Ramp metering is a traffic engineering idea that requires cars entering a freeway to stop for a certain period of time before joining the traffic flow. The theory is that ramp metering controls the number of cars on the freeway and the number of cars accessing the freeway, resulting in a freer flow of cars, which ultimately results in faster travel times. To test whether ramp metering is effective in reducing travel times, engineers conducted an experiment in which a section of freeway had ramp meters installed on the on-ramps. The response variable for the study was speed of the vehicles. A random sample of 15 cars on the highway for a Monday at 6 p.m. with the ramp meters on and a second random sample of 15 cars on a different Monday at 6 p.m. with the meters off resulted in the following speeds (in miles per hour).
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Does there appear to be a difference in the speeds?
A.Yes, the Meters Off data appear to have higher speeds.
B.Yes, the Meters On data appear to have higher speeds.
C.No, the box plots do not show any difference in speeds.
Are there any outliers?
A.Yes, there appears to be a high outlier in the Meters On data.
B.No, there does not appear to be any outliers.
C.Yes, there appears to be a low outlier in the Meters On data.
D.Yes, there appears to be a high outlier in the Meters Off data.
Are the ramp meters effective in maintaining a higher speed on the freeway? Use the alphaαequals=0.01 0.01 level of significance. State the null and alternative hypotheses. Choose the correct answer below.
Determine the P-value for this test.
Choose the correct conclusion
A researcher wanted to determine if carpeted rooms contain more bacteria than uncarpeted rooms. The table shows the results for the number of bacteria per cubic foot for both types of rooms.
State the null and alternative hypotheses. Let population 1 be carpeted rooms and population 2 be uncarpeted rooms.
Determine the P-value for this hypothesis test.(round to 3 decimals)
State the appropriate conclusion. Choose the correct answer below.
The data is
Carpeted: 15.3,12.9,10.2,6.9,15.6,12.7,10.6,14.6
Uncarpeted;8.7,10,11.2,10.7,14,6.9,6.4,11.1
In: Statistics and Probability
Do a two-sample test for equality of means assuming unequal
variances. Calculate the p-value using Excel.
(a-1) Comparison of GPA for randomly chosen
college juniors and seniors:
x¯1x¯1 = 4.75, s1 = .20, n1
= 15, x¯2x¯2 = 5.18, s2 = .30,
n2 = 15, α = .025, left-tailed
test.
(Negative values should be indicated by a minus sign. Round
down your d.f. answer to the nearest whole number and
other answers to 4 decimal places. Do not use "quick" rules for
degrees of freedom.)
| d.f. = | |
| t-calculated = | |
| p-value = | |
| t-critical = | |
(a-2) Based on the above data choose the correct
decision.
Do not reject the null hypothesis
Reject the null hypothesis
(b-1) Comparison of average commute miles for
randomly chosen students at two community colleges:
x¯1x¯1 = 25, s1 = 5, n1 =
22, x¯2x¯2 = 33, s2 = 7, n2
= 19, α = .05, two-tailed test.
(Negative values should be indicated by a minus sign. Round
down your d.f. answer to the nearest whole number and
other answers to 4 decimal places. Do not use "quick" rules for
degrees of freedom.)
| d.f. = | |
| t-calculated = | |
| p-value = | |
| t-critical = | +/- |
(b-2) Based on the above data choose the correct
decision.
Reject the null hypothesis
Do not reject the null hypothesis
(c-1) Comparison of credits at time of graduation
for randomly chosen accounting and economics students:
x¯1x¯1 = 150, s1 = 2.8, n1
= 12, x¯2x¯2 = 143, s2 = 2.7,
n2 = 17, α = .05, right-tailed
test.
(Negative values should be indicated by a minus sign. Round
down your d.f. answer to the nearest whole number and
other answers to 4 decimal places. Do not use "quick" rules for
degrees of freedom.)
| d.f. = | |
| t-calculated = | |
| p-value = | |
| t-critical = | |
(c-2) Based on the above data choose the correct
decision.
Reject the null hypothesis
Do not reject the null hypothesis
In: Statistics and Probability
Do a two-sample test for equality of means assuming unequal
variances. Calculate the p-value using Excel.
