Randomly sample 50% of cases from cohort study C and place them in the cells of Table 4 below. If the sample of cases (or controls) is random it will maintain the same ratio of exposed to unexposed among cases and non-cases that is present in cohort C. Next, determine how many controls will be required in table 4 in order to have 1 control for each case. There are two ways to sample the required number of controls from cohort C. First sample “controls 1” from all persons who entered the cohort (column 2 of Table 3), prior to knowledge of disease status. Then, sample “controls 2” from all persons who did not develop the disease during follow-up (Column 4 of Table 3). Although, it’s not realistic, retain two decimal places in the numbers of controls. As your samples of controls must also be random, they should also maintain the same ratio of exposed to unexposed that is present among potential controls in Cohort C.
|
Table 3. (4 points) Cohort Study C. Provide an answer for all 9 of the shaded cells. Make the calculation and show an appropriate number of decimal places. For example, 100/500 is not an acceptable answer, calculate the final number. Odds of disease = individuals with disease/individuals without disease. |
|||||
|
Total |
Disease |
Disease |
Risk of |
Odds of |
|
|
Present |
Absent |
Disease |
Disease |
||
|
Exposed |
1,500 |
1350 |
150 |
||
|
Unexposed |
28,500 |
8550 |
19,950 |
||
|
Total |
30,000 |
9900 |
20,100 |
||
|
Risk Ratio = |
Odds ratio = |
||||
|
Population Attributable Risk (PAR) use formula for cohort = |
|||||
|
Table 4. (5 points) Sampling two studies from Cohort C. Provide an answer for all 13 of the shaded cells. Show 2-4 decimal places. |
|||
|
Cases |
Control 1 |
Control 2 |
|
|
Exposed |
|||
|
Unexposed |
|||
|
Total |
|||
|
CC Study 1: OR 1 (Control 1) = |
|||
|
CC Study 2: OR 2 (Control 2) = |
|||
|
Study 1: Population Attributable Risk (PAR) using ca/co formula = |
|||
|
Study 2: Population Attributable Risk (PAR), Using ca/co formula = |
|||
In: Biology
1. Find the local maxima of the function:
(1) f(x,y) = xy, subject to the constraint that
x+y-1=0. Result should be 1/4.
2. Find the local minima of the functions:
(1) f(x,y) = x^2+y^2, subject to the constraint that xy-3=0. Result should be 6.
(2) f(x,y) = x^2+4xy+y^2, subject to the constraint that x-y-6=0. Result should be -18.
In: Math
A medical researcher wishes to try three different techniques to lower the blood pressure of patients with high blood pressure. The subjects are randomly selected and assigned to one of three groups. Group 1 is given medication, Group 2 is given an exercise program, and Group 3 is assigned a diet program. At the end of six weeks, each subject's blood pressure is recorded
| group1 | group 2 | group 3 | |
| 13 | 8 | 6 | |
| 12 | 6 | 4 | |
| 11 | 4 | 3 | |
| 15 | 7 | 5 | |
| 9 | 5 | 4 | |
| 8 | 0 | 2 | |
| mean | 11.33 | 5 | 4 |
| sd | 2.58 | 2.83 | 1.41 |
1) Identify the null hypothesis. 2) Identify the alternative hypothesis. 3) Determine the value of the F test statistic. 4) Identify the P-value. 5) State a conclusion about the null hypothesis. 6) State a final conclusion that addresses the original claim. 7. Write the result in APA format
In: Statistics and Probability
|
Year |
Cash Flow |
|
1 |
$ 950 |
|
2 |
1,040 |
|
3 |
1,130 |
|
4 |
1,075 |
If the discount rate is 10 percent, what is the present value of these cash flows? What is the present value at 18%? What is the present value at 24%? Do not round intermediate calculations and round your final answer to 2 decimal places.
|
Year |
Cash Flow |
|
1 |
$ 940 |
|
2 |
1,090 |
|
3 |
1,340 |
|
4 |
1,405 |
If the discount rate is 8 percent, what is the future value of the cash flows in year 4? If the discount rate is 11 percent, what is the future value of the cash flows in year 4? If the discount rate is 24 percent, what is the future value of the cash flows in year 4? Do not round intermediate calculations and round your final answer to 2 decimal places.
