1.-The following data were obtained for a randomized block design involving five treatments and three blocks: SST = 570, SSTR = 390, SSBL = 95. Set up the ANOVA table. (Round your value for F to two decimal places, and your p-value to three decimal places.)
| Source of Variation |
Sum of Squares |
Degrees of Freedom |
Mean Square |
F | p-value |
|---|---|---|---|---|---|
| Treatments | 390 | 4 | 97.5 | ||
| Blocks | 95 | 2 | 47.5 | ||
| Error | 85 | 8 | 10.625 | ||
| Total | 570 | 14 |
Find the value of the test statistic. (Round your answer to two decimal places.)___
Find the p-value. (Round your answer to three decimal places.)
p-value = ___
An automobile dealer conducted a test to determine if the time in minutes needed to complete a minor engine tune-up depends on whether a computerized engine analyzer or an electronic analyzer is used. Because tune-up time varies among compact, intermediate, and full-sized cars, the three types of cars were used as blocks in the experiment. The data obtained follow.
| Analyzer | |||
|---|---|---|---|
| Computerized | Electronic | ||
| Car | Compact | 51 | 43 |
| Intermediate | 55 | 44 | |
| Full-sized | 62 | 45 | |
Use α = 0.05 to test for any significant differences.
Find the value of the test statistic. (Round your answer to two decimal places.)___
Find the p-value. (Round your answer to three decimal places.)
p-value = ___
In: Statistics and Probability
2. You deposit $1000. How much will you have under each of the following conditions?
a) 8 percent compounded semi-annually for two years
b) 8 percent compounded quarterly for two years
c) 8 percent compounded monthly for two years
In: Finance
Let x be a random variable that represents the level of glucose in the blood (milligrams per deciliter of blood) after a 12 hour fast. Assume that for people under 50 years old, x has a distribution that is approximately normal, with mean μ = 50 and estimated standard deviation σ = 12. A test result x < 40 is an indication of severe excess insulin, and medication is usually prescribed.
(a) What is the probability that, on a single test, x
< 40? (Round your answer to four decimal places.)
(b) Suppose a doctor uses the average x for two tests
taken about a week apart. What can we say about the probability
distribution of x? Hint: See Theorem 6.1.
The probability distribution of x is not normal. The probability distribution of x is approximately normal with μx = 50 and σx = 8.49. The probability distribution of x is approximately normal with μx = 50 and σx = 12. The probability distribution of x is approximately normal with μx = 50 and σx = 6.00.
What is the probability that x < 40? (Round your answer
to four decimal places.)
(c) Repeat part (b) for n = 3 tests taken a week apart.
(Round your answer to four decimal places.)
(d) Repeat part (b) for n = 5 tests taken a week apart.
(Round your answer to four decimal places.)
(e) Compare your answers to parts (a), (b), (c), and (d). Did the
probabilities decrease as n increased?
Yes No
Explain what this might imply if you were a doctor or a nurse.
The more tests a patient completes, the stronger is the evidence for excess insulin. The more tests a patient completes, the weaker is the evidence for excess insulin. The more tests a patient completes, the stronger is the evidence for lack of insulin. The more tests a patient completes, the weaker is the evidence for lack of insulin.
In: Statistics and Probability
| 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
PLEASE SHOW WORKING SOLUTION
13. Previous research states, "no evidence currently exists supporting or refuting the use of electric fans during heat waves" in terms of mortality and illness. Counterintuitively, Public Health guidelines suggest not using fans during hot weather, with some research reporting the potential of fans accelerating body heating.
You decide to research further this seemingly contradictory guidance, hypothesizing that the true population average core body temperature amidst higher ambient temperature and humidity levels while using an electric fan is different than 90.1 degrees Fahrenheit (°F) and you set the level of significance at 2.5% for your formal hypothesis test. You randomly sample 50 participants based on your research funding and for 45 minutes, the study participants sit in a chamber maintained at a temperature of 108°F (i.e., 42 degrees Celsius) and a relative humidity of 70%. After the first 45 minute warming period, for each participant you place a personal sized electric fan 3 feet away with its airflow directed at a given participant's chest area, and the participants relax in this position for the next 45 minutes. At the end of this 45minute fan period, you record the core body temperature of all participants. The following table comprises the data you collect.
