BA 275
Winter 2007
Midterm Examination #2
B
BA 275 Quantitative Business Methods
Winter 2007
Midterm Examination #2
February 26, 2007
Answer Key Version B
Name (please print)
Section (circle one)
12noon – 1:50pm
2:00 – 3:50pm
4:00 – 5:50pm
*** 5 points will be deducted if either your name or section is not clearly shown. ***
*** No credit will be given if I cannot find your answer within the answer box. ***
*** Points will be deducted if multiple answers appear within the answer box. ***
*** 5 points will be deducted if an index card is used but not turned in with the exam. ***
Question 1. The Survey of Study Habits and Attitudes (SSHA) is a psychological test that
measures the motivation, attitude toward school, and study habits of students. The mean score
for U.S. college students is about 115, and the population standard deviation is about 30. A
teacher suspects that older students (30 years or older) have better attitudes toward school and
wishes to test H0:  = 115 vs. Ha:  > 115. To test the hypothesis, she decides to use a sample of
n = 30, and to reject the null hypothesis H0 if the sample mean X > 128.
A. (2 points) Find the probability of a Type I error.
128  115
P( reject H0 given H0 is true ) = P( X > 128 given  = 115 ) = P( Z >
) = P(
=
30 / 30
Z > 2.37 ) = 1 – 0.9911 = 0.0089.
B. (2 points) Find the probability of a Type II error when  = 135.
P( fail to reject H0 when H0 is false ) = P( X < 128 given  = 135 ) = P( Z <
128  135
=
) = P( Z < –1.28 ) = 0.1003.
30 / 30
C. (2 points) Find the power of the test when  = 135.
Power = 1 –  = 1 – 0.1003 = 0.8997
D. (2 points) Suppose the sample mean x turns out to be 131.2. Find the p-value of the test.
P( observing X 131.2 or more extreme when H0 is true ) = P( X >131.2 given  =
131.2  115
p-value
115 ) = P( Z >
) = P( Z > 2.96 ) = 1 – 0.9985 = 0.0015.
30 / 30
Hsieh, P.-H.
1
BA 275
Winter 2007
Midterm Examination #2
B
Question 2. High blood pressure, if untreated, can lead to increased risk of stroke and heart
attack. A common definition of hypertension is diastolic blood pressure of 90 or more.
2 – A. (2 points) State the null and alternative hypotheses for a physician who checks your blood
pressure.
 = 90
H0 :
A patient has no high blood pressure
 > 90
Ha :
A patient has high blood pressure, so a physician checks to see if
hypertension can be confirmed.
2 – B. (2 points) Which of the following describes Type I error?
A) A patient does not have high blood pressure and the blood pressure exam does not show
hypertension.
B) A patient has high blood pressure and the blood pressure exam shows hypertension.
C) A patient does not have high blood pressure but the blood pressure exam shows
hypertension.
D) A patient has high blood pressure but the blood pressure exam fails to show hypertension.
Type I error = reject H0 given H0 is true = a patient does not have high
Answer
blood pressure but the exam shows hypertension. False alarm.
2 – C. (2 points) Which of the following describes Type II error?
A) A patient does not have high blood pressure and the blood pressure exam does not show
hypertension.
B) A patient has high blood pressure and the blood pressure exam shows hypertension.
C) A patient does not have high blood pressure but the blood pressure exam shows
hypertension.
D) A patient has high blood pressure but the blood pressure exam fails to show hypertension.
Type II error = fail to reject H0 when H0 is false = a patient has high
Answer
blood pressure but the exam says no.
2 – D. (2 points) What are the consequences of Type I and II errors?
A) Type I error leads to unneeded treatment and Type II error leads to untreated hypertension.
B) Both Type I and II errors lead to untreated hypertension.
C) Both Type I and II errors lead to unneeded treatment.
D) Type I error leads to untreated hypertension and Type II error leads to unneeded treatment.
Type I error leads to unneeded treatment and Type II error leads to
Answer
untreated hypertension.
Hsieh, P.-H.
2
BA 275
Winter 2007
Midterm Examination #2
B
Question 3. The mean yield of corn in the United States is about 135 bushels per acre. A survey
of 25 farmers this year gives a sample mean yield of x =138.8 bushels per acre. We want to
know whether this is good evidence that the national mean this year is not 135 bushels per acre.
Assume that the farmers surveyed are a simple random sample from the population of all
commercial corn growers and that the standard deviation of the yield in this population is  = 10
bushels per acre.
3 – A. (2 points) What are the null and alternative hypotheses?