In: Finance
Below are some data from the land of milk and honey
| Year | Price of Milk | Quantity of Milk | Price of Honey | Quantity of Honey |
| 2013 (Base yr) | $1 | 100 quarts | $2 | 50 quarts |
| 2014 | $1 | 200 | $2 | 100 |
| 2015 | $2 | 200 | $4 | 100 |
Compute nominal GDP, real GDP, and the GDP deflator for each year, using the information from the above table. The base year is 2013. Consider an economy that produces only chocolate bars. In year 1, the quantity produced is 3 bars and the price is $4. In year 2, the quantity produced is 4 bars and the price is $5. In year 3, the quantity produced is 5 bars and the price is $6. Year 1 is the base year. What is nominal GDP for each of these three years? What is real GDP for each of these years? What is the GDP deflator for each of these years? What is the percentage growth rate of real GDP from year 2 to year 3? What is the inflation rate as measured by the GDP deflator from year 2 to year 3?
In: Economics
Express the equation of the plane in explicit form
a. d = [1, -3, -5]; e = [-2, 1, -1] and f = [5, -1, -3] Express the
equation of the plane through d, e, and f explicitly
b. d = [-1, 7, -6]; e = [2, -1, 3] and f = [4, -2, 9] Express the
equation of the plane through d, e, and f explicitly
In: Advanced Math
| Date | Close |
| 7/3/2017 | 898.700012 |
| 7/5/2017 | 911.710022 |
| 7/6/2017 | 906.690002 |
| 7/7/2017 | 918.590027 |
| 7/10/2017 | 928.799988 |
| 7/11/2017 | 930.090027 |
| 7/12/2017 | 943.830017 |
| 7/13/2017 | 947.159973 |
| 7/14/2017 | 955.98999 |
| 7/17/2017 | 953.419983 |
| 7/18/2017 | 965.400024 |
| 7/19/2017 | 970.890015 |
| 7/20/2017 | 968.150024 |
| 7/21/2017 | 972.919983 |
| 7/24/2017 | 980.340027 |
| 7/25/2017 | 950.700012 |
| 7/26/2017 | 947.799988 |
| 7/27/2017 | 934.090027 |
| 7/28/2017 | 941.530029 |
| 7/31/2017 | 930.5 |
| 8/1/2017 | 930.830017 |
| 8/2/2017 | 930.390015 |
| 8/3/2017 | 923.650024 |
| 8/4/2017 | 927.960022 |
| 8/7/2017 | 929.359985 |
| 8/8/2017 | 926.789978 |
| 8/9/2017 | 922.900024 |
| 8/10/2017 | 907.23999 |
| 8/11/2017 | 914.390015 |
| 8/14/2017 | 922.669983 |
| 8/15/2017 | 922.219971 |
| 8/16/2017 | 926.960022 |
| 8/17/2017 | 910.97998 |
| 8/18/2017 | 910.669983 |
| 8/21/2017 | 906.659973 |
| 8/22/2017 | 924.690002 |
| 8/23/2017 | 927 |
| 8/24/2017 | 921.280029 |
| 8/25/2017 | 915.890015 |
| 8/28/2017 | 913.809998 |
| 8/29/2017 | 921.289978 |
| 8/30/2017 | 929.570007 |
| 8/31/2017 | 939.330017 |
| 9/1/2017 | 937.340027 |
| 9/5/2017 | 928.450012 |
| 9/6/2017 | 927.809998 |
| 9/7/2017 | 935.950012 |
| 9/8/2017 | 926.5 |
| 9/11/2017 | 929.080017 |
| 9/12/2017 | 932.070007 |
| 9/13/2017 | 935.090027 |
| 9/14/2017 | 925.109985 |
| 9/15/2017 | 920.289978 |
| 9/18/2017 | 915 |
| 9/19/2017 | 921.809998 |
| 9/20/2017 | 931.580017 |
| 9/21/2017 | 932.450012 |
| 9/22/2017 | 928.530029 |
| 9/25/2017 | 920.969971 |