|
Subject |
Core Body |
|
1 |
108.8 |
|
2 |
109.0 |
|
3 |
110.3 |
|
4 |
108.6 |
|
5 |
110.3 |
|
6 |
109.0 |
|
7 |
107.7 |
|
8 |
109.1 |
|
9 |
107.9 |
|
10 |
108.7 |
|
11 |
108.1 |
|
12 |
109.3 |
|
13 |
109.6 |
|
14 |
109.5 |
|
15 |
109.3 |
|
16 |
109.9 |
|
17 |
108.6 |
|
18 |
110.3 |
|
19 |
109.5 |
|
20 |
108.9 |
|
21 |
108.5 |
|
22 |
110.0 |
|
23 |
110.5 |
|
24 |
110.0 |
|
25 |
108.6 |
|
26 |
109.9 |
|
27 |
110.4 |
|
28 |
110.9 |
|
29 |
109.7 |
|
30 |
108.2 |
|
31 |
108.8 |
|
32 |
109.7 |
|
33 |
108.6 |
|
34 |
109.8 |
|
35 |
111.4 |
|
36 |
109.0 |
|
37 |
108.8 |
|
38 |
109.1 |
|
39 |
108.9 |
|
40 |
108.1 |
|
41 |
108.3 |
|
42 |
109.8 |
|
43 |
110.4 |
|
44 |
110.9 |
|
45 |
107.9 |
|
46 |
111.6 |
|
47 |
109.5 |
|
48 |
109.1 |
|
49 |
108.7 |
|
50 |
109.2 |
Per Step 3 of the 5-Steps to Hypothesis Testing, choose the appropriate decision rule.
Select one:
a. Reject H0 if z = +2.576 and z > -2.241
b. Reject H0 if t = -1.960 and t < -1.960
c. Accept H1 if z ≤ -1.645 or t ≥ +1.645
d. Reject H0 if z ≤ -2.241 or z ≥ +2.241
In: Statistics and Probability
A company has a 13% WACC and is considering two mutually exclusive investments (that cannot be repeated) with the following cash flows:
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| Project A | -$300 | -$387 | -$193 | -$100 | $600 | $600 | $850 | -$180 |
| Project B | -$400 | $131 | $131 | $131 | $131 | $131 | $131 | $0 |
What is each project's NPV? Round your answer to the nearest cent. Do not round your intermediate calculations.
Project A $
Project B $
What is each project's IRR? Round your answer to two decimal places.
Project A %
Project B %
What is each project's MIRR? (Hint: Consider Period 7 as the end of Project B's life.) Round your answer to two decimal places. Do not round your intermediate calculations.
Project A %
Project B %
| Discount Rate | NPV Project A | NPV Project B |
| 0% | $ | $ |
| 5 | ||
| 10 | ||
| 12 | ||
| 15 | ||
| 18.1 | ||
| 23.54 |
What is each project's MIRR at a WACC of 18%? Round your answer to two decimal places. Do not round your intermediate calculations.
Project A %
Project B %
In: Finance
A company has a 13% WACC and is considering two mutually exclusive investments (that cannot be repeated) with the following cash flows:
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| Project A | -$300 | -$387 | -$193 | -$100 | $600 | $600 | $850 | -$180 |
| Project B | -$400 | $132 | $132 | $132 | $132 | $132 | $132 | $0 |
What is each project's NPV? Round your answer to the nearest cent. Do not round your intermediate calculations.
Project A $
Project B $
What is each project's IRR? Round your answer to two decimal places.
Project A %
Project B %
What is each project's MIRR? (Hint: Consider Period 7 as the end of Project B's life.) Round your answer to two decimal places. Do not round your intermediate calculations.
Project A %
Project B %
| Discount Rate | NPV Project A | NPV Project B |
| 0% | $ | $ |
| 5 | ||
| 10 | ||
| 12 | ||
| 15 | ||
| 18.1 | ||
| 23.86 |
What is each project's MIRR at a WACC of 18%? Round your answer to two decimal places. Do not round your intermediate calculations.
Project A %
Project B %
In: Finance
he results of ANOVA test are summarized in Table 1.
