Exhibit 1:
We have the following information about number of violent crimes (Y) and the number of police personnel (X) for a certain year for a sample of three metropolitan areas. We also know the following statistics: SST = 1250, SSE = 937.5
|
Crime (Y) |
Police Personnel (X) |
|
300 |
5000 |
|
325 |
3000 |
|
350 |
4000 |
Question 6
To answer this question, refer to Exhibit 1 in question 1.
What does value of r2 tell you?
| A. |
25 percent of variation in crime is explained by the number of police personnel. |
|
| B. |
50 percent of variation in crime is explained by the number of police personnel. |
|
| C. |
75 percent of variation in crime is explained by the number of police personnel. |
|
| D. |
None of the above |
Question 7
To answer this question, refer to Exhibit 1 in question 1.
The coefficient of correlation is (to 2 decimal places)
| A. |
0.87 |
|
| B. |
-0.87 |
|
| C. |
0.5 |
|
| D. |
-0.5 |
Question 8
To answer this question, refer to Exhibit 1 in question 1.
What is the estimate of the standard error of the overall regression (to 2 decimal places)?
|
10.91 |
||
|
30.62 |
||
|
45.88 |
||
|
55.67 |
Question 9
To answer this question, refer to Exhibit 1 in question 1.
What is the estimate of the standard error of slope estimate (to 3 decimal places)?
|
0.001 |
||
|
0.015 |
||
|
0.022 |
||
|
0.053 |
Question 10
To answer this question, refer to Exhibit 1 in question 1.
Is police personnel a significant variable affecting crime in the above data?
|
No because we cannot reject the null the slope is 0. |
||
|
Yes because we can reject the null the slope is 0. |
||
|
Need more information to answer the question |
In: Math
The ability to scale up renewable energy and in particular wind power and speed is dependent on the ability to forecast its short term availability soman et al (2010( describe different methods for wind power forecasting (the quote is slightly edited for brevity)
Persistence method: this method is also known as ‘naïve predictor. Its is assumed that the wind speed at time t + tetat will be the same as it was at time t. unbelievably it is more accurate than most of the physical and statistical methods for very short to short term forecasts.
Physical approach: physical systems. Use parameterizations based on a detailed physical description of the atmosphere
Statistical approach: the statistical approach is based on training with measurement data and uses difference between the predicted and the actual wind speeds in immediate past to tune model parameters. It is easy to model, inexpensive, predefined mathematical model and rather it is based on patterns
Hybrid approach: In general, the combination of different approaches such as mixing physical and statistical approaches or combining short term and medium term models etc is referred to as a hybrid approach.
In: Math
You take a quiz with 6 multiple choice questions. After you studied, you estimated that you would have about an 80% chance of getting any individual question right. What are your chances of getting them all right? The random numbers below represent a simulation with 20 trials. Let 0-7 represent a correct answer and let 8-9 represent an incorrect answer.
1, 6 5 6 2
0 5
2, 4 7 3 1
6 6
3, 5 2 9 6
3 2
4, 8 0 1 6
0 8
5, 3 4 3 3
4 4
6, 2 9 1 7
3 0
7, 6 5 9 6
8 3
8, 8 6 4 4
2 7
9, 6 1 1 8
2 6
10, 5 3 0 3
8 6
11, 0 2 8 1
3 2
12, 6 8 6 0
0 4
13, 9 9 4 6
1 8
14, 1 7 3 2
5 1
15, 7 6 6 1
4 5
16, 3 5 3 1
4 5
17, 0 2 7 7
3 1
18, 3 6 1 6
1 0
19, 8 4 6 7
1 3
20, 5 3 5 2
0 9
In: Math
A local university found it could classify its students into one of three general categories: morning students (42%), afternoon students (33%), and evening students. 37% of the morning students, 23% of the afternoon students, and 16% of the evening students live on campus. What is the probability a non-evening student does not live on campus?
In: Math
In: Math
You wish to test the following claim (HaHa) at a significance
level of α=0.001α=0.001. For the context of this problem,
μd=μ2−μ1μd=μ2-μ1 where the first data set represents a pre-test and
the second data set represents a post-test.
Ho:μd=0Ho:μd=0
Ha:μd≠0Ha:μd≠0
You believe the population of difference scores is normally
distributed, but you do not know the standard deviation. You obtain
pre-test and post-test samples for n=229n=229 subjects. The average
difference (post - pre) is ¯d=3.3d¯=3.3 with a standard deviation
of the differences of sd=19.4sd=19.4.
What is the critical value for this test? (Report answer accurate
to three decimal places.)
critical value = ±±
What is the test statistic for this sample? (Report answer accurate
to three decimal places.)
test statistic =
The test statistic is...
This test statistic leads to a decision to...