| 9/26/2017 | 924.859985 |
| 9/27/2017 | 944.48999 |
| 9/28/2017 | 949.5 |
| 9/29/2017 | 959.109985 |
| 10/2/2017 | 953.27002 |
| 10/3/2017 | 957.789978 |
| 10/4/2017 | 951.679993 |
| 10/5/2017 | 969.960022 |
| 10/6/2017 | 978.890015 |
| 10/9/2017 | 977 |
| 10/10/2017 | 972.599976 |
| 10/11/2017 | 989.25 |
| 10/12/2017 | 987.830017 |
| 10/13/2017 | 989.679993 |
| 10/16/2017 | 992 |
| 10/17/2017 | 992.179993 |
| 10/18/2017 | 992.809998 |
| 10/19/2017 | 984.450012 |
| 10/20/2017 | 988.200012 |
| 10/23/2017 | 968.450012 |
| 10/24/2017 | 970.539978 |
| 10/25/2017 | 973.330017 |
| 10/26/2017 | 972.559998 |
| 10/27/2017 | 1019.27002 |
| 10/30/2017 | 1017.109985 |
| 10/31/2017 | 1016.640015 |
| 11/1/2017 | 1025.5 |
| 11/2/2017 | 1025.579956 |
| 11/3/2017 | 1032.47998 |
| 11/6/2017 | 1025.900024 |
| 11/7/2017 | 1033.329956 |
| 11/8/2017 | 1039.849976 |
| 11/9/2017 | 1031.26001 |
| 11/10/2017 | 1028.069946 |
| 11/13/2017 | 1025.75 |
| 11/14/2017 | 1026 |
| 11/15/2017 | 1020.909973 |
| 11/16/2017 | 1032.5 |
| 11/17/2017 | 1019.090027 |
| 11/20/2017 | 1018.380005 |
| 11/21/2017 | 1034.48999 |
| 11/22/2017 | 1035.959961 |
| 11/24/2017 | 1040.609985 |
| 11/27/2017 | 1054.209961 |
| 11/28/2017 | 1047.410034 |
| 11/29/2017 | 1021.659973 |
| 11/30/2017 | 1021.409973 |
| 12/1/2017 | 1010.169983 |
| 12/4/2017 | 998.679993 |
| 12/5/2017 | 1005.150024 |
| 12/6/2017 | 1018.380005 |
| 12/7/2017 | 1030.930054 |
| 12/8/2017 | 1037.050049 |
| 12/11/2017 | 1041.099976 |
| 12/12/2017 | 1040.47998 |
| 12/13/2017 | 1040.609985 |
| 12/14/2017 | 1049.150024 |
| 12/15/2017 | 1064.189941 |
| 12/18/2017 | 1077.140015 |
| 12/19/2017 | 1070.680054 |
| 12/20/2017 | 1064.949951 |
| 12/21/2017 | 1063.630005 |
| 12/22/2017 | 1060.119995 |
| 12/26/2017 | 1056.73999 |
| 12/27/2017 | 1049.369995 |
| 12/28/2017 | 1048.140015 |
| 12/29/2017 | 1046.400024 |
| 1/2/2018 | 1065 |
| 1/3/2018 | 1082.47998 |
| 1/4/2018 | 1086.400024 |
| 1/5/2018 | 1102.22998 |
| 1/8/2018 | 1106.939941 |
| 1/9/2018 | 1106.26001 |
| 1/10/2018 | 1102.609985 |
| 1/11/2018 | 1105.52002 |
| 1/12/2018 | 1122.26001 |
| 1/16/2018 | 1121.76001 |
| 1/17/2018 | 1131.97998 |
| 1/18/2018 | 1129.790039 |
| 1/19/2018 | 1137.51001 |
| 1/22/2018 | 1155.810059 |
| 1/23/2018 | 1169.969971 |
| 1/24/2018 | 1164.23999 |
| 1/25/2018 | 1170.369995 |
| 1/26/2018 | 1175.839966 |
| 1/29/2018 | 1175.579956 |
| 1/30/2018 | 1163.689941 |
| 1/31/2018 | 1169.939941 |
| 2/1/2018 | 1167.699951 |
| 2/2/2018 | 1111.900024 |
| 2/5/2018 | 1055.800049 |
| 2/6/2018 | 1080.599976 |
| 2/7/2018 | 1048.579956 |
| 2/8/2018 | 1001.52002 |
| 2/9/2018 | 1037.780029 |
| 2/12/2018 | 1051.939941 |
| 2/13/2018 | 1052.099976 |
| 2/14/2018 | 1069.699951 |
| 2/15/2018 | 1089.52002 |
| 2/16/2018 | 1094.800049 |
| 2/20/2018 | 1102.459961 |
| 2/21/2018 | 1111.339966 |
| 2/22/2018 | 1106.630005 |
| 2/23/2018 | 1126.790039 |