(a-1) Comparison of GPA for randomly chosen
college juniors and seniors:
x¯1x¯1 = 4.75, s1 = .20, n1
= 15, x¯2x¯2 = 5.18, s2 = .30,
n2 = 15, α = .025, left-tailed
test.
(Negative values should be indicated by a minus sign. Round
down your d.f. answer to the nearest whole number and
other answers to 4 decimal places. Do not use "quick" rules for
degrees of freedom.)
| d.f. = | |
| t-calculated = | |
| p-value = | |
| t-criticaln = | |
(a-2) Based on the above data choose the correct
decision.
Do not reject the null hypothesis
Reject the null hypothesis
(b-1) Comparison of average commute miles for
randomly chosen students at two community colleges:
x¯1x¯1 = 25, s1 = 5, n1 =
22, x¯2x¯2 = 33, s2 = 7, n2
= 19, α = .05, two-tailed test.
(Negative values should be indicated by a minus sign. Round
down your d.f. answer to the nearest whole number and
other answers to 4 decimal places. Do not use "quick" rules for
degrees of freedom.)
| d.f. = | |
| t-calculated = | |
| p-value = | |
| t-critical = | |
(b-2) Based on the above data choose the correct
decision.
Reject the null hypothesis
Do not reject the null hypothesis
(c-1) Comparison of credits at time of graduation
for randomly chosen accounting and economics students:
x¯1x¯1 = 150, s1 = 2.8, n1
= 12, x¯2x¯2 = 143, s2 = 2.7,
n2 = 17, α = .05, right-tailed
test.
(Negative values should be indicated by a minus sign. Round
down your d.f. answer to the nearest whole number and
other answers to 4 decimal places. Do not use "quick" rules for
degrees of freedom.)
| d.f. = | |
| t-calculated = | |
| p-value = | |
| t-critical = | |
(c-2) Based on the above data choose the correct
decision.
Reject the null hypothesis
Do not reject the null hypothesis
In: Statistics and Probability
In conxt to supplychain, Faw Motors, Inc., was incorporated in Volkswagen on July 01, 2003. It has 4 plants across the China that design, manufacture, and market earth moving, construction, and materials handling equipment. It also manufactures engines for earthmoving vehicles and tractor-trailers.
Faw Motors products are distributed worldwide. Net income last year totaled $350,000,000. Faw Motors has developed a “Transportation Quality” program in order to reduce shipping damages to its equipment and to ensure its just-in-time production and inventory system. The program consists of two parts. The first part ensures proper lifting and tie-down provisions by working with engineers in the design process. The second part focuses on internal practices to prepare the product for shipment.
The chief transportation quality engineer has developed a carrier certification program for both inbound and outbound freight. The program establishes standards requiring the carrier to adhere to 100 percent performance. Use of fewer certified carriers increases the amount of business given to each one. The price is obtained through competitive bidding. It is a function of the travel distance and the weight and density of the shipment.
At the present time, Faw Motors is considering one of three carriers to add to its list of certified carriers.
‘Carrier X’ has 10,000 trucks and a claim rate of 1.5 percent payment to revenue. The company’s pickup/delivery time meets the industry average of four days to transport from Beijing to Hong Kong.
‘Carrier Y’ implements a quality program for its 9,000 trucks to meet on time delivery. It has a 1 percent claim rate.
‘Carrier Z’ has 9,500 trucks and an excellent safety record, but it has not met the average pickup/delivery time. Its claim rate is 1 percent. (See Exhibit A for price estimates.)
EXHIBIT A
Price Estimatesper ton-miles (PPTM):
Carrier X: PPTM $1.05
Carrier Y: PPTM$1.15
Carrier Z: PPTM$0.95
Requirement:
a) Develop a checklist of items that should be considered when selecting a carrier.
b) What are the advantages of certifying the carriers?
c) Is price the most important factor in evaluating carriers? Justify your answer with an example.
d) What are the key factors regarding Faw’s carrier needs?
