One late Friday afternoon your obnoxious boss comes into your office as you are about to leave, and shows you 26 observations that he believes to be related. Y, he believes, is the dependent variable and X1 is the independent variable. He also thinks there is a 2nd order polynomial relationship in the data ( Y = B1X1 +B2X12 + B0 ), and, as you casually view the data, you tend to agree. He insists that the determination of B1, B2 and B0 is far more important than your Friday afternoon gathering of young-urban-millennial-professionals (YUMPS) at a local watering hole. So, you perform the analysis using the scatter diagram and Trendline tool in Excel. Then you quietly exit for the YUMPS gathering.
Now, it is Monday morning. You want to use the assumption of a 2nd order polynomial to find the exact values of B1, B2 and B0 . (Hint: generate a new variable that fits the assumed polynomial model and then use the regression tool in Excel)
a) Use a regression tool in Excel to determine the exact values of B1, B2 and B0 .
b) Is the regression model significant? Use alpha 0.05.
Data:
| X1 | X2 | Y |
| 68,067 | 4,633,116,489 | 1,598,278,294 |
| 70,103 | 4,914,430,609 | 1,695,432,796 |
| 76,370 | 5,832,376,900 | 2,011,698,722 |
| 86,686 | 7,514,462,596 | 2,592,259,645 |
| 86,759 | 7,527,124,081 | 2,597,646,596 |
| 91,805 | 8,428,158,025 | 2,907,666,308 |
| 92,306 | 8,520,397,636 | 2,940,058,192 |
| 93,731 | 8,785,500,361 | 3,030,806,599 |
| 100,913 | 10,183,433,569 | 3,512,923,429 |
| 102,199 | 10,444,635,601 | 3,603,809,834 |
| 109,399 | 11,968,141,201 | 4,129,345,241 |
| 113,430 | 12,866,364,900 | 4,439,623,027 |
| 118,133 | 13,955,405,689 | 4,814,663,900 |
| 122,820 | 15,084,752,400 | 5,203,806,895 |
| 123,417 | 15,231,755,889 | 5,255,000,229 |
| 123,054 | 15,142,286,916 | 5,224,742,756 |
| 127,860 | 16,348,179,600 | 5,640,704,277 |
| 132,868 | 17,653,905,424 | 6,091,148,092 |
| 131,160 | 17,202,945,600 | 5,935,115,092 |
| 132,132 | 17,458,865,424 | 6,022,925,591 |
| 132,583 | 17,578,251,889 | 6,064,201,852 |
| 136,859 | 18,730,385,881 | 6,462,049,499 |
| 140,562 | 19,757,675,844 | 6,816,147,332 |
| 144,594 | 20,907,424,836 | 7,212,915,055 |
| 141,493 | 20,020,269,049 | 6,906,705,120 |
| 139,465 | 19,450,486,225 | 6,710,038,170 |
In: Math
Starting salaries of 95 college graduates who have taken a
statistics course have a mean of $44,915. Suppose the distribution
of this population is approximately normal and has a standard
deviation of $10,596.
Using an 85% confidence level, find both of the following:
(NOTE: Do not use commas nor dollar signs in your answers.)
(a) The margin of error:
(b) The confidence interval for the mean μ:
In: Math
Consider the sample space of all people living in the United States, and within that sample space, the following two events.
??=people who are college graduates=people who are unemployedA=people who are college graduatesB=people who are unemployed
Suppose the following statements describe probabilities regarding these two events. Which of the statements describe conditional probabilities? Select all that apply.
The probability is 4.1% that a college graduate is unemployed.
Twenty‑six point two percent of unemployed people are college graduates.
There is a 6.1% chance that a randomly chosen person is unemployed.
One point six percent of people living in the United States are unemployed college graduates.
Of people living in the United States, 39.4% are college graduates.
There is a 43.9% probability that a person is a college graduate or unemployed.
In: Math
You may need to use the appropriate appendix table or technology to answer this question.
A 95% confidence interval for a population mean was reported to be 151.29 to 160.71. If
σ = 15,
what sample size was used in this study? (Round your answer to the nearest integer.)
In: Math
In a random sample of 32 criminals convicted of a certain crime, it was determined that the mean length of sentencing was 61 months, with a standard deviation of 12 months. Construct and interpret a 90% confidence interval for the mean length of sentencing for this crime.
We can be 90% confident that the mean length of sentencing for the crime is between ______ and ______ months.
In: Math
Below are monthly rents paid by 30 students who live off campus. 700 700 700 900 670 540 660 1,000 710 590 690 625 530 710 620 630 820 885 570 590 730 660 680 470 700 770 790 810 690 685
(a) Using Excel, find the mean, median, mode, and standard deviation. (Round your answers to 2 decimal places.)
