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Unformatted text preview: STAT 350 – Spring 2009
Exam 2 SOLUTION
Last Name: ________________________ First Name: ____________________ Your Seat: __________ Section Time (circle): 10:30 11:30 4:30 Note:
• You are responsible for upholding the Honor Code of Purdue University. This includes
protecting your work from other students.
• Show your work on all questions. Unsupported work will not receive full credit. Credit will not
be given for dumb luck. Showing work includes defining any random variable or event you use
in the solution. Showing work also includes identifying any named distribution you are using the
solution and the values of all relevant parameters. You do not need to show work for multiple
choice questions.
• Decimal answers should be exact or to at least four significant digits.
• Unless otherwise stated, assume the significance level for any hypothesis test is 0.05.
• Standard Normal (Z) and values/probabilities must be taken from the tables provided.
Probabilities, pvalues, critical values, etc., for χ2, t, and F distributions must also be taken from
the tables provided, unless this information is available on SAS output provided, in which case
the values from SAS must be used.
• You are allowed the following aids: a onepage (8.5×11 inch) cheat sheet, a scientific calculator,
and pencils.
• Turn off and put away your cell phone before the exam begins! Question Points Possible Points Missed
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31
2
3 20 4 18 5 16 6 5 Total STAT 350 Exam 2 10 100 Page 1 of 1 1. Apples and oranges were compared for the quantities of fiber. The results of PROC TTEST are
given below.
The TTEST Procedure
Statistics Variable fruit N fiber
fiber
fiber apple
orange
Diff (12) 15
18 Lower CL
Mean Mean Upper CL
Mean Lower CL
Std Dev Std Dev Upper CL
Std Dev Std Err 45.849
53.703
27.14 52.716
65.751
13.03 59.583
77.798
1.0702 9.0781
18.18
15.859 12.4
24.227
19.782 19.555
36.32
26.299 3.2016
5.7104
6.9158 TTests
Variable Method Variances fiber
fiber Pooled
Satterthwaite Equal
Unequal DF t Value Pr > t 31
26.2 1.88
1.99 0.0689
0.0570 Equality of Variances
Variable Method fiber Folded F Num DF Den DF F Value Pr > F 17 14 3.82 0.0150 a. (6 pts; 3 pts each part) Is it appropriate to assume equal variances (i.e., σapple = σorange)?
(i) Just answer "yes" or "no". No
(ii) Give the pvalue you used to determine this. 0.0150 b. (15 pts; 5 pts each part) For the following questions, assume that the null hypothesis is
µapple  µorange = 0. Be sure your answers below are consistent with your answer to part (a).
(i) Assuming the alternative hypothesis was µapple  µorange < 0, what is the appropriate pvalue?
0.0570/2 = 0.0285
(ii) Assuming the alternative hypothesis was µapple  µorange > 0, what is the appropriate pvalue?
1 – 0.0570/2 = 0.9715
(iii) Assuming the alternative hypothesis was µapple  µorange ≠ 0, what is the appropriate pvalue?
0.0570 STAT 350 Exam 2 Page 2 of 2 1. (continued)
c. (10 pts) You have just picked an apple. Give a 95% prediction interval for the quantity of fiber
in this apple.
52.716 ± 2.145(12.4) 1 + 1
15 → (25.2467, 80.1863) 2. (10 pts; 5 pts each part) Basics of hypothesis testing
a. You are testing the null hypothesis μ = 50 against the alternative hypothesis μ ≠ 50 at the α =
0.10 significance level. You have 10 observations (x1, x2, . . . , x10). Give the critical value for
your test and sketch the rejection region.
df = 10 – 1 = 9
tcrit = ±1.833 (Table IV)
The rejection region is the area ≤1.833 and the area ≥+1.833
b. You are testing the null hypothesis μ = 50 against the alternative hypothesis μ ≠ 50. You have
16 observations (x1, x2, . . . , x16). The value of your test statistic is 1.8. What is the pvalue for
this test?
df = 16 – 1 = 15
pvalue = (0.046)×2 = 0.092 (Table VI) STAT 350 Exam 2 Page 3 of 3 3. A clinical trial was conducted to test the effectiveness of DeChol, a new medication intended to
reduce cholesterol. 400 subjects, aged 50 to 65, volunteered to participate. Subjects were randomly
assigned to take either DeChol or a placebo (though they did not know which they are on). The trial
lasted 5 years. In addition to measuring cholesterol levels of the subjects, many other variables were
recorded (e.g., blood pressure, weight) including any medical complaints (e.g., dry mouth,
headaches) or medical conditions or complications (e.g., cancers, heart attacks, death) for the
purpose of finding sideeffects. Below is a table showing the number of subjects in each treatment
group that either had or did not have a heart attack during the clinical trial.
Heart Attack No Heart Attack
12
188
DeChol
Placebo
4
196
a. (8 pts) Fill in the table below with the expected counts.
Heart Attack No Heart Attack
8
192
DeChol
placebo
8
192
b. (6 pts) What is the value of the appropriate test statistic? (12 − 8 )
χ=
2 8 2 ( 4 − 8)
+
8 2 (188 − 192 )
+
192 2 (196 − 192 )
+
192 2 = 4.16667 c. (2 pts) What is the critical value for this test (α = 0.05)?
df = (21)(21) = 1
Critical Value = 3.84
d. (4 pts) Based on your analysis above, what is your conclusion? Circle the single most
appropriate choice from the options below.