H0 :
 = 135
Ha :
 ≠ 135
3 – B. (2 points) Calculate and report the z statistic (i.e., z score).
138.8  135
Z=
 1.90
Answer
10 / 25
3 – C. (2 points) Define the rejection region or provide the p-value of the test. Assume  = 5%.
Rejection region: reject the null hypothesis if z > 1.96 or z < –1.96
Answer
p-value = P( Z > 1.90 or Z < –1.90 ) = 0.0287 + 0.0287 = 0.0574
3 – D. (2 points) Based on the z statistic or the p-value in Questions 3 – B and 3 – C, do we have
statistically significant evidence to say that the population mean is not 135 bushels per acre?
A). Yes, we do and H0 should be rejected.
B). No, we do not and H0 should be rejected.
C). Yes, we do and H0 should not be rejected.
D). No, we do not and H0 should not be rejected.
Z statistic 1.90 is NOT in the rejection region and the p-value 0.0574 is
Answer
greater than  = 5%. No strong evidence and cannot reject H0.
Hsieh, P.-H.
3
BA 275
Winter 2007
Midterm Examination #2
B
Question 4. A U.S. dime is supposed to weigh 2.268 grams. A random sample of 10 circulated
dimes showed a mean weight of 2.256 grams with a sample standard deviation of 0.026 grams.
Does the sample provide statistically significant evidence for us to claim that the mean weight of
all circulated dimes lower than the specification?
4 – A. (2 points) State the null and alternative hypotheses.
H0 :
 = 2.268
Ha :
 < 2.268
4 – B. (2 points) Which of the following statements regarding the p-value of the test is true?
A). p-value < 0.01
B). 0.01 < p-value < 0.025
C). 0.025 < p-value < 0.05
D). 0.05 < p-value < 0.10
E). p-value > 0.10
Population standard deviation is unknown and the sample size is small,
2.256  2.268
calculate the t statistic t 
 1.46 . Given that t score
0
.
026
/
10
Answer
= –1.46 and the degrees of freedom is n – 1 = 9, one would reject H0 if 
= 0.10 but will not reject H0 if  = 0.05. Hence the p-value must be
between 0.05 and 0.10.
4 – C. (2 points) Given  = 5%, what is the statistical conclusion of the test? (circle one)
Reject H0
Do not reject H0
Question 5. When the background music was slow, the mean amount of bar purchases for a
sample of 20 restaurant patrons was $30.47 with a standard deviation of $15.10. When the
background music was fast, the mean amount of bar purchases for a sample of 15 patrons in the
same restaurant was $21.62 with a standard deviation of $9.50. The owner would like to know if
the true mean amount of bar purchases is higher when the background music is slow, that is, H0:
slow = fast vs. Ha: slow > fast.
5 – A. (2 points) Calculate and report the t statistic (i.e., t score).
30.47  21.62
t=
15.10 2 9.50 2

20
15
 2.12
5 – B. (2 points) Calculate and report the degrees of freedom used in testing H0: slow = fast vs.
Ha: slow > fast?
Degrees of freedom =
Hsieh, P.-H.
Minimum of 20 – 1 or 15 – 1 = 14
4
BA 275
Winter 2007
Midterm Examination #2
B
Question 6. (2 points) Which of the following statements is true?
A). The correlation coefficient measures the fraction of outliers that appear in a scatterplot.
B). The correlation coefficient is a unitless number and must always lie between –1.0 and +1.0,
inclusive.
C). If r is the correlation between X and Y, then –r is the correlation between Y and X.
D). The correlation coefficient equals the proportion of times two variables lie on a straight line.
E). The correlation coefficient will be +1.0 only if all the data lie on a perfectly horizontal
straight line.
The correlation coefficient is a unitless number and must always lie
between –1.0 and +1.0, inclusive.
Answer
Question 7. In a study of 1991 model cars, a researcher found that the fraction of the variation
in the price of cars which was explained by the least-squares regression on horsepower was
about 0.64. For the cars in this study, the correlation between the price of the car and its
horsepower was found to be positive.
7 – A. (2 points) What is the actual value of the correlation?
R-square = r 2 = the fraction of the variation in the dependent variable Y
that has been explained by the independent variable X = 0.64.
Answer
Correlation = R  square  r 2  r  0.8
7 – B. (2 points) Suppose the researcher obtained the following equation for the least-squares
line: price = –6677 + 175  horsepower. Based on the equation, how much a 1991 model car
with horsepower equal to 250 would cost? A point estimate is sufficient. There is no need to
calculate margin of error.
Answer
Hsieh, P.-H.
Estimated price = –6677 + 175  250 = 37073
5