Practice Final Examination
(With Answers)
Statistics 515
Spring Semester 2002
E. A. Pena's Class
Part I: (20 points) Basic Concepts: Explain briefly what each of the following
terms/phrases mean, or what their importance is.
1. Which statistical hypothesis typically correspond to the "research hypothesis"?
Answer: The alternative hypothesis. The null hypothesis usually coincides with the
statusquo.
2. In statistical hypothesis testing, which type of error is considered to be more serious?
Answer: The type I error, which is committed when the null hypothesis is rejected
when in reality it is correct, is the more serious type of error. This is the reason why
we set an upper limit (the level of significance) to the probability of committing this
error.
3. Science Magazine reported that the mean listening time of 7-month-old infants
exposed to a three-syllable sentence (e.g., "ga ti ti") is 9 seconds. Set up the null and
alternative hypotheses for testing the claim.
Answer: NULL: Mean listening time is 9 seconds. ALTERNATIVE: Mean listening
time is not equal to 9 seconds.
4. What is the level of significance of a test.
Answer: It is the (maximum) probability of committing a type I error.
5. How is the p-value used in making decisions in hypothesis testing?
Answer: It is the probability under the null hypothesis of observing the observed
value of the test statistic, or more extreme values. Equivalently, it is the lowest
significance level such that the null hypothesis will be rejected for the data at hand.
6. How are the probabilities of a Type I and a Type II error related for a fixed sample
size?
1
Answer: They are inversely related. When one is decreased, the other increases.
(Such is life … no free lunch).
7. Why is it that we could not "accept" the null hypothesis, but instead simply conclude
that we "fail to reject the null hypothesis"?
Answer: It is because we did not control the probability of committing a type II
error, which is committed when we do not reject the null when in reality it is false.
In contrast, since we specify the level of significance, we have controlled the
probability of committing a type I error.
8. What does the regression coefficient  in the simple linear regression model
represent?
Answer: It represents the change in the mean value of the response variable (Y) per
unit change in the predictor variable (X).
9. In simple linear regression, what is the idea behind the least-squares principle for
obtaining the coefficients in the regression model?
Answer: The idea is to minimize the distance between the observed Y-values and the
predicted Y-values. The distance is measured by taking the sum of the squared
residuals, where the residual is the difference between the observed Y-value and the
predicted value.
10. In simple linear regression, as well as in a one-way analysis of variance, which
quantity serves as an estimator of the common variance 
Answer: The mean square error (MSE) is the estimator of the variance.
2
Part II: Problem Solving and Interpretations.
1. (20 points) Environmental Science and Technology reported on a study of
contaminated soil in The Netherlands. Seventy-two 400-gram soil specimens were
sampled, dried, and analyzed for the contaminant cyanide. The cyanide concentration
[in milligrams per kilogram (mg/kg) of soil] of each soil specimen was determined
using an infrared microscopic method. The sample resulted in a mean cyanide level of
84 mg/kg and a standard deviation of S = 80 mg/kg. Perform a test of the null
hypothesis that the true mean cyanide level in The Netherlands exceeds 100 mg/kg.
Use a level of significance of 0.05.
a) State the hypotheses.
H0 (Null): Mean cyanide level >= 100 mg/kg.
H1 (Alternative): Mean cyanide level < 100 mg/kg.
b) State your decision rule.
Reject Ho if Z = (XBAR - 100)/(S/SQRT(n)) < -1.645.
c) Compute your test-statistic.
Z = (84 - 100)/(80/SQRT(72)) = -1.70.
d) State your decision.
Since -1.70 < 1.645 then we reject the null hypothesis.
e) State your conclusion with regards to the practical problem considered.
We are 95% confident that the mean cyanide level is less than 100 mg/kg.
3
2. (20 points) The Cleveland Casting Plant is a large, highly automated producer of gray
and nodular iron automotive castings for Ford Motor Company. One process variable
of interest to Cleveland Casting is the pouring temperature of the molten iron. The
pouring temperatures (in degrees Fahrenheit) for a random sample of ten crankshafts
produced at Cleveland Casting are listed below. The target setting for the pouring
temperature is 2,550 degrees. Assuming the process is stable, conduct a test to
determine whether the true mean pouring temperature differs from the target setting.
2543
2541
2544
2620
2560
2559
2562
2553
2552
2553
For this data set, the sample mean equals 2558.7 and the sample standard deviation is
22.7452.
a) State the hypotheses.
H0 (Null): Mean pouring temperature is 2550 degrees.
H1 (Alternative): Mean pouring temperature is different from 2550 degrees.
b) State your decision rule.
Reject H0 if |T| > t9;.025 = 2.262, where T = (XBAR - 2550)/(S/SQRT(n))
c) Compute your test-statistic.
