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Stat 303 - Section 503 - Practice Exam III
- A university professor teaches two sections of the exact same course. Recently, he gave the
same 30 point quiz to both sections. What type of test should he run to see if the two sections are
doing about the same?
A. You can’t prove H 0
: μ 1
= μ 2
, the section averages are the same, so he can’t test this claim.
B. large 2 sample t -test (an approximate z - test) since there were 30 questions on both exams
C. a paired t -test, pairing the students by class standing, ı.e., the best in one section with the best
in the other
D. a pooled t -test since the standard deviations of the two sections would be the same and the
pooled t -test has more power
E. As long as there are at least 30 students in each section, he could do a 2 sample or pooled t -
test depending on how close the standard deviations were.
- What is the advantage of using a paired t -test (Case 10) over either 2 sample t -tests (Cases 8 or
A. You only need half as many observations (smaller sample size).
B. You have more power (easier to detect a difference).
C. You have more degrees of freedom (less conservative test).
D. All of the above are advantages to the paired t -test.
E. Exactly two of the above are advantages to the paired t -test.
- Twelve people were given 10 minutes to memorize a list of 20 nonsense words. They were
asked to list as many of the words as possible after 1 hour had passed and then asked again 23
hours later (the next day). The average after 1 hour was 15 words and the average after 1 day was
only 13 words. What method should be used to determine whether the average number of words
recalled after 1 hour exceeds the average after 1 day?
A. a one sample t -test comparing the sample average after 1 day with 15 assuming the data is
approximately normal
B. a two sample z -test of a proportions using the percent of words recalled after 1 hour and 1 day
assuming the data is approximately normal
C. a paired t -test pairing by the 12 people assuming the data is approximately normal
D. a pooled t -test since the standard deviations should be about the same assuming the data is
approximately normal
E. a 2 sample t -test since we can’t assume anything
- We want to test whether eating a good breakfast (or lunch if it’s an afternoon class) improves
the mental performance of students on exams. If we decided to give a sample of 50 students an
exam without letting them eat and then the next day giving them a similar exam (but having
taken the first exam wouldn’t help them on the next one) after they all ate, what type of test
procedure we use?
A. a one sample t -test since we don’t know the true standard deviation
B. a one sample test of proportions using the success rate as the sample proportion
C. a two sample test of proportions comparing the success rates
D. a two sample test of mean ( t -test) comparing the exam averages
E. a paired t -test pairing the exam scores by student
- We want to test if there is sufficient evidence to say that the percentage of heart disease
victims for smokers is different for men and women. How should we gather the data?
A. take a sufficiently large random sample of men and women, have them smoke for 10 years,
and calculate the proportion with heart disease
B. take two sufficiently large random samples of men and women smokers and calculate the
proportion with heart disease
C. take two sufficiently large random samples of men and women smokers and calculate the time
until the onset of heart disease
D. take two sufficiently large random samples of men and women smokers and calculate the
proportion with a family history of heart disease
E. take two sufficiently large random samples of men and women smokers and force them to
stop smoking since it also causes heart disease
- Suppose you are interested in the caloric content of hamburgers. You took a random sample
from 3 different chains and calculated the following confidence intervals: Chain 1: (248.9,307.8);
Chain 2: (263.5,309.3); Chain 3: (307.1,349.7). Which of the following would be the most
appropriate conclusion?
A. The true mean caloric content for Chain 3 is more than that of Chain 1 or 2.
B. The true mean caloric content for Chain 3 is more than that of Chain 1 but not Chain 2.
C. The true mean caloric content is different for all three chains.
D. The true mean caloric content is possibly the same for all three chains.
E. We cannot determine any conclusion since we don’t have a method for testing three means.
- Which of the following tests require the data to be normal?
A. a one sample z -test of a proportion
B. a one sample t -test of a mean with n = 50
C. a paired t -test for the difference of means with a total of 50 observations
D. All of the above need the data to be normal.
E. Only two of the above need the data to be normal.
- When should you use a significance level of 1% instead of 5%?
A. when you want to keep the chance of making a Type I error low.
B. when you want to keep the chance of making a Type II error low.
C. when you want to keep the chance of making a Type II error high.
D. when you want to keep the power of the test thigh.
E. Exactly two of the above are good reasons.
- Which of the following tests would be the most powerful?
(assuming everything else about the tests is the same)
A: n=100 α=.
B: n=100 α=.