As such, the final conclusion is that...
In: Math
Let us suppose that some article studied the probability of death due to burn injuries. The identified risk factors in this study are age greater than 60 years, burn injury in more than 40% of body-surface area, and presence of inhalation injury. It is estimated that the probability of death is 0.003, 0.03, 0.33, or 0.84, if the injured person has zero, one, two, or three risk factors, respectively. Suppose that three people are injured in a fire and treated independently. Among these three people, two people have one risk factor and one person has three risk factors. Let the random variable X denote number of deaths in this fire. Determine the cumulative distribution function for the random variable.
Round your answers to five decimal places (e.g. 98.76543).
F(x)= with x < 0
F(x)= with 0 <= x < 1
F(x)= with 1 <= x < 2
F(x)= with 2 <= x < 3
F(x)= with 3 <= x
In: Math
Hypothesis Testing for a Single Population
Times for a Table
Dominick Aldo owns and operates Carolina’s which is an Italian restaurant in New York. The file Single Population.xlsx is shown below and contains the amount of time that table times varied from table to table.
A) Test the hypothesis that the average table time exceeds 98 minutes using 0.05α=.
B) What is the p-value? Interpret the results.
Be sure that your project shows the following steps
1.Null and alternative hypothesis
2.Determine which distribution to use for the test statistic.
3.Using data provided, calculate necessary sample statistics.
4.Draw a conclusion and interpret the decision.
70 80 100 90 75 60 70 75 110 90 100 75 60 75 80 75 165 60 75 90 110 95 110 150 50 70 90 110 165 60 145 80 115 75 50 90 110 90 100 110 110 85 70 145 120 130 80 90 105 105 100 95 80 100 120 130 75 90 70 125 80 90 95 120 150 195 70 80 110 80 80 85 90 150 60 90 135 170 85 90 120 105 60 70 50 80 100 90 135 120
In: Math
Please note that for all problems in this course, the standard cut-off (alpha) for a test of significance will be .05, and you always report the exact power unless SPSS output states p=.000 (you’d report p<.001). Also, remember that we divide the p value in half when reporting one-tailed tests with 1 – 2 groups.
|
Problem Set 2: (8 pts) Research Scenario: Does distraction and/or amount of details affect the ability of people to make good decisions? In this fictitious scenario, researchers used a within-subjects design. Participants (N=15) were given four different scenarios based on amount of details (4 or 14) and distraction level (no distraction or distraction), and were asked to make an objective decision at the end of each scenario. Objective decision was the dependent variable and was quantified numerically using an interval scale of measurement. Each participant provided four objective decisions – one for each condition. Assume the data is parametric. Select and conduct the most appropriate statistical test to determine whether distraction and/or amount of details affect people’s ability to make good decisions. Hint: since this is within subjects, each level for each factor will have its own column of data, so you will have 4 columns of 15 rows of data in your SPSS data view. You will analyze two factors (“Distraction” and “Details”) and each factor has 2 levels. Please label your columns “NoDistract4”, “NoDistract14”, “Distract4”, and “Distract14”.
|
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In: Math
Consider the following gasoline sales time series data. Click on the datafile logo to reference the data.
| Week | Sales (1000s of gallons) |
| 1 | 17 |
| 2 | 21 |
| 3 | 19 |
| 4 | 24 |
| 5 | 18 |
| 6 | 15 |
| 7 | 21 |
| 8 | 19 |
| 9 | 22 |
| 10 | 19 |
| 11 | 15 |
| 12 | 23 |
a. Using a weight of 1/2 for the most recent observation,1/3 for the second most recent observation, and 1/6 third the most recent observation, compute a three-week weighted moving average for the time series (to 2 decimals). Enter negative values as negative numbers.
Week |
Time-Series Value |
Weighted Moving Average Forecast |
Forecast Error |
(Error)2 |
||
| Total | ||||||
b. Compute the MSE for the weighted moving
average in part (a).
MSE =
Do you prefer this weighted moving average to the unweighted
moving average? Remember that the MSE for the unweighted moving
average is 14.39 .
Prefer the unweighted moving average here; it has a - Select your
answer -greatersmallerItem 42 MSE.
c. Suppose you are allowed to choose any
weights as long as they sum to 1. Could you always find a set of
weights that would make the MSE at least as small for a weighted
moving average than for an unweighted moving average?