In: Economics
Mountain Distribution has decided to analyze the profitability of five new customers. The company has the following activities:
|
Activity |
Cost Driver Rate |
|
Order taking |
$80 per purchase order |
|
Customer visits |
$80 per customer visit |
|
Deliveries |
$4.00 per delivery mile travelled |
|
Product handling |
$0.85 per case sold |
|
Expedited deliveries |
$335 per expedited delivery. |
It buys bottled water at $12.20 per case and sells to retail customers at a list price of $14.50
per case. Data pertaining to the five customers are:
|
Customer |
|||||
|
P |
Q |
R |
S |
T |
|
|
Cases sold |
2,160 |
8,820 |
60,800 |
31,900 |
4,200 |
|
List selling price |
$14.50 |
$14.50 |
$14.50 |
$14.50 |
$14.50 |
|
Actual selling price |
$14.50 |
$14.22 |
$13.40 |
$14.02 |
$13.02 |
|
Number of purchase orders |
16 |
26 |
34 |
26 |
34 |
|
Number of customer visits |
3 |
5 |
8 |
3 |
5 |
|
Number of deliveries |
14 |
28 |
64 |
38 |
30 |
|
Miles travelled per delivery |
20 |
5 |
4 |
10 |
48 |
|
Number of expedited deliveries |
0 |
0 |
0 |
0 |
3 |
Requirement
|
1. |
Compute the customer-level operating income of each of the five retail customers now being examined (P, Q, R, S, and T). Comment on the results. |
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|
2. |
What insights are gained by reporting both the list selling price and the actual selling price for each customer? |
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|
3. What factors should Mountain Distribution consider in deciding whether to drop one or more of the five customers? Requirement 1. Compute the customer-level operating income of each of the five retail customers now being examined (P, Q, R, S, and T). Comment on the results. Begin by computing the customer-level operating income of each customer. (Enter all balances including zero balances. Use parentheses or a minus sign when entering operating losses. Round all answers to the nearest whole dollar.)
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In: Accounting
In a study of high-achieving high school graduates, the authors
of a report surveyed 837 high school graduates who were considered
"academic superstars" and 436 graduates who were considered "solid
performers." One question on the survey asked the distance from
their home to the college they attended.
Assuming it is reasonable to regard these two samples as random
samples of academic superstars and solid performers nationwide, use
the accompanying data to determine if it is reasonable to conclude
that the distribution of responses over the distance from home
categories is not the same for academic superstars and solid
performers. Use
α = 0.05.
| Distance of College from Home (in miles) | |||||
|---|---|---|---|---|---|
| Student Group | Less than 40 |
40 to 99 |
100 to 199 |
200 to 399 |
400 or More |
| Academic Superstars | 159 | 158 | 143 | 151 | 226 |
| Solid Performers | 104 | 94 | 82 | 67 | 89 |
State the null and alternative hypotheses.
a). H0: The proportions
falling into the distance categories are not all the same for the
two student groups.
Ha: The proportions falling into the
distance categories are the same for the two student groups.
b). H0: Student group and
distance of college from home are not independent.
Ha: Student group and distance of
college from home are independent.
c). H0: The proportions
falling into the distance categories are the same for the two
student groups.
Ha: The proportions falling into the
distance categories are not all the same for the two student
groups.
d). H0: Student group and
distance of college from home are independent.
Ha: Student group and distance of
college from home are not independent.
Calculate the test statistic. (Round your answer to two decimal
places.)
χ2 =
What is the P-value for the test? (Round your answer to
four decimal places.)
P-value =
What can you conclude?
Do not reject H0. There is not enough evidence to conclude that there is an association between student group and distance of college from home.
Reject H0. There is convincing evidence to conclude that the proportions falling into the distance categories are not all the same for the two student groups.
Do not reject H0. There is not enough evidence to conclude that the proportions falling into the distance categories are not all the same for the two student groups.
Reject H0. There is convincing evidence to conclude that there is an association between student group and distance of college from home.
In: Statistics and Probability
Do a two-sample test for equality of means assuming unequal
variances. Calculate the p-value using Excel.
(a-1) Comparison of GPA for randomly chosen
college juniors and seniors:
x⎯⎯1x¯1 = 4.05, s1 = .20,
n1 = 15, x⎯⎯2x¯2 = 4.35, s2
= .30, n2 = 15, α = .025, left-tailed
test.
(Negative values should be indicated by a minus sign. Round
down your d.f. answer to the nearest whole number and
other answers to 4 decimal places. Do not use "quick" rules for
degrees of freedom.)
| d.f. | |
| t-calculated | |
| p-value | |
| t-critical | |
(a-2) Based on the above data choose the correct
decision.
