Use the data and Excel to answer this question. It contains the United States Census Bureau’s estimates for World Population from 1950 to 2014. You will find a column of dates and a column of data on the World Population for these years. Generate the time variable t. Then run a regression with the Population data as a dependent variable and time as the dependent variable. Have Excel report the residuals.
(a) Based on the ANOVA table and t-statistics, does the regression appear significant?
(b) Calculate the Durbin-Watson Test statistic. Is there a serial correlation problem with the data? Explain.
(d) What affect might your answer in part (b) have on your conclusions in part (a)?
| Year | Population |
| 1950 | 2,557,628,654 |
| 1951 | 2,594,939,877 |
| 1952 | 2,636,772,306 |
| 1953 | 2,682,053,389 |
| 1954 | 2,730,228,104 |
| 1955 | 2,782,098,943 |
| 1956 | 2,835,299,673 |
| 1957 | 2,891,349,717 |
| 1958 | 2,948,137,248 |
| 1959 | 3,000,716,593 |
| 1960 | 3,043,001,508 |
| 1961 | 3,083,966,929 |
| 1962 | 3,140,093,217 |
| 1963 | 3,209,827,882 |
| 1964 | 3,281,201,306 |
| 1965 | 3,350,425,793 |
| 1966 | 3,420,677,923 |
| 1967 | 3,490,333,715 |
| 1968 | 3,562,313,822 |
| 1969 | 3,637,159,050 |
| 1970 | 3,712,697,742 |
| 1971 | 3,790,326,948 |
| 1972 | 3,866,568,653 |
| 1973 | 3,942,096,442 |
| 1974 | 4,016,608,813 |
| 1975 | 4,089,083,233 |
| 1976 | 4,160,185,010 |
| 1977 | 4,232,084,578 |
| 1978 | 4,304,105,753 |
| 1979 | 4,379,013,942 |
| 1980 | 4,451,362,735 |
| 1981 | 4,534,410,125 |
| 1982 | 4,614,566,561 |
| 1983 | 4,695,736,743 |
| 1984 | 4,774,569,391 |
| 1985 | 4,856,462,699 |
| 1986 | 4,940,571,232 |
| 1987 | 5,027,200,492 |
| 1988 | 5,114,557,167 |
| 1989 | 5,201,440,110 |
| 1990 | 5,288,955,934 |
| 1991 | 5,371,585,922 |
| 1992 | 5,456,136,278 |
| 1993 | 5,538,268,316 |
| 1994 | 5,618,682,132 |
| 1995 | 5,699,202,985 |
| 1996 | 5,779,440,593 |
| 1997 | 5,857,972,543 |
| 1998 | 5,935,213,248 |
| 1999 | 6,012,074,922 |
| 2000 | 6,088,571,383 |
| 2001 | 6,165,219,247 |
| 2002 | 6,242,016,348 |
| 2003 | 6,318,590,956 |
| 2004 | 6,395,699,509 |
| 2005 | 6,473,044,732 |
| 2006 | 6,551,263,534 |
| 2007 | 6,629,913,759 |
| 2008 | 6,709,049,780 |
| 2009 | 6,788,214,394 |
| 2010 | 6,858,584,755 |
| 2011 | 6,935,999,491 |
| 2012 | 7,013,871,313 |
| 2013 | 7,092,128,094 |
| 2014 | 7,169,968,185 |
Thanks id advance! Will try to rate the answer ASAP. Please show your process too :)
In: Statistics and Probability
NOTE THAT
((This should be done by R studio !))
