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BA275 / Winter 2006 / Exam #2 / 80 Points
For multiple choice problems circle the letter corresponding to the BEST answer provided.
For calculation problems, SHOW ALL WORK. Credit is given for the method, not the answer.
For short answer questions, keep your answers concise, and to the point.
1. The weights of artichoke shipping crates have a mean of 40.0 pounds and a standard deviation of 2.80
pounds.
1a. (5 points) A random sample of 32 artichoke shipping crates is selected. What is the probability
that the sample mean is greater than 41 points. Draw a picture of the probability distribution that
you use to find the answer, and show the area that corresponds to your answer. Also, find the mean
and the standard deviation of this probability distribution.
Answer:
 X    40

2.8
X 

 .4950
n
32
For a sample mean of 41:
Z
X   X 41  40

 2.02
X
.4950
P( X  41)  P( Z  2.02)  .5  .4783  .0217
1b. (3 points) If the sample size had been 4, rather than 32, carefully explain why it would not
possible to find probabilities for the sample mean.
Answer: Because we don't know that the population is normally distributed, and because the
sample size is not large enough for the Central Limit Theorem to apply, we do not know the
shape of the sampling distribution of the mean. If we do not know the shape of the
distribution, we cannot find probabilities.
2. (3 points) Which of the following describes a Type I error:
a. We reject the null hypothesis, when the null hypothesis is true.
***correct answer
b. We fail to reject the null hypothesis, when the null hypothesis is false.
c. We reject the null hypothesis even though the P-Value is greater than α.
d. We fail to reject the null hypothesis, even though the P-Value is less than α.
3. (3 points) Which of the following describes a Type II error:
a. We reject the null hypothesis, when the null hypothesis is true.
b. We fail to reject the null hypothesis, when the null hypothesis is false. ***correct answer
c. We reject the null hypothesis even though the P-Value is greater than α.
d. We fail to reject the null hypothesis, even though the P-Value is less than α.
4. The quality control manager at a light bulb factory needs to estimate the average life of a large shipment
of light bulbs. It is known that the population standard deviation of bulb life is 75 hours. A random sample
of 50 light bulbs from the shipment has a sample mean of 340 hours.
4a. (5 points) Find the 94% confidence interval for the average life of light bulbs in the large
shipment.
Answer:
Since we know the population standard deviation (σ), we use a Z statistic:
X  Z


n
75
340  1.88
 340  19.94 hours
50
2
4b. (4 points) Use an equation (a probability statement) to explain what this confidence interval
means.
Answer: P(340 – 19.94 <  < 340 + 19.94) = .94
4c. (4 points) Which of the following interpretations of this confidence interval is correct?
a. 94% of the sampled light bulbs had lives within the confidence interval.
b. We are 94% confident that the average life of all the light bulbs in the large shipment is within
the confidence interval. ***correct answer
c. Of all the light bulbs in the large shipment, 94% of them will have lives within the confidence
interval.
d. We are 94% certain that the average life of the sampled light bulbs is within the confidence
interval.
e. 6% of the light bulbs in the large shipment have lives that are outside the confidence interval.
5. (4 points) The purchasing director for an industrial parts factory is investigating the possibility of
purchasing a new type of injection molding machine. He determines that the new machine will be bought
if there is convincing evidence that the new machine produces a smaller proportion of defective parts that
the old, existing machine. Samples of 200 parts are gathered from each machine. For the old machine the
sample proportion of faulty parts is .040. For the new machine sample, the sample proportion of faulty
parts is .036. Identify the hypotheses (null and alternative) that would allow the purchasing manager to
make this decision.
Answer:
H0: pnew  pold,
HA: pnew < pold
Alternatively,
H0: pnew - pold  0,
HA: pnew - pold < 0
6. An engineer is evaluating whether or not to upgrade a machine component from the current all-metal
component, to a new, polymer component. Because the upgrade is expensive, the upgrade will be made
only if there is convincing evidence that the new, polymer component has a longer, average service life.
Extensive experience with the existing, all-metal component indicates that the average service life is 200
hours. The engineer sampled 46 polymer components, and found a sample mean service life of 215 hours
and a sample standard deviation of 21 hours.
6a. (2 points) For an alternative hypothesis of: H1: µ > 200, state the null hypotheses.
Answer: H0: μ ≤ 200
6b. (3 points) Using = 1.0%, identify the critical value(s) for this hypothesis test.
Answer: The critical value of t ( = .01, dof = 45) = 2.412
6c. (3 points) Calculate the test statistic for this hypothesis test.
Answer:
Test statistic:
t
X   0 209  200