Do not reject the null hypothesis
Reject the null hypothesis
(b-1) Comparison of average commute miles for
randomly chosen students at two community colleges:
x⎯⎯1x¯1 = 19, s1 = 5, n1 =
22, x⎯⎯2x¯2 = 25, s2 = 7,
n2 = 19, α = .05, two-tailed
test.
(Negative values should be indicated by a minus sign. Round
down your d.f. answer to the nearest whole number and
other answers to 4 decimal places. Do not use "quick" rules for
degrees of freedom.)
| d.f. | |
| t-calculated | |
| p-value | |
| t-critical | +/- |
(b-2) Based on the above data choose the correct
decision.
Reject the null hypothesis
Do not reject the null hypothesis
(c-1) Comparison of credits at time of graduation
for randomly chosen accounting and economics students:
x⎯⎯1x¯1 = 144, s1 = 2.8, n1
= 12, x⎯⎯2x¯2 = 143, s2 = 2.7,
n2 = 17, α = .05, right-tailed
test.
(Negative values should be indicated by a minus sign. Round
down your d.f. answer to the nearest whole number and
other answers to 4 decimal places. Do not use "quick" rules for
degrees of freedom.)
| d.f. | |
| t-calculated | |
| p-value | |
| t-critical | |
(c-2) Based on the above data choose the correct
decision.
Do not reject the null hypothesis
Reject the null hypothesis
In: Statistics and Probability
NORMAL DISTRIBUTION
1A.
A company has a policy of retiring company cars; this policy
looks at number of miles driven, purpose of trips, style of car and
other features. The distribution of the number of months in service
for the fleet of cars is bell-shaped and has a mean of 58 months
and a standard deviation of 4 months. Using the 68-95-99.7 rule,
what is the approximate percentage of cars that remain in service
between 66 and 70 months?
Do not enter the percent symbol.
ans = %
1B.
The Acme Company manufactures widgets. The distribution of
widget weights is bell-shaped. The widget weights have a mean of 45
ounces and a standard deviation of 10 ounces.
Use the Standard Deviation Rule, also known as the Empirical
Rule.
Suggestion: sketch the distribution in order to answer these
questions.
a) 95% of the widget weights lie between and
b) What percentage of the widget weights lie between 15 and 65
ounces? %
c) What percentage of the widget weights lie above 35
? %
1C.
If the distribution of weight of newborn babies in Maryland is
normally distributed with a mean of 3.56 kilograms and a standard
deviation of 0.68 kilograms, find the weights that correspond to
the following z-scores. Round your answers to the nearest tenth, if
necessary.
(a) z = -1.2
kilograms
(b) z = 0.94
kilograms
1D.
A doctor measured serum HDL levels in her patients, and found
that they were normally distributed with a mean of 64.7 and a
standard deviation of 3.6. Find the serum HDL levels that
correspond to the following z-scores. Round your answers to the
nearest tenth, if necessary.
(a) z = -1.25
(b) z = 1.54
1E.
The average resting heart rate of a population is 88 beats per
minute, with a standard deviation of 13 bpm. Find the z-scores that
correspond to each of the following heart rates. Round your answers
to the nearest hundredth, if necessary.
(a) 116 bpm
z =
(b) 73 bpm
z =
1F.
The widths of platinum samples manufactured at a factory are
normally distributed, with a mean of 1.2 cm and a standard
deviation of 0.5 cm. Find the z-scores that correspond to each of
the following widths. Round your answers to the nearest hundredth,
if necessary.
(a) 2 cm
z =
(b) 1 cm
z =
In: Statistics and Probability
A regional express delivery service company recently conducted a study to investigate the relationship between the cost of shipping a package ($), the package weight (in pound) and the distance shipped (in miles). Twenty packages were randomly selected from among the large number received for shipment, and a detailed analysis of the shipping cost was conducted for each package. The data for this sample observations are given in the file Assignment 4 S1 2020.XLS.