Q: Upload your data as a CSV in R studio, then do
any
cleaning or convert needed for example convert the date in your
table
from character to date and NA identifiers . After
do all these, run a summary statistics
|
Year |
REX |
OilP |
Food exports (% of merchandise exports) |
Ores and metals exports (% of merchandise exports) |
|
1980 |
239.5433424 |
35.52 |
0.09638294 |
0.060083757 |
|
1981 |
240.3102173 |
34 |
0.094079554 |
0.024360528 |
|
1982 |
245.3895131 |
32.38 |
0.128489839 |
0.025668368 |
|
1983 |
242.8677506 |
29.04 |
.. |
.. |
|
1984 |
238.0284197 |
28.2 |
.. |
.. |
|
1985 |
221.878717 |
27.01 |
0.259787311 |
0.116943755 |
|
1986 |
169.6457184 |
13.53 |
.. |
.. |
|
1987 |
144.1934823 |
17.73 |
.. |
.. |
|
1988 |
134.5212315 |
14.24 |
1.371078529 |
0.732151804 |
|
1989 |
136.0536024 |
17.31 |
1.374888969 |
0.834330299 |
|
1990 |
125.5311345 |
22.26 |
0.713126234 |
0.491007478 |
|
1991 |
125.8812467 |
18.62 |
0.526384845 |
0.242750346 |
|
1992 |
118.7733668 |
18.44 |
1.074388363 |
0.548851562 |
|
1993 |
122.2521688 |
16.33 |
0.982275388 |
0.429968062 |
|
1994 |
117.8952881 |
15.53 |
0.673955645 |
0.346686956 |
|
1995 |
114.1213899 |
16.86 |
0.810242733 |
0.567217625 |
|
1996 |
116.3114665 |
20.29 |
0.632336949 |
0.304958406 |
|
1997 |
121.4661302 |
18.86 |
.. |
.. |
|
1998 |
127.1948915 |
12.28 |
1.114818605 |
0.507089276 |
|
1999 |
121.9490893 |
17.44 |
0.930990348 |
0.262574488 |
|
2000 |
123.200674 |
27.6 |
0.538501429 |
0.147164016 |
|
2001 |
125.2424379 |
23.12 |
0.558465111 |
0.201693533 |
|
2002 |
121.5455166 |
24.36 |
0.628539417 |
0.223275991 |
|
2003 |
111.1523893 |
28.1 |
0.835851768 |
0.182707717 |
|
2004 |
103.4682918 |
36.05 |
0.7405123 |
0.172800798 |
|
2005 |
100.5070052 |
50.59 |
0.620831971 |
0.137293785 |
|
2006 |
98.93290899 |
61 |
0.64203501 |
0.219532433 |
|
2007 |
95.96813741 |
69.04 |
0.838923226 |
0.283587719 |
|
2008 |
93.62494305 |
94.1 |
0.744029125 |
0.221986187 |
|
2009 |
100.1652448 |
60.86 |
1.407633083 |
0.232499732 |
|
2010 |
100 |
77.38 |
1.155876888 |
0.154654215 |
|
2011 |
96.57013945 |
107.46 |
0.898301922 |
0.122271232 |
|
2012 |
99.61967144 |
109.45 |
0.860627792 |
0.138455596 |
|
2013 |
102.3680362 |
105.87 |
0.878931429 |
0.403127249 |
|
2014 |
105.3894897 |
96.29 |
1.006265279 |
0.769034983 |
|
2015 |
118.5851177 |
49.49 |
1.798068624 |
1.307540253 |
R ONLY !!
In: Computer Science
In the following problem, check that it is appropriate to use
the normal approximation to the binomial. Then use the normal
distribution to estimate the requested probabilities.
Ocean fishing for billfish is very popular in the Cozumel region of
Mexico. In the Cozumel region about 48% of strikes (while trolling)
resulted in a catch. Suppose that on a given day a fleet of fishing
boats got a total of 26 strikes. Find the following probabilities.
(Round your answers to four decimal places.)
(a) 12 or fewer fish were caught
(b) 5 or more fish were caught
(c) between 5 and 12 fish were caught
In the following problem, check that it is appropriate to use
the normal approximation to the binomial. Then use the normal
distribution to estimate the requested probabilities.
It is estimated that 3.4% of the general population will live past
their 90th birthday. In a graduating class of 730 high school
seniors, find the following probabilities. (Round your answers to
four decimal places.)
(a) 15 or more will live beyond their 90th birthday
(b) 30 or more will live beyond their 90th birthday
(c) between 25 and 35 will live beyond their 90th birthday
(d) more than 40 will live beyond their 90th birthday
Suppose the heights of 18-year-old men are approximately normally distributed, with mean 73 inches and standard deviation 2 inches.
(a) What is the probability that an 18-year-old man selected at
random is between 72 and 74 inches tall? (Round your answer to four
decimal places.)
(b) If a random sample of eleven 18-year-old men is selected, what
is the probability that the mean height x is between 72and
74 inches? (Round your answer to four decimal places.)
(c) Compare your answers to parts (a) and (b). Is the probability
in part (b) much higher? Why would you expect this?
The probability in part (b) is much higher because the standard deviation is larger for the x distribution.
The probability in part (b) is much higher because the mean is smaller for the x distribution.
The probability in part (b) is much higher because the standard deviation is smaller for the x distribution.
The probability in part (b) is much lower because the standard deviation is smaller for the x distribution.The probability in part (b) is much higher because the mean is larger for the x distribution.
In: Statistics and Probability
1. Which of the following are examples of price discrimination? Explain your answers.
a. A cell phone carrier offers unlimited calling on the weekends for all of its customers.
b. Tickets to the student section for all basketball games are $5
c. A restaurant offers a 20% discount for customers who order dinner between 4 and 6 p.m.
d. A music store has a half-price sale on last year's guitars.
e. A well-respected golf instructor charges each customer a fee just under the customer's maximum willingness to pay for lessons.
In: Economics
At two different branches of a department store, pollsters randomly sampled 100 customers at store 1 and 80 customers at store 2, all on the same day. At store 1, the average amount purchased was $41.25 per customer with a sample standard deviation of $24.25. At store 2 the average amount purchased was $45.75 with a sample standard deviation of $34.76.
a) Construct a 95% confidence interval for the mean amount purchased per customer in Store 1 and Store 2.
b) Construct a 95% confidence interval for the difference between the means of purchases per customer of the two stores.