In: Math
Case Study II: Ski Right
After retiring as a physician, Bob Guthrie became an avid downhill skier on the steep slopes of the Utah Rocky Mountains. As an amateur inventor, Bob was always looking for something new. With the recent deaths of several celebrity skiers, Bob knew he could use his creative mind to make skiing safer and his bank account larger. He knew that many deaths on the slopes were caused by head injuries. Although ski helmets have been on the market for some time, most skiers considered them boring and basically ugly. As a physician, Bob knew that some type of new ski helmet was the answer.
Bob’s biggest challenge was to invent a helmet that was attractive, safe, and fun to wear. Multiple colors, using the latest fashion designs, would be a must. After years of skiing, Bob knew that many skiers believed that how you looked on the slopes was more important than how you skied. His helmets would have to look good and fit in with current fashion trends. But attractive helmets were not enough. Bob had to make the helmets fun and useful. The name of the new ski helmet, Ski Right, was sure to be a winner. If Bob could come up with a good idea, he believed that there was a 20% chance that the market for the Ski Right helmet would be excellent. The chance of a good market should be 40%. Bob also knew that the market for his helmet could be only average (30% chance) or even poor (10% chance).
The idea of how to make ski helmets fun and useful came to Bob on a gondola ride to the top of a mountain. A busy executive on the gondola ride was on his cell phone, trying to complete a complicated merger. When the executive got off the gondola, he dropped the phone, and it was crushed by the gondola mechanism. Bob decided that his new ski helmet would have a built-in cell phone and an AM/FM stereo radio. All the electronics could be operated by a control pad worn on a skier’s arm or leg.
Bob decided to try a small pilot project for Ski Right. He enjoyed being retired and didn’t want a failure to cause him to go back to work. After some research, Bob found Progressive Products (PP). The company was willing to be a partner in developing the Ski Right and sharing any profits. If the market was excellent, Bob would net $5,000. With a good market, Bob would net $2,000. An average market would result in a loss of $2,000, and a poor market would mean Bob would be out $5,000.
Another option for Bob was to have Leadville Barts (LB) make the helmet. The company had extensive experience in making bicycle helmets. PP would then take the helmets made by LB and do the rest. Bob had a greater risk. He estimated that he could lose $10,000 in a poor market or $4,000 in an average market. A good market for Ski Right would result in a $6,000 profit for Bob, and an excellent market would mean a $12,000 profit.
A third option for Bob was to use TalRad (TR), a radio company in Tallahassee, Florida. TR had extensive experience in making military radios. LB could make the helmets, and PP could do the rest. Again, Bob would be taking on greater risk. A poor market would mean a $15,000 loss, and an average market would mean a $10,000 loss. A good market would result in a net profit of $7,000 for Bob. An excellent market would return $13,000.
Bob could also have Celestial Cellular (CC) develop the cell phones. Thus, another option was to have CC make the phones and have PP do the rest of the production and distribution. Because the cell phone was the most expensive component of the helmet, Bob could lose $30,000 in a poor market. He could lose $20,000 in an average market. If the market was good or excellent, Bob would see a net profit of $10,000 or $30,000, respectively.
Bob’s final option was to forget about PP entirely. He could use LB to make the helmets, CC to make the phones, and TR to make the AM/FM stereo radios. Bob could then hire some friends to assemble everything and market the finished Ski Right helmets. With this final alternative, Bob could realize a net profit of $55,000 in an excellent market. Even if the market were just good, Bob would net $20,000. An average market, however, would mean a loss of $35,000. If the market was poor, Bob would lose $60,000.
Discussion Questions
1. What do you recommend?
2. What is the opportunity loss for this problem?
3. Compute the expected value of perfect information.
In: Math
A. Define the following terms:
-Correlation
-Causation
B. How do we know when 2 variables are correlated?
C. Explain the following: “Correlation does not equal Causation”
D. What requirements must be met to satisfy Causation?
In: Math
A health psychologist conducted an experiment in which participants watched a film that either did or did not include a person being injured because of not wearing a seatbelt. A week later, as part of a seemingly different study, these same participants reported how important they thought it was to wear seatbelts (higher scores = greater importance). The 16 participants who had seen the injury film gave a mean rating of 8.9, with a standard deviation of 2.1. The 16 participants in the control condition had a mean of 7.0 with a standard deviation of 2.4.
In: Math
Consider the continuous uniform distribution.
a. Why would you use this to model the distribution of ages of people between the ages of 0 and 50 years in the United States?
b. Why would this likely not be a good distribution to use beyond about 50 years of age?
c. Is this distribution characterized as a PMF or PDF? Explain.
d. What is the mean age? Show your calculation.
e. What is the standard deviation? Show your calculation.
f. What is the probability that a person will be exactly 30 years of age? Explain your answer.
g. What is the probability that a person will be between ages 30 and 31? Show your calculation.
In: Math
A process used in filling bottles with soft drink results in net weights that are normally distributed, with a mean of 2 liters and a standard deviation of 0.05 liter. Bottles filled to less than 95% of the listed net weight can make the manufacturer subject to penalty by the state office of consumer affairs; bottles filled above 2.10 liters may cause excess spillage upon opening.