T = (2558.7 - 2550)/(22.7452/SQRT(10)) = 1.21.
d) State your decision.
Since 1.21 < 2.262, then we fail to reject the null hypothesis.
e) State your conclusion with regards to the practical problem considered.
Based on the data, and at the 5% level of significance, we cannot conclude that the
mean pouring temperature is different from 2550, so we cannot conclude that the
process is out of order.
4
3. (20 points) Marine biochemists at the University of Tokyo studied the properties of
crustacean striated muscles (The Journal of Experimental Zoology). The main purpose of
the experiment was to compare the biochemical properties of fast and slow muscles of
crayfish. Using crayfish obtained from a local supplier, the researchers excised twelve
fast-muscle fiber bundles and tested each fiber bundle for uptake of calcium. Twelve
slow-muscle fiber bundles were excised from a second sample of crayfish, and calcium
uptake was measured.
A summary of the sample statistics associated with the calcium uptake (in moles per
milligram) for these two groups is provided below.
Descriptive Statistics
Group
n
Sample Mean
Fast Muscle
Slow Muscle
12
12
.57
.37
Sample
Standard
Deviation
.104
.035
Based on this information, compare the population means of the calcium uptake for the
fast and slow-muscle groups. In particular, test the null hypothesis that the two means are
identical.
In performing your test you may assume that the population distribution of the calcium
uptakes for each group is normally distributed, and that the two populations have equal
variances.
Also, use a 5% level of significance. Again you may answer this question by following
the steps below.
a) State the hypotheses.
NULL: Mean calcium uptakes for the fast and slow-muscle groups are identical.
ALT: Mean calcium uptakes for the fast and slow-muscle groups are different.
b) State your decision rule.
Decision Rule: Reject the null hypothesis if |T| > t22;.025 = 2.074, where
T = (XBAR1-XBAR2)/{Sp[SQRT(1/n1 + 1/n2)]} where Sp is the pooled standard
deviation.
c) Compute your test-statistic.
Sp2 = {(12-1)(.104)2 + (12-1)(.035)2}/(12 + 12 - 2) = .0060
Sp = .0776
5
T = (.57 - .37)/[.0776 SQRT(1/12 + 1/12)] = 6.31
d) State your decision.
Since 6.31 > 2.074, then we reject the null hypothesis of equal means.
e) State your conclusion with regards to the practical problem considered.
Based on the data, we can conclude that there is a difference between the mean
calcium uptakes of slow- and fast-muscle groups, with the fast-muscle groups
having a higher mean calcium uptake.
6
4. (30 points) The quality of the orange juice produced by a manufacturer (e.g.,
Tropicana) is constantly monitored. There are numerous sensory and chemical
components that combine to make the best tasting orange juice. There is a measure of
"sweetness" of an orange juice, with the higher the value of this "sweetness" measure, the
better the orange juice. In order to study the relationship between the "sweetness" and a
chemical measure such as the amount of water soluble pectin (parts per million), in 24
production runs, the sweetness and the pectin level were measured..
6.0
5.9
Sweetness
5.8
5.7
5.6
5.5
5.4
5.3
5.2
200
300
400
PectinLevel
A scatterplot of these 24 pairs of values is provided above.
A simple linear regression analysis with Sweetness as response or dependent variable and
PectinLevel as predictor or independent variable was fitted using Minitab. The output of
this analysis is given below.
Regression
The regression equation is
y = 6.25 - 0.00231 x
Predictor
Constant
x
Coef
6.2521
-0.0023106
S = 0.2150
StDev
0.2366
0.0009049
R-Sq = 22.9%
T
26.42
-2.55
P
0.000
0.018
R-Sq(adj) = 19.4%
Analysis of Variance
Source
Regression
Residual Error
Total
DF
1
22
23
SS
0.30140
1.01693
1.31833
MS
0.30140
0.04622
F
6.52
P
0.018
7
a) By examining the scatterplot, describe the type of relationship between PectinLevel
and Sweetness. For instance, is there a negative type of relationship?
There is a negative linear (almost) relationship between PectinLevel and Sweetness.
b) Based on the simple linear regression analysis, what are the least-squares estimates of
 and ?
The estimate of  is 6.2521.
The estimate of  is -.0023.
c) Provide an interpretation for the value of b, the estimate of .
The value of b = -.0023 means that for a change of one unit in the Pectin Level, the
mean Sweetness will change by the amount of -.0023.
d) For testing the hypothesis that  = 0 (that is, there is no linear relationship between
PectinLevel and Sweetness), what will be your conclusion at the 5% level of
significance? Indicate the information you are using to make your conclusion.