In: Statistics and Probability
At two different branches of a department store, pollsters randomly sampled 100 customers at store 1 and 80 customers at store 2, all on the same day. At store 1, the average amount purchased was $41.25 per customer with a sample standard deviation of $24.25. At store 2 the average amount purchased was $45.75 with a sample standard deviation of $34.76.
a) Construct a 95% confidence interval for the mean amount purchased per customer in Store 1 and Store 2.
b) Construct a 95% confidence interval for the difference between the means of purchases per customer of the two stores.
In: Statistics and Probability
Customers using a self-service soda dispenser take an average of
12 ounces of soda with an SD of 4 ounces. Assume that the amount
would be normally distributed.
What is the probability that a randomly selected customer takes
over 8 ounces of soda?
0.25
0.84
0.95
0.75
0.66
What is the probability that a randomly selected customer takes
between 14 to 16 ounces of soda?
0.17
0.24
0.14
0.15
0.85
How many of the next 100 customers will take an average of less
than 12.48 ounces?
54
70
33
45
5
In: Math
ohnston Adhesives Company makes three widely used industrial adhesives: A101, A204, and B216. Sales and production information for each of the three adhesives are shown in the following table. Most of Johnston’s customers ask for a special blend of the three products, which improves heat-resistance. The additional separable processing requires additional time and materials, and the price is increased accordingly, as shown in the table. Assume that Johnston produces only for specific customer orders, so there is no beginning or ending inventory. Assume also that all of Johnston’s customers requested the heat-resistant version of the products so that all production required additional separable processing. Total joint cost for the three products is $3,425,000.
| A101 | A204 | B216 | |||||||||
| Gallons sold | 202,000 | 188,000 | 135,000 | ||||||||
| Final sales price per gallon | $ | 14 | $ | 8 | $ | 9 | |||||
| Price at split-off | 11 | 4 | 5 | ||||||||
| Separable processing cost | $ | 607,000 | $ | 115,000 | $ | 550,000 | |||||
Required:
1. Calculate the unit product cost and total gross margin for each of the three product lines using the following methods: (a) physical measure method, (b) sales value at split-off method, (c) the net realizable value method, and (d) the constant gross margin percentage method. (Round intermediate calculations and cost per unit answers to 4 decimal places. Round your final answers to whole dollar amounts. Negative amounts should be indicated with a minus sign.)
In: Accounting
7-45
Johnson adhesives company makes three widely used industrial adhesives: A101, A 204 and B216 sales and production information for each of the three adhesives are shown in the following table. Most of johnsons customers ask for a special blend of the three products which improves heat-resistance. The aditional separable processing requires additional time and materials, and the price is increased accordingly as shown in the table. Assume that johnson produces only for specific customer orders, so there is no beginning or ending inventory. Assume also that all of johnsons customers requested the heat resistant version of the product, so that all production required additional separable processing. Total joint cost for the three products is $3,500,000
Galloons sold 175,000 135,000 115,000 A101 A204 B216
Price/gal (after addit'l processing) 14 10 12
price at split-off 10 5 10
separable processing cost 550,000 125,000 625,000
required: Using 4 or more decimal points avoid rounding error, calculate the product cost and gross margin for each of the the three product lines using the following methods:
a.) physical unit method b.) sales value at split off method c.) the net realizable value method
2.) which of the four methods do you think would be preffered in this case? Why?
In: Accounting
What are the internal control weaknesses in the following company's revenue cycle? And what are the potential impacts of these weaknesses on the organization? And give some suggested specific internal controls?
Customers place orders on the company’s website, via email, or by telephone. Due to a renewed interest in music, this year sales have increased significantly. All sales are on credit, with 17% of credit sales in the last 12 months needing to be written off as uncollectible. This included several large online orders to first-time customers who denied ordering or receiving the merchandise.
Customer orders are picked and sent to the warehouse, where they are placed near the loading dock in alphabetical sequence by customer name. The loading dock is used both for outgoing shipments to customers as well as receiving incoming deliveries. There are ten to twenty incoming deliveries every day, from a large number of sources.
The increased volume of sales orders has resulted in a large number of errors where customers have been sent the wrong items. There have also been delays in shipping as items that supposedly were in stock could not be found in the warehouse. Although a perpetual inventory is maintained, there has been no physical count of inventory for over two years. When an item is missing, the warehouse staff notes the information down in a log book. At the end of the week, the warehouse staff uses the log book to update the inventory records.
The system is configured to prepare the sales invoice only after shipping employees enter the actual quantities sent to a customer, thereby ensuring that customers are billed only for items actually sent and not for anything on back order. Terms of trade are payment within 21 days of invoicing, with a 2% discount offered for payments made within 5 days. Approximately 50% of Strings long standing repeat customers pay within the 5 days.
In: Finance