What proportion of the bottles will contain
Between 1.90 and 2.0 liters?
At least 2.10 liters?
Ninety-nine percent of the bottles would be expected to contain at least how much soft drink?
The central 40% of bottles will lye between what two liters?
The machine is considered a bargain if it is unlikely to require major repair before the sixth year.
Find the probability that a major repair occurs after 6 years.
Find the probability that a major repair occurs in the first year.
Does the machine seem to be a bargain?
Find the median time before a major repair.
Note: For (a), (b) and (d), don’t forget your concluding statements! For (c), you must give an explanation.
In: Math
1. Parkruns are free 5km timed runs usually run on weekends in different cities, towns and suburbs around the world. It is suspected that they are competitive and hence faster than 5km family fun runs which require payment for registration. Parkrun finishers do not get medals whereas all family fun run finishers get medals. The summary statistics below are from two random samples of 10 park runners and 10 5km family fun runners in Bellville. The races were run on the same route and on 2 separate days but with identical weather conditions. The summary statistics are for the recorded finishing times (in minutes) of the runners. park runners family fun runners Sample Mean: 20.58 25.67 Sample standard deviation : 3.5 5.7 (a) Is there evidence that the mean finishing time differs between the park runners and family fun runners? Perform an appropriate statistical test. (b) Construct a 95% confidence interval for the difference between the population means of the two groups. Compare your results to your conclusions in (a). (c) What assumption(s) are necessary for performing the hypothesis tests and constructing the confidence interval above? Hint: Use a 5% significance level for the F-test but report a p-value for the t-test
In: Math
Ch.8 #5
Consider the following time series data.
| Quarter | Year 1 | Year 2 | Year 3 |
| 1 | 4 | 6 | 7 |
| 2 | 2 | 3 | 6 |
| 3 | 3 | 5 | 6 |
| 4 | 5 | 7 | 8 |
1) Use a multiple regression model with dummy variables as follows to develop an equation to account for seasonal effects in the data. Qtr1 = 1 if Quarter 1, 0 otherwise; Qtr2 = 1 if Quarter 2, 0 otherwise; Qtr3 = 1 if Quarter 3, 0 otherwise.
If required, round your answers to three decimal places. For subtractive or negative numbers use a minus sign even if there is a + sign before the blank. (Example: -300) If the constant is "1" it must be entered in the box. Do not round intermediate calculation.
Value = ________ + __________ Qtr1 + ___________ Qtr2 + ___________ Qtr3
2) Compute the quarterly forecasts for next year based on the model you developed in part (b). If required, round your answers to three decimal places. Do not round intermediate calculation.
Quarter 1 forecast _____________
Quarter 2 forecast_____________
Quarter 3 forecast_____________
Quarter 4 forecast_____________
3) Use a multiple regression model to develop an equation to account for trend and seasonal effects in the data. Use the dummy variables you developed in part (b) to capture seasonal effects and create a variable t such that t = 1 for Quarter 1 in Year 1, t = 2 for Quarter 2 in Year 1,… t = 12 for Quarter 4 in Year 3.
If required, round your answers to three decimal places. For subtractive or negative numbers use a minus sign even if there is a + sign before the blank. (Example: -300)
Value = __________ + __________ Qtr1 + __________ Qtr2 + ___________ Qtr3 + ________ t
4) Compute the quarterly forecasts for next year based on the model you developed in part (d).
Quarter 1 forecast _____________
Quarter 2 forecast_____________
Quarter 3 forecast_____________
Quarter 4 forecast_____________
5) Is the model you developed in part (b) or the model you developed in part (d) more effective?
| If required, round your intermediate calculations and final answer to three decimal places. |
| Model developed in part (b) | Model developed in part (d) | |
| MSE |
Justify your answer.
In: Math
This exercise has two parts. Please respond to both. The calculation formulas and procedures for both parts are included in the mini-lecture titled “Rates and Percentages” Part One The Mayors of two small towns that adjoin a large community have engaged in an argument on which of their cities is safer, with respect to its crime rate. City A has a residential population of 123,000 people and reported 185 violent crimes. City B has a residential population of 84,000 people and reported 135 violent crimes. Answer the following questions. 1. What is the reported crime rate per 100,000 residents in both cities? 2. Based on your responses to the previous question, which city has the lowest crime rate? Part Two Last year. 135 vehicles were reported stolen in the City of Bigton. This year, 185 vehicles were reported stolen in the City of Bigton. What is the percent change in reported stolen vehicles from last year to this year in the City of Bigton?
In: Math
1. The waiting times (in minutes) of a random sample of 21 people at a bank have a sample standard deviation of 3.5 minutes. Construct a confidence interval for the population variance sigma squared and the population standard deviation sigma. Use a 99 % level of confidence. Assume the sample is from a normally distributed population. What is the confidence interval for the population variance sigma squared?
2.You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 55 home theater systems has a mean price of $113.00. Assume the population standard deviation is $15.20.
In: Math