 2.9067
S
21
46
n
6d. (5 points) Draw a picture of the t-distribution for the test statistic (assuming that the null
hypothesis is true). Show and label the critical value(s), the test statistic, the rejection region, and
the area corresponding to the P-Value.
6e. (3 points) In the context of the problem, state the conclusion(s) of the hypothesis test.
Answer: Because the test statistic is greater than the critical value, we reject the null hypothesis
and conclude the alternative hypothesis. “At a 5% level of significance, there is sufficient
statistical evidence to conclude that the average service life of the new, plastic component is
greater than 200 hours.
6f. (2 points) Carefully explain whether the P-Value for this hypothesis test is greater than or less
than 1%?
Answer: Since we rejected the null hypothesis, the P-Value must be less than alpha = 1%.
7. The following hypothesis test was carried out:
H0:  = 0.0,
HA:  ≠ 0.0
STATGRAPHICS provided the following as part of the analysis:
Power Curve
Power
alpha = 0.1, sigma = 1.82
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-1.2 -1 -0.8 -0.6 -0.4 -0.2 0
0.2 0.4 0.6 0.8
1
1.2
True Mean
7a. (3 points) If the true population mean = 0.20, the probability of making a Type II error is:
a. .00 ≤ probability < .08
b. .08 ≤ probability < .18
c. .18 ≤ probability < .28
d. .28 ≤ probability < .38
e. .38 ≤ probability < .53
f. .53 ≤ probability ≤ 1.00
*** correct answer
7b. (3 points) If the true population mean = 0.20, the probability of making the correct decision is:
a. .00 ≤ probability < .08
b. .08 ≤ probability < .18
c. .18 ≤ probability < .28
***correct answer
d. .28 ≤ probability < .38
e. .38 ≤ probability < .53
f. .53 ≤ probability ≤ 1.00
8. (5 points) The following hypothesis test was carried out:
It is known that the population standard deviation (σ) = 15.
A random sample of n = 100 had a sample mean of 88.
Find the P-Value for this hypothesis test.
H0: µ≥;
µ
Answer: Since we know the population standard deviation, we use a Z statistic:
Test Statistic:
Z
X   0 88  90

 1.3333

15
n
100
P-Value = P(Z < -1.333) = .5-.4082 = .0918
9. (5 points) A researcher calculated the following confidence interval for a population mean: [21.0, 29.0].
The confidence interval was based on a sample of size 45, a sample mean of 25.0, and a known population
standard deviation of 10.417. What α did the researcher use to calculate the confidence interval?
Answer:
e  Z
2

n
; 4.0  Z
10.417
; Z  2.5758
2
2
45
α/2 = .005, so α = .01
10. An independent testing agency has been contracted to determine whether an experimental (new)
formulation of gasoline improves fuel economy. The original (old) formulation is tested in 200 cars giving
a sample mean of 18.5 mpg and a standard deviation of 4.6 mpg. The experimental formulation is tested in
100 cars, giving a sample mean of 19.34 mpg and a standard deviation of 5.2 mpg. The (partial) results of
a hypothesis test ( = .05) are shown below:
Hypothesis Tests
---------------Sample means = 18.5 and 19.34
Sample standard deviations = 4.6 and 5.2
Sample sizes = 200 and 100
Null Hypothesis: difference between means = 0.0
Alternative: less than
Computed t statistic = -1.4266
P-Value = 0.0773714
(Equal variances assumed).
10a. (4 points) Restate the hypotheses in the context of the problem. In other words, restate the null
and alternative hypotheses in terms of :
µOLD: “average mpg using the old formulation”, and
µNEW: “average mpg using the new formulation”.
Answer:
H 0 : OLD   NEW  0
H1 : OLD   NEW  0
OR
H 0 : OLD   NEW
H1 : OLD   NEW
10b. (4 points) Using an α of 5.0%, use the reported P-Value to state the conclusions of the
hypothesis test in the context of the problem.
Answer: Because the P-Value is greater than α = .05, we cannot reject the null hypothesis.
"There is insufficient statistical evidence to conclude that the average mpg using the new
experimental formulation is greater than the average mpg using the old, original formulation.
10c. (3 points) If the average mpg is exactly the same with either formulation, what is the
probability that we would get a sample result where:
X OLD  X NEW  0.84
Answer: This is just the P-Value: P-Value = 0.0773714
11. (4 points) A mutual fund manager would like to investigate whether companies that are “employee
friendly” provide higher stock returns. She identifies 50 "employee friendly" firms, and measures the 5year stock return for each. For each of the 50 "employee friendly" firms, she then identifies a "nonemployee friendly" industry peer (a similar company in the same industry), and measures the 5-year stock
return for each of those firms. She then carries out the following hypothesis test:
H0: µ1 - µ2≤
µ1 - µ2
where µ1 : the average 5-year stock return of employee friendly firms,
µ2the average 5-year stock return of non-employee friendly firms
Using the same data (5-year stock returns for the same 100 firms) briefly describe an alternative
approach to this hypothesis test that would provide a more powerful test.
Answer: This is an example of a “related samples” situation. This allows us to define a new
variable such as:
DIFFERENCE = return of employee friendly firm – return of non-employee friendly
industry peer
This would give us 50 observations for the variable DIFFERENCE.
We would then just do a one sample hypothesis test such as:
H0: µDIFFERENCE ≤0.0
H1: µDIFFERENCE > 0.0
The rejection of the null hypothesis would allow the same conclusion as the other approach,
but this is a more powerful test… when the null hypothesis is false, it is more likely that we
would reject the null hypothesis in the related samples approach.