Based on the p-value of .018 associated with the t-value of -2.55, we can conclude
that Pectin Level is a significant predictor of the Sweetness level. You could also
obtain this same conclusion by looking at the analysis of variance table where the pvalue is also .018.
e) What will be the estimate of the common standard deviation ?
The estimate of the common standard deviation  is the square root of the MSE and
this equals .2150.
Using the "fitted line" option in Minitab, the 95% confidence band and prediction interval
were also generated. These are shown in the plot that follows.
8
Regression Plot
Y = 6.25207 - 2.31E-03X
R-Sq = 22.9 %
Sweetness
6.0
5.5
Regression
5.0
95% CI
95% PI
200
300
400
PectinLevel
f) Based on these plots, if a new production line produced a Pectin Level equal to 300,
what will be a 95% confidence interval for the mean Sweetness of the orange juice?
A 95% confidence interval for the mean Sweetness of the orange juice when the
Pectin Level is 300 is between (approximately) 5.46 to 5.75. These values are
obtained from the red curves at PectinLevel of 300.
g) What will be a 95% prediction interval for the exact value of the Sweetness of this
orange juice with Pectin Level of 300?
A 95% prediction interval for Pectin Level of 300 goes from (approximately) 5.1 to
6.0. These values are obtained from the blue curves when PectinLevel is 300.
g) The coefficient of determination of the fitted simple linear regression was 22.9%.
Based on this value, how would you assess the ability of Pectin Level to explain the
variation in the Sweetness measure? Is it high or is it low?
The coefficient of determination of 22.9% indicates that 22.9% of the total
variability in the Y-values (the sweetness) can be explained through the predictor
variable which is the Pectin Level. The value is not high, so as a predictor of
Sweetness, the Pectin Level may not be very good.
9
5. (20 points) The Journal of Hazardous Materials published the results of a study of the
chemical properties of three different types of hazardous organic solvents used to clean
metal parts: aromatics, choloalkanes, and esters. One variable studied was sorption rate,
measured as mole percentage. Independent samples of solvents from each type were
tested and their sorption rates were recorded. Summary statistics for the three groups are
provided below.
Descriptive Statistics
Variable
Aromatic
Chloroal
Esters
N
9
8
15
Variable
Aromatic
Chloroal
Esters
Mean
0.9422
1.006
0.3300
Minimum
0.6500
0.430
0.0600
Median
0.9500
1.015
0.3400
Maximum
1.1500
1.580
0.6100
TrMean
0.9422
1.006
0.3292
Q1
0.8050
0.635
0.1000
StDev
0.1683
0.401
0.2076
SE Mean
0.0561
0.142
0.0536
Q3
1.0900
1.377
0.5300
Overlaid boxplots for the three groups is also given below.
Aromatics
1.5
1.0
0.5
0.0
Aromatics
Chloroalkanes
Esters
To determine whether the population mean sorption rate for the three groups are
identical, a one-way analysis of variance was performed using Minitab. The output of this
analysis is provided below.
10
One-way Analysis of Variance
Analysis of Variance
Source
DF
SS
Factor
2
3.3054
Error
29
1.9553
Total
31
5.2607
Level
Aromatic
Chloroal
Esters
MS
1.6527
0.0674
N
Mean
StDev
9
8
15
0.9422
1.0063
0.3300
0.1683
0.4010
0.2076
Pooled StDev =
0.2597
F
24.51
P
0.000
Individual 95% CIs For Mean
Based on Pooled StDev
----+---------+---------+---------+(----*-----)
(------*-----)
(----*----)
----+---------+---------+---------+0.30
0.60
0.90
1.20
Based on the description of the problem and the Minitab output, answer the following
questions.
a) What will be your null hypothesis and your alternative hypothesis.
NULL: The (population) mean sorption rates for the three groups are identical.
ALT: At least two of the three (population) mean sorption rates are different.
b) How many levels do you have in your factor? What are they?
There are three levels. The levels are Aromatic, Choloalkanes, and Esters
c) What will be your estimate of the common variance of the three populations?
The common variance is estimated by the MSE, which is 0.0674.
d) What will be your conclusion with regards to your hypothesis, and what is the basis
of your conclusion?
Since the p-value in the analysis of variance table is 0, then we will reject the null
hypothesis and conclude that at least two of the mean sorption rates are different.
e) Which population mean would you conclude is different from the other two?
By examining the confidence intervals for the three mean sorption rates, we note
that the intervals for the aromatic and choloalkanes overlap, and these intervals do
not overlap with the interval for the ester group. Therefore, we could conclude that
the mean sorption rate of the ester group is different from the means of the
aromatic and choloalkanes groups.
11