- Select your answer -YesNoItem 43
In: Math
Kamini, a student of the 1-Year post graduate program at the International School of Business and Design is trying to establish the relationship between compensation (in Rs. Lakh) and years of work experience. She collected data from 9 students who have been placed and fitted a regression equation with Compensation (in Rs. Lakh) as the dependent variable and Years of experience as the independent variable. The Excel output is given below (with some missing values):
|
SUMMARY OUTPUT |
||||||
|
Regression Statistics |
||||||
|
Multiple R |
||||||
|
R Square |
||||||
|
Adjusted R Square |
0.67224 |
|||||
|
Standard Error |
1.262251 |
|||||
|
Observations |
9 |
|||||
|
ANOVA |
||||||
|
df |
SS |
MS |
F |
Significance F |
||
|
Regression |
1 |
17.40811 |
0.004177 |
|||
|
Residual |
7 |
11.15294 |
1.593277 |
|||
|
Total |
8 |
38.88889 |
||||
|
Coefficients |
Standard Error |
t Stat |
P-value |
Lower 95% |
Upper 95% |
|
|
Intercept |
11.91765 |
1.542901 |
0.000114 |
8.269266 |
15.56603 |
|
|
years of experience |
0.290431 |
0.004177 |
0.525005 |
1.898524 |
||
Answer the following questions based on the above.
1. What is the value of Regression Sum of Squares?
2. What is the 95% confidence interval for the slope?
3. What is the estimated compensation for a person with 8 years of experience?
4. What is the coefficient of correlation between Compensation and Years of experience?
5. What is the R2 for the above regression equation?
6. What is the t-value corresponding to the intercept?
7. Interpret the value 0.004177 under the column “P-Value”
8. What is the expected compensation for a person with no work experience?
9. The above output provides 95% confidence interval for the intercept. What is the lower limit for the 90% confidence interval for the intercept?
10. The above output provides 95% confidence interval for the intercept. What is the upper limit for the 90% confidence interval for the intercept?
In: Math
Let us suppose that some article investigated the probability of corrosion of steel reinforcement in concrete structures. It is estimated that the probability of corrosion is 0.19 under specific values of half-cell potential and concrete resistivity. The risk of corrosion in five independent grids of a building with these values of half-cell potential and concrete resistivity. Let the random variable X denote number of grids with corrosion in this building. Determine the cumulative distribution function for the random variable X.
Round your answers to five decimal places (e.g. 98.76543).
f(x)= with x < 0
f(x)= with 0 <= x < 1
f(x)= with 1 <= x < 2
f(x)= with 2 <= x < 3
f(x)= with 3 <= x < 4
f(x)= with 4 <= x < 5
f(x)= with 5 <= x
In: Math
Studies have examined changes over time in the annual global temperature based on planet-wide recordings. To make temperatures at different locations comparable, "temperature anomalies" are computed locally by comparing the local annual sea surface temperature average with the local temperature reference, the 1951-1980 average. The analysis showed that, in each of several time periods, the distribution of local seasonal temperature anomalies was approximately Normal. Because temperature anomalies are computed relative to the 1951-1980 reference period, summer temperature N(0,1). Decades later, summer temperature in the northern hemisphere over the 2005-2015 period followed approximately the N(1.6, 1.3) distribution. (a) Draw both distributions on the same graph, indicating the mean and standard deviation of each curve. (Select the graph that best matches the graph you drew. Make sure that the means and standard deviations on the legend match the curves.) 13 (b) In the reference period, standardized summer temperature anomalies greater than 3 were considered to be extreme heat events. Based on the proposed Normal model, what percent of local summer temperature anomalies in the northern hemisphere were extreme heat events in the 1951-1980 reference period? (Enter your answer rounded to one decimal place.) percent: (c) Based on the proposed Normal model, what percent of local summer temperature anomalies in the northern hemisphere between 2005 and 2015 were extreme heat events? (Enter your answer rounded to one decimal place.) percent (d) Based on the recording stations at numerous worldwide locations, 14.5% of temperature anomalies in the northern hemisphere were extreme heat events between 2005 and 2015, compared with 0.1% in the reference period of 1951 to 1980 Compare the actual values to the ones you obtained using the proposed Normal models. O Both values found using the Normal models are very close to the actual values. Neither value found using the Normal models is very close to the actual value. The value found for 1951-1980 using the Normal model is very close to the actual value, but not the one for 2005-201:5 O The value found for 2005-2015 using the Normal model is very close to the actual value, but not the one for 1951-1980.
In: Math
explain how to detect assignable causes of variation using probability paper
In: Math
Please use the skills you learned in section 9.2 for this assignment. For this activity, you will be creating a confidence interval for the average number of hours of TV watched. Last semester, my MAT 152 online students asked the question, “How many hours of screen time do you have in a typical week?” Please use the data they collected to answer the following questions. Data: 21, 2, 28, 30, 18, 21, 25, 20, 25, 14, 21, 25, 50, 39, 46, 20, 35, 45, 37, 46
Are there any outliers in this data set?
What calculator test (1-PropZInt, Z-Interval, or T-Interval), Excel, or StatCrunch function will you use to find the confidence interval?
In: Math