The test statistic (4.1667) is greater than the critical value (3.84), therefore we REJECT the null
hypothesis of independence.
Because this data was based on an EXPERIMENT (random assignment of the drug/placebo), we
can conclude the difference is CAUSED by the treatment (DeChol vs. placebo)
There is no statistically significant association between the treatment group and the
whether or not a person has a heart attack. In other words, there is no reason to believe
taking DeChol alters ones risk of having a heart attack.
There is a statistically significant association between the treatment group and the whether
or not a person has a heart attack, but we cannot conclude that the difference in heart attack
rates between the two groups is caused by DeChol.
We can conclude that taking DeChol causes a statistically significant change in the
likelihood of having a heart attack. STAT 350 Exam 2 Page 4 of 4 4. (18 pts; 3 pts each part) Multiple Choice. For each part, circle the single most appropriate answer
from the options provided.
a. Assume that you are testing H0: μ = 5 versus Ha: μ ≠ 5 at α = 0.05 and you rejected the null
hypothesis. If you had used the same data to instead test H0: μ = 5 versus Ha: μ > 5 at α = 0.05,
would you have also rejected the null hypothesis?
Note that you do not know whether x was <5 or >5, so we do not know if the test statistic is
positive or negative.
definitely yes
definitely no
maybe
b. You want to test the null hypothesis that μ = 25 against the alternative hypothesis that μ ≠ 25.
The 95% CI for μ is (23.2, 47.6). Therefore, the pvalue of your test must be greater than 0.05.
See Lecture K, page # 14
definitely yes
definitely no
maybe
c. In hypothesis testing, the pvalue is the probability that the null hypothesis is true.
See Lecture K, page # 4
True
False
d. When conducting a ttest, a paired ttest can be used as long as there are the same number of
observations in both groups.
The data must be inherently paired
True
False
f. You are testing the null hypothesis μ1  μ2 = 0 against the alternative hypothesis μ1  μ2 ≠ 0 at the
α = 0.01 significance level. Your pvalue is 0.0345.
the pvalue is greater than the significance level(alpha), therefore we fail to reject the null
You conclude that the two group means are not significantly different
You conclude that the two group means are significantly different
g. If you conduct a 1way ANOVA with only 2 treatment levels, this is equivalent to
See Lecture L, page # 1617
conducting an independent sample Satterthwaite ttest
conducting an independent sample pooled ttest
conducting a paired ttest STAT 350 Exam 2 Page 5 of 5 5. Kristen is a food scientist studying methods of reducing lactose content in ice cream. In her
experiment, she used lactase (an enzyme that breaks down lactose) was taken from 4 different
microbial sources (Kluyveromyces fragilis, K. lactis, Aspergillus niger, and A. oryzae). She also
used ice cream with 2 different levels of butterfat (10% and 15%). She made 3 batches of ice cream
for each combination of lactase source and butterfat content. One of the measurements she was
interested in was the texture of the product. Volunteers each tasted each batch of ice cream and
rated the texture on a scale of 1 to 20 (20 being the best). Kristen took the average score for each
batch as the measure of texture. The SAS results from the ANOVA are given below. Some
information has been omitted and replaced with asterisks (*).
The omitted values have been replaced for the solution (in blue)
The ANOVA Procedure
Class Level Information
Class Levels Values lactase 4 an ao kf kl butterfat 2 10 15 Number of Observations Read
Number of Observations Used 24
24 The ANOVA Procedure
Dependent Variable: texture DF Sum of
Squares Mean Square F Value Pr > F Model 7 588.5330292 84.0761470 21.14 <.0001 Error 16 63.6276667 3.9767292 Corrected Total 23 652.1606958 Source RSquare Coeff Var Root MSE texture Mean 0.902436 19.26816 1.994174 10.34958 Source
lactase
butterfat
lactase*butterfat STAT 350 Exam 2 DF Anova SS Mean Square F Value Pr > F 3
1
3 25.8171458
543.8776042
18.8382792 8.6057153
543.8776042
6.2794264 2.16
136.77
1.58 0.1322
<.0001
0.2335 Page 6 of 6 5 (continued) Was there a significant lactase effect?
a. (7 pts) Give the value of the appropriate test statistic. = MS (lactase) SS (lactase) (4 − 1) 25.817 3 8.6057
=
=
=
= 2.164
MS ( Error )
SS (error ) 16
63.628 16 3.9767 b. (3 pts) Give the degrees of freedom for this test statistic.
numerator df = 3, denominator df = 16
c. (2 pts) Give the critical value for this test (α = 0.05). 3.24 d. (4 pts) Based on your analysis above, what is your conclusion? Circle one of the options below.
The test statistic (2.164) is less than the critical value (3.24), therefore fail to reject null
There is NO statistically significant difference in the mean texture among the 4 sources of
lactase studied
The mean texture for each lactase source is significantly different from each of the other
lactase sources.
At least one of the lactase sources has a mean texture that is significantly different from the
other lactase sources.
6. (5 pts) Assume you have conducted an analysis of variance (ANOVA) and a multiple comparison
procedure is warranted. When is it appropriate to use Dunnett’s procedure instead of Tukey’s?
You must only be concerned with comparing the other treatment groups to a control (or baseline)
treatment – not comparing the other treatments to each other.
Note – it is not enough to say that there is a control (or that you wish to compare the other groups to
the control) – you can still use Tukey’s test when there is a control and Tukey’s will tell you if the
other treatment are different than the control. STAT 350 Exam 2 Page 7 of 7 ...
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