| 2/26/2018 | 1143.75 |
| 2/27/2018 | 1118.290039 |
| 2/28/2018 | 1104.72998 |
| 3/1/2018 | 1069.52002 |
| 3/2/2018 | 1078.920044 |
| 3/5/2018 | 1090.930054 |
| 3/6/2018 | 1095.060059 |
| 3/7/2018 | 1109.640015 |
| 3/8/2018 | 1126 |
| 3/9/2018 | 1160.040039 |
| 3/12/2018 | 1164.5 |
| 3/13/2018 | 1138.170044 |
| 3/14/2018 | 1149.48999 |
| 3/15/2018 | 1149.579956 |
| 3/16/2018 | 1135.72998 |
| 3/19/2018 | 1099.819946 |
| 3/20/2018 | 1097.709961 |
| 3/21/2018 | 1090.880005 |
| 3/22/2018 | 1049.079956 |
| 3/23/2018 | 1021.570007 |
| 3/26/2018 | 1053.209961 |
| 3/27/2018 | 1005.099976 |
| 3/28/2018 | 1004.559998 |
| 3/29/2018 | 1031.790039 |
| 4/2/2018 | 1006.469971 |
| 4/3/2018 | 1013.409973 |
| 4/4/2018 | 1025.140015 |
| 4/5/2018 | 1027.810059 |
| 4/6/2018 | 1007.039978 |
| 4/9/2018 | 1015.450012 |
| 4/10/2018 | 1031.640015 |
| 4/11/2018 | 1019.969971 |
| 4/12/2018 | 1032.51001 |
| 4/13/2018 | 1029.27002 |
| 4/16/2018 | 1037.97998 |
| 4/17/2018 | 1074.160034 |
| 4/18/2018 | 1072.079956 |
| 4/19/2018 | 1087.699951 |
| 4/20/2018 | 1072.959961 |
| 4/23/2018 | 1067.449951 |
| 4/24/2018 | 1019.97998 |
| 4/25/2018 | 1021.179993 |
| 4/26/2018 | 1040.040039 |
| 4/27/2018 | 1030.050049 |
| 4/30/2018 | 1017.330017 |
| 5/1/2018 | 1037.310059 |
| 5/2/2018 | 1024.380005 |
| 5/3/2018 | 1023.719971 |
| 5/4/2018 | 1048.209961 |
| 5/7/2018 | 1054.790039 |
| 5/8/2018 | 1053.910034 |
| 5/9/2018 | 1082.76001 |
| 5/10/2018 | 1097.569946 |
| 5/11/2018 | 1098.26001 |
| 5/14/2018 | 1100.199951 |
| 5/15/2018 | 1079.22998 |
| 5/16/2018 | 1081.77002 |
| 5/17/2018 | 1078.589966 |
| 5/18/2018 | 1066.359985 |
| 5/21/2018 | 1079.579956 |
| 5/22/2018 | 1069.72998 |
| 5/23/2018 | 1079.689941 |
| 5/24/2018 | 1079.23999 |
| 5/25/2018 | 1075.660034 |
| 5/29/2018 | 1060.319946 |
| 5/30/2018 | 1067.800049 |
| 5/31/2018 | 1084.98999 |
| 6/1/2018 | 1119.5 |
| 6/4/2018 | 1139.290039 |
| 6/5/2018 | 1139.660034 |
| 6/6/2018 | 1136.880005 |
| 6/7/2018 | 1123.859985 |
| 6/8/2018 | 1120.869995 |
| 6/11/2018 | 1129.98999 |
| 6/12/2018 | 1139.319946 |
| 6/13/2018 | 1134.790039 |
| 6/14/2018 | 1152.119995 |
| 6/15/2018 | 1152.26001 |
| 6/18/2018 | 1173.459961 |
| 6/19/2018 | 1168.060059 |
| 6/20/2018 | 1169.839966 |
| 6/21/2018 | 1157.660034 |
| 6/22/2018 | 1155.47998 |
| 6/25/2018 | 1124.810059 |
| 6/26/2018 | 1118.459961 |
| 6/27/2018 | 1103.97998 |
| 6/28/2018 | 1114.219971 |
| 6/29/2018 | 1115.650024 |
| 7/2/2018 | 1127.459961 |
Project 3 instructions
Based on Larson & Farber: sections 5.2–5.3
Using the provided data. Assume that the closing prices of the stock form a normally distributed data set. This means that you need to use Excel to find the mean and standard deviation. Then, use those numbers and the methods you learned in sections 5.2–5.3 of the course textbook for normal distributions to answer the questions. Do NOT count the number of data points.
