3
points)
The points lie close to a straight line so the mileages are probably normally distributed.
B. The goal is to determine by a test of hypothesis whether the cars equipped with radial
tires give better fuel economy than those equipped with belted tires. The Minitab output
follows.
Paired T-Test and CI: Radial Tires, Belted Tires
Paired T for Radial Tires - Belted Tires
Radial Tires
Belted Tires
Difference
N
12
12
12
Mean
5.750
5.608
0.1417
StDev
1.053
0.994
0.1975
SE Mean
0.304
0.287
0.0570
95% lower bound for mean difference: 0.0393
T-Test of mean difference = 0 (vs > 0): T-Value = 2.48
P-Value = 0.015
(1) What is the null and the alternative hypothesis for this test? (2 points)
H0 µ R = µ B
H1 µ R > µ B
(2) Besides the fact that the differences are normally distributed what other important
characteristic of this data set must be included in the assumptions? (2 points)
Paired data
(3) What is the value of the test statistic and the P value? (2 points)
T= 2.48 pvalue =.015
(4) At α =.05 would you conclude that radial tires give better fuel economy than belted
tires? Explain.(3 points)
Yes because the pvalue is .015<.05
(5) At α =.01 would you conclude that radial tires give better fuel economy than belted
tires? Explain.(3 points)
No because the pvalue is .015>.01
(6) Find a 99% lower confidence bound on the mean of the differences between radial
and belted tires and explain how your result is or is not consistent with the p value given
in the computer output. (7 points)
LCB = .1417 ! 2.718(.1975) / 12
LCB = !.0132
The one sided confidence interval (-.0132,∝) does contain zero.The p value was .015 so
the hypothesis would not be rejected at α=.01.