Table 1. Shows the results of ANOVA for three different procedures
| Source | Sum of Square | d.f | Mean Square | F |
|---|---|---|---|---|
| Between | 160.13 | 80.97 | 9.17 | |
| Within (error) | 104.80 | 8.73 | ||
| Total | 264.93 | 14 |
The degrees of freedom for between and within are:
Select one:
A. 1 and 13 respectively
B. 2 and 12 respectively
C. 3 and 11 respectively
D. 4 and 10 respectively
QUESTION 42
Not yet answered
Marked out of 2.00
Flag question
Question text
A research group claims by taking a special vitamin, a weight lifter can increase his strength. After two weeks of training, supplemented with vitamin, they tested again. Test the effectiveness of the regiment at α = 0.05. Assume that the variable is normally distributed. The alternative hypothesis is :
Select one:
a. H0: µD ≥ 0
b. H0: µD = 0
c. H0: µD ≠ 0
d. H0: µD ≤ 0
QUESTION 43
Question text
What is nP0? NEED TO FIND THEM N AND po
Select one:
A. 1
B. no answer
C. n
D. 0
QUESTION 44
Not yet answered
Marked out of 2.00
Flag question
Question text
The data for a sample of 100 gave the regression line equation for age and blood pressure is Ÿ = 100 + 0.96X, and the standard error is 5. The 95% confidence interval of the prediction of the blood pressure of a person who is 43 years old showed that the lower confidence limit is :
Select one:
A. 125.67
B. 141.34
C. 142.34
D. 131.48
QUESTION 45
Not yet answered
Marked out of 2.00
Flag question
Question text
A researcher wants to investigate if there is a difference in the rates of hotel room in two cities. A sample of 50 were selected from each city, the average hotel room in the first city is RM88.42 and in the second city is RM80.61 and the standard deviation are RM5.62 and RM4. The null hypothesis for the difference between the means is
Select one:
A. µ1 - µ2 ≤ 0
B. µ1 - µ2 = 0
C. µ1 - µ2 ≥ 0
D. µ1 - µ2 ≠ 0
QUESTION 46
Not yet answered
Marked out of 2.00
Flag question
Question text
When a distribution is bell-shaped approximately what percentage of data values will fall within one standard deviation of the mean?
Select one:
A. 95%
B. 68%
C. 99.7%
D. 50%
QUESTION 47
Not yet answered
Marked out of 2.00
Flag question
Question text
A repair team is responsible for a stretch of oil pipeline 2 miles long. The distance (in miles) at which any fracture occurs can be represented by a uniformly distributed random variable f(x) = 0.5
What is the probability that any given fracture occurs between 0.5 mile and 1.5 miles along this stretch pipeline?
Select one:
A. 0.2
B. 0.5
C. 0.1
D. 0.7
QUESTION 48
Not yet answered
Marked out of 2.00
Flag question
Question text
In an advertisement, a retail store stated that its employees averaged nine years of service. The distribution is shown here.
| Number of Employees | Years of Service |
|---|---|
| 8 | 2 |
| 2 | 6 |
| 3 | 10 |
Using the weighted mean, the correct average is .........
Select one:
A. 4.5
B. 3.5
C. 5.4
D. 5.3
QUESTION 49
Not yet answered
Marked out of 2.00
Flag question
Question text
The data for a sample of 100 gave the regression line equation for age and blood pressure is Ÿ = 100 + 0.96X, and the standard error is 5. The 95% confidence interval of the prediction of the blood pressure of a person who is 43 years old showed that the upper confidence limit is :
Select one:
A. 159.08
B. 149.09
C. 151.08
D. 155.08
QUESTION 50
Not yet answered
Marked out of 2.00
Flag question
Question text
The variance for the data values “ 87, 85, 80, 78, 86, 90” is :
Select one:
A. 4.1
B. 12
C. 85
D. 17.1
In: Statistics and Probability
10. The natural unemployment rate in the United States has varied over the last 50 years. According to the Congressional Budget Office, the natural rate was 5.5% in 1960, rose to about 6.5% in the 1970s, and had declined to about 4.8% by 2000. What do you think might have caused this variation?
11. Suppose the Fed begins carrying out an expansionary monetary policy in order to close a recessionary gap. Relate what happens during the next two phases of the inflation-unemployment cycle to the maxim “You can fool some of the people some of the time, but you can’t fool all of the people all of the time.”
In: Economics
Dr. Al Maisari is a veterinarian who sees only dogs and
cats.
In each appointment, he may or may not give the animal a
vaccine.
The two-way frequency table summarizes Dr. Al Maisari's 50
appointments last week.
| Vaccine | no vaccine | |
| dog | 14 | 6 |
| cat | 11 | 19 |
Let dog be the event that a randomly chosen appointment (from the table) involved a dog.
Let vaccine be the event that a randomly chosen appointment (from the table) included a vaccine.
Find the following probabilities. Write your answers as decimals.
p(dog)=
p(vaccine dog)=
p(vaccine l dog)=
In: Statistics and Probability