General Questions. 
 How should you interpret the statement "the 95% confidence
interval for the mean is (8.9, 12.7)"?
 As the level of confidence (a/2) goes
up, does the sample size go up or down?
 As the underlying level of variance goes up, does the sample size
go up or down?
 What are the five steps of a statistical test? (commit this to memory!)
 Define what is meant by a Type I error. (Commit this to memory also!)
 Define what is meant by a Type II error.
 Define what is meant by the Power of a test.
 With a twotailed test, is the rejection region composed of one
region or two?
 What does an OC curve tell us?
 What is a pvalue? (see section 5.6).
 What is the differences between the ztest and the ttest?
 The shape of the tdistribution is indexed by one parameter. What
is this parameter referred to as?
 Using the table of percentage points for the Student's t distribution
given on page 1093, compute the following. ( n is the degrees of freedom).
 The value of t where P(T>t  n=26)=.05 (i.e. the 5% critical
value for a tdistribution with 26 degrees of freedom.)
 P(T>3.169  n=10) = ?

For
students in agriculture and environmental fields. 
 A oil company advertises that its new brand of tractor fuel significantly
decreases fuel consumption in heavy use. You have been charged with
determining whether the company's claim is valid. Your records indicated
that the average fuel consumption for a standard configuration of
tractor/plow is 10 liters per hour under a common standard plowing
protocol (i.e. specified speed, depth of plow and soil conditions).
You wish to determine if, using the new fuel, average fuel use is
less than 10 l/hr. You decide to test this hypothesis by using a standard
tractor/plow configuration and repeated measurements over similar
field conditions. Initially you will do 15 of these replications,
three per day for five days. At the end of the week, you will be testing
a null hypothesis that the average fuel consumption is 10 l/h against
an alternative that consumption is less than 10 l/h. Assuming the
true standard deviation in consumption is roughly (s=)
1.5 l/h, and the test is being done at a Type I error rate of (a=)
0.05, what would the statistical power be for significant differences
of +.5, +1.0 and +1.5 l/h (D) less than
the expected 10 l/h.
 You decide to run the experiment with 15 replications, with the
results produced in the following table. With these data, test the
null hypothesis that nicotine content is 10 l/d against the alternative
that the average is less than 10 l/d. (Note: The data table below
can be directly cut and pasted into a spreadsheet or stat package
 i.e. no html codes are imbedded.)
Replication Consumption
1 8.99
2 9.16
3 10.20
4 9.82
5 11.54
6 9.48
7 10.65
8 9.71
9 11.43
10 9.97
11 9.11
12 8.24
13 8.44
14 8.50
15 11.34
 You are encouraged by the number of replicates with consumption
less than 9.7 l/d and think that the true average consumption might
really be 9.7 l/d. You want to do more sampling to get more definitive
results. How many more replicates would you need to determine that
the true population mean was really 9.7 l/d (the specific alternative
hypothesis) instead of 10 l/d (the null hypothesis) if the true standard
deviation is (s =) 1.0 l/d. The test is
to be performed at a Type I error probability of (a=)
0.05 and we wish 85% power for the test (i.e. Type II error probability
of b=0.15). [Hint: page 221]
 For comparison purposes, redo the test in step 2 using the Sign
Test. How do the results compare?

For
students in engineering fields. 
 A copper mine manager is interested in installing new equipment
that is expected to increase production. In the current process, average
daily production varies by plus or minus 50 tons per day with an average
of about 50 tons per day. The manager can perform a trial run with
the new equipment for (n=) 15 days. At the end he will be testing
a null hypothesis that the average is 50 t/d against an alternative
that production is greater than 50 t/d. Assuming the true standard
deviation of the process is roughly (s=)
100/4=25 t/d, and the test were being done at a Type I error rate
of (a=) 0.05, what would the power
be for significant differences of +10, +20 and +40 t/d (D)
greater than the expected 50 t/d.
 The manager decided to run the experiment over 15days with the
results produced in the following table. With these data, test the
null hypothesis that the average production is 50 tons per day against
the alternative that the average is greater than 50 tons. (Note: The
data table below can be directly cut and pasted into a spreadsheet
or stat package  i.e. no html codes are imbedded.)
Day Yield
1 57.8
2 58.3
3 50.3
4 38.5
5 47.9
6 157.0
7 38.6
8 140.2
9 39.3
10 138.7
11 49.2
12 139.7
13 48.3
14 59.2
15 49.7
 The mine manager is encouraged by the four days where production
was over 70 and thinks that the true production mean might really
be 70 t/d and wants to continue the trial to get more definitive results.
How many days would be required to determine that the true population
mean was really 70 t/d (the specific alternative hypothesis) instead
of 50 t/d (the null hypothesis) if the true standard deviation is
(s =) 40 t/d. The test is to be performed
at a Type I error probability of (a=) 0.05
and we wish 80% power for the test (i.e. Type II error probability
of b=0.20). [Hint: page 221]
 For comparison purposes, redo the test in step 2 using the Sign
Test. How do the results compare?