C: n=1000 α=.
D: n=1000 α=.
E: These tests have the same power
F: You cannot tell from the information given.
- Luis wants to test the following hypothesis:
H
0
: μ ≥ 50 vs. H A
: μ < 50
Toaster 6 18 22
Toaster 7 15 14
With α=.10 test whether the Pop-Tarts are more burnt at Brent’s house then at Mason’s house.
Helps:
Standard deviation of one toaster at Brent’s house: 2.
Standard deviation of one toaster at Mason’s house: 3.
Pooled standard deviation: 3.
Matched Pairs standard deviation: 2.
- George Bush Sr. mentions on T.V. that the average age of a student at Texas A&M is 23
years old. To test his hypothesis, you ask 3 randomly chosen A&M students what their ages are,
and use α=.
Assume the ages of students at A&M are normally distributed.
The ages were : 22 years old, 28 years old, and 24 years old.
Fill in the following values: (You will not be penalized for math errors, if I can tell your equation
is correct – for example for the standard deviation – so write it out!)
n =
x
μ 0
df
s x
Now do all 7 steps of hypothesis testing
1. E
2. B
3. C
4. E
5. B
6. D
7. C
8. A
- Power is the probability of NOT making a type II error. In other words, it is the probability of being
able to tell when the NULL is false. A larger sample size is always good – if you have more data, then it
is easier to tell when the null is false, so n=1000 is more powerful than n=
Lowering alpha is nice because it lowers the probability of a type I error. What is the tradeoff? You
increase beta – you are more likely to make a type II error. Making more type II errors means less power.
Thus a smaller alpha means smaller power. The most power will be when alpha is the larger value,.
- If the hypothesized mu (50) is in the confidence interval, then we accept the null. If it is not, then we
reject the null. This is the same as saying if the p-value is greater than alpha we accept the null, if it is
not then we reject the null.
We accepted the null at 95% confidence,
We rejected the null at 92% confidence,
Since this is a one-tailed test, the alpha level corresponding to the 95% confidence interval is the
probability in the upper tail of the test. Since there is 5% in both tails, the alpha level would be .05/2 =
.
So if alpha was .05/2=.025 we would accept, if alpha was .08/2=.04 then we would reject.
Thus the p-value for the hypothesis test must be somewhere between those two values,
.025 < p-value <.
Stddev(average) = Stddev(x)/sqrt(n)
.23 = 3/sqrt(n)
n=(3/.23)
2
=170.
So Morgan needs 171 syringes.
This problem should look familiar! Now for an extra challenge, what if this was the standard error of
{proportions, two proportions, xbar1-xbar2, etc}?
H 0
: п W
≥ п M
H A
: п W
< п M
(claim)
α=.05 (because it wasn’t given in the problem)
n* п = 102 for women, 132 for men
n*(1- п) = 18 for women, 18 for men
So we can use the CLT
W m
W M
W m
n n
x x
p
n
p p
n
p p
p p
Z
P
P P P P
W M
where
H 0
: μ d
≤ 0
H A
: μ d
> 0 (claim. Brent thinks subtracting will give him a positive number)
α=.
The toaster times are normal, so we can use the t-test.
6
n
s
x
t
D
D D
μ
This is a greater than test, so we want the probability to the right of the test statistic. The t-table tells the
lower tail probability, so we want one minus that probability.
The tail probability is between 0.1 and 0.15. So the p-value in the direction of the rejection region is
between .85 and .9. Looking at the picture this makes sense, we are no where near the rejection region.
We cannot reject the null hypothesis. Brent’s outlets are not carrying more voltage than Mason’s (we
should investigate whether his outlets get LESS voltage than Mason’s)
n = 3
x = (22+28+24)/3 = 24.
μ 0
= 23
df = 3-1 = 2
s x
=
2 2 2
At this point, whatever you get for s x
will be right, because the formula is right.
2 2 2
H 0
: μ = 23
H A
: μ ≠ 23
α=.
The ages are normally distributed, so we can use a t-test
2
t
n
s
x
t
df
μ
The t-value of .94 is between .817 and 1.061 on the t-table for 2 degrees of freedom, so the tail-
probability is between .25 and .2. This is a two-sided test, so we need to double the probability to get the
p-value.
.4 < p-value <.
Since .25 > .01 we cannot reject the null hypothesis.
So Bush was right. (Again), it appears that the average age of students at Texas A&M is about 23 years
old.