Complete this portion of the assignment within a single Excel file. Show your work or explain how you obtained each of your answers. Answers with no work and no explanation will receive no credit.
Show all work
1) If a person bought 1 share of Google stock within the last year, what is the probability that the stock on that day closed at less than the mean for that year? Hint: You do not want to calculate the mean to answer this one. The probability would be the same for any normal distribution. (5 points)
2) If a person bought 1 share of Google stock within the last year, what is the probability that the stock on that day closed at more than $825? (5 points)
3a) If a person bought 1 share of Google stock within the last year, what is the probability that the stock on that day closed within $50 of the mean for that year? (5 points)
3b) If a person bought 1 share of Google stock within the last year, what is the probability that the stock on that day closed at less than $700 per share. (5 points)
3c) At what prices would Google have to close in order for it to be considered statistically unusual? You will have a low and high value. Use the definition of unusual from the course textbook that is measured as a number of standard deviations. (5 points)
4) What are Quartile 1, Quartile 2, and Quartile 3 in this data set? Use Excel to find these values. This is the only question that you must answer without using anything about the normal distribution. (5 points)
5) Is the normality assumption that was made at the beginning valid? Why or why not? Hint: Does this distribution have the properties of a normal distribution as described in the course textbook? Real data sets are never perfect, however, it should be close. One option would be to construct a histogram like you did in Project 1 to see if it has the right shape. Something in the range of 10 to 12 classes is a good number. (5 points)
There are also 5 points for miscellaneous items like correct date range, correct mean, correct SD, etc.
In: Statistics and Probability
For the function
a) f(x)=x^3-9x^2+23x-15
b)f(x)=(x+3)^2(2x+1)(x-1)
c)f(x)=-(x^2-6x+9)(x^2-x-6)
Find:
1) the zeros
2) the y-intercept
3) left-right end behavior
4) the sketch of the graph
In: Math
For this problem you must write the functions in a recursive manner (i.e. the function must call itself) – it is not acceptable to submit an iterative solution to these problems.
A. Complete the recursive function gcd(m, n) that calculate the greatest common denominator of two numbers with the following rules:
# If m = n, it returns n
# If m < n, it returns gcd(m, n-m)
# If m > n, it returns gcd(m-n, n) #
def gcd(m,n):
return None # Replace this with your implementation
B. Complete the following function that uses recursion to find and return the max (largest) value in the list u.
# find_max([1, 7, 4, 5] returns 7 # find_ max ([1, 7, 4, 5, 9, 2] returns 9 # def find_max(u):
return None # Replace this with your implementation
C. Complete the following recursive function that returns the zip of two lists u and v of the same length. Zipping the lists should place the first element from each into a new array, followed by the second elements, and so on (see example output).
# zip([1, 2, 3], [4, 5, 6]) returns [1, 4, 2, 5, 3, 6] # def zip(u, v):
return None # Replace this with your implementation
D. Complete the following recursive function that removes all occurrences of the number x from the list nums.
# remove_number(5, [1, 2, 3, 4, 5, 6, 5, 2, 1]) returns [1, 2, 3, 4, 6, 2, 1] # def remove_number(x, nums):
return None # Replace this with your implementation
E. Write a recursive function removeLetter(string, letter) that takes a string and a letter as input, and recursively removes all occurrences of that letter from the string. The function is case sensitive.
Some example test cases are below:
>>> removeLetter("test string", "t")
es sring
>>> removeLetter("mississipi", "i")
mssssp
>>> removeLetter("To be or not to be is a question.", "t")
To be or no o be is a quesion.
In PyCharm
In: Computer Science
For the eight reactions done in test tubes, explain your observations in terms of mechanism (SN1 or SN2) , expected intermediates (carbocation or a single step conversion) and product. When they react with Sodium iodine in acetone and Silver nitrate in ethanol. 1. 1-Cholorbutane 2. 1-bromobutane 3. 2-Chlorobutane 4. 2- Chloro-2-Methyl Propane 5. Bromobutane 6. 2-Bromobutane 7. 2-Bromo-2-Metylpropane 8. 1-Chloro-2-Methyl Propane
In: Chemistry