For
students in toxicology and health science fields. 
 A tobacco company advertises that the average nicotine content of
its cigarettes is at most 14 milligrams. You have been charged with
determining whether the company's claim is valid. If the average nicotine
is greater than 14.75 milligrams the company could be in trouble with
government oversight agencies. You decide to test 20 cigarettes selected
at random, one per day over 20 days of production. At the analysis
step, you will be testing a null hypothesis that the average is 14
mg against an alternative that nicotine content is greater than 14
mg. Assuming the true standard deviation in nicotine content is roughly
(s=) .8 mg, and the test were being done
at a Type I error rate of (a=) 0.05, what
would the power be for significant differences of +.2, +.4
and +.75 mg (D) greater than the expected
14 mg.
 You decide to run the experiment on 25 cigarettes, with the results
produced in the following table. With these data, test the null hypothesis
that nicotine content is 14 mg against the alternative that the average
is greater than 14 mg. (Note: The data table below can be directly
cut and pasted into a spreadsheet or stat package  i.e. no html codes
are imbedded.)
Cigarette Nicotine
1 13.96
2 15.33
3 15.01
4 14.46
5 14.71
6 15.35
7 16.03
8 15.22
9 14.35
10 15.56
11 15.87
12 14.82
13 13.82
14 14.19
15 13.84
16 14.33
17 13.79
18 15.32
19 12.50
20 14.17
21 14.07
22 14.58
23 13.28
24 16.60
25 13.37
 You are concerned by the number of cigarettes with nicotine levels
over 14.5 and think that the true production mean might really be
14.5 mg. You want to do more sampling to get more definitive results.
How many more cigarettes would you need to determine that the true
population mean was really 14.5 mg (the specific alternative hypothesis)
instead of 14 mg (the null hypothesis) if the true standard deviation
is (s =) .90 mg. The test is to be performed
at a Type I error probability of (a=) 0.05
and we wish 90% power for the test (i.e. Type II error probability
of b=0.10). [Hint: page 221]
 For comparison purposes, redo the test in step 2 using the Sign
Test. How do the results compare?

For
students in community development, education and social services fields. 
 The University is concerned that married graduate students may be
spending a large proportion of their graduate stipends on health care
costs. When figuring out the total amount of stipends to offer, University
officials factor in about $500 per year for health care expenses.
You have been charged with determining, using a sample survey, if
this number is a good estimate for average health care costs. If the
average cost greater than $500, the University might increase stipends
appropriately. As a first step, you decide to question 40 married
male graduate students in the College of Agriculture, selected at
random from the spring enrollment list. At the analysis step, you
will be testing a null hypothesis that the average health care cost
is $500 against an alternative that costs are greater than $500. Assuming
the true standard deviation in costs is roughly (s=)
$150, and the test is being done at a Type I error rate of (a=)
0.05, what would the power be for significant differences of
+$50, +$75 and +$100 mg (D) greater than
the expected $500.
 You decide to run the study with 40 students, with the results produced
in the following table. With these data, test the null hypothesis
that average costs is $500 against the alternative that the average
is greater than $500. (Note: The data table below can be directly
cut and pasted into a spreadsheet or stat package  i.e. no html codes
are imbedded.)
Person Expenditure
1 417
2 788
3 605
4 475
5 577
6 496
7 585
8 512
9 482
10 569
11 505
12 516
13 408
14 696
15 575
16 702
17 596
18 627
19 444
20 500
21 534
22 181
23 338
24 604
25 519
26 606
27 482
28 551
29 550
30 494
31 356
32 551
33 426
34 541
35 511
36 758
37 328
38 438
39 520
40 480
 These data seem to indicate that true health care costs might be
closer to $550 than to $500. You want to do more sampling to get more
definitive results. How many more students would you need to determine
that the true population mean was really $550 (the specific alternative
hypothesis) instead of $500 (the null hypothesis) if the true standard
deviation is (s =) $110. The test is to
be performed at a Type I error probability of (a=)
0.05 and we wish 90% power for the test (i.e. Type II error probability
of b=0.10). [Hint: page 221]
 For comparison purposes, redo the test in step 2 using the Sign
Test. How do the results compare