a. Estimate a simple linear regression model involving shipping cost and package weight. Interpret the slope coefficient of the least squares line as well as the computed value of ? 2 . [4 marks]
b. Add another explanatory variable–distance shipped–to the regression model in part a. Estimate and interpret this expanded model. How does the ? 2value for this multiple regression model compare to that of the simple regression model estimated in part a? [5 marks]
c. Use the F test to determine the overall significance of the regression relationship for the expanded model. What is the conclusion at the 0.01 level of significance? [4 marks]
d. Use the t test to determine the significance of each independent variable. What is the conclusion for each test at the 0.01 level of significance? [4 marks]
Data
| Cost of Shipment and Potentially Relevant Data | ||
| Cost_of_Shipment | Package_Weight | Distance_Shipped |
| $3.30 | 4.10 | 95 |
| $2.00 | 0.30 | 160 |
| $11.00 | 5.10 | 240 |
| $2.60 | 5.90 | 47 |
| $1.90 | 4.50 | 53 |
| $8.00 | 3.50 | 250 |
| $15.50 | 7.00 | 260 |
| $5.00 | 2.40 | 209 |
| $1.00 | 0.60 | 100 |
| $4.40 | 0.75 | 280 |
| $6.00 | 6.20 | 115 |
| $1.70 | 1.10 | 90 |
| $14.50 | 6.50 | 240 |
| $14.00 | 7.50 | 190 |
| $9.20 | 6.60 | 160 |
| $1.10 | 2.70 | 45 |
| $12.10 | 8.10 | 160 |
| $1.50 | 0.70 | 80 |
| $8.00 | 4.40 | 202 |
| $3.90 | 3.20 | 145 |
| $4.40 | 0.75 | 280 |
| $16.50 | 7.20 | 280 |
| $15.50 | 7.00 | 250 |
| $14.00 | 7.50 | 190 |
| $3.30 | 4.10 | 95 |
| $2.20 | 1.50 | 160 |
| $11.00 | 5.10 | 240 |
| $1 | 0.6 | 100 |
| $4 | 0.75 | 280 |
| $2 | 0.7 | 80 |
| $8 | 4.4 | 202 |
| $2 | 4.5 | 52 |
| $8.00 | 3.2 | 240 |
| $15.50 | 7.6 | 270 |
| $5.00 | 2.5 | 211 |
| $1.00 | 7 | 98 |
| $8.00 | 4.4 | 202 |
| $3.90 | 3.2 | 145 |
| $4.40 | 0.75 | 280 |
| $5.00 | 2.4 | 209 |
In: Statistics and Probability
Do a two-sample test for equality of means assuming unequal
variances. Calculate the p-value using Excel.
(a-1) Comparison of GPA for randomly chosen
college juniors and seniors:
x¯1x¯1 = 4, s1 = .20, n1 =
15, x¯2x¯2 = 4.25, s2 = .30,
n2 = 15, α = .025, left-tailed
test.
(Negative values should be indicated by a minus sign. Round
down your d.f. answer to the nearest whole number and
other answers to 4 decimal places. Do not use "quick" rules for
degrees of freedom.)
| d.f. | |
| t-calculated | |
| p-value | |
| t-critical | |
(a-2) Based on the above data choose the correct
decision.
Do not reject the null hypothesis
Reject the null hypothesis
(b-1) Comparison of average commute miles for
randomly chosen students at two community colleges:
x¯1x¯1 = 17, s1 = 5, n1 =
22, x¯2x¯2 = 21, s2 = 7, n2
= 19, α = .05, two-tailed test.
(Negative values should be indicated by a minus sign. Round
down your d.f. answer to the nearest whole number and
other answers to 4 decimal places. Do not use "quick" rules for
degrees of freedom.)
| d.f. | |
| t-calculated | |
| p-value | |
| t-critical | +/- |
(b-2) Based on the above data choose the correct
decision.
Do not reject the null hypothesis
Reject the null hypothesis
(c-1) Comparison of credits at time of graduation
for randomly chosen accounting and economics students:
x¯1x¯1 = 141, s1 = 2.8, n1
= 12, x¯2x¯2 = 138, s2 = 2.7,
n2 = 17, α = .05, right-tailed
test.
(Negative values should be indicated by a minus sign. Round
down your d.f. answer to the nearest whole number and
other answers to 4 decimal places. Do not use "quick" rules for
degrees of freedom.)
| d.f. | |
| t-calculated | |
| p-value | |
| t-critical | |
(c-2) Based on the above data choose the correct
decision.
Do not reject the null hypothesis
Reject the null hypothesis
In: Statistics and Probability