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Homework answers / question archive / the random variable Y has a Normal distribution with mean 60 and standard deviation 4 find P( 61 < Y < 70) In a random sample of 100 taken from over 55,000 fatal traffic accident reports, alcohol is the apparent cause in 39 reports
The regression equation is
NOX = - 11.6 + 0.000114 Humidity + 0.00177 Temperature + 0.422 Pressure
Predictor Coef SE Coef T P
Constant -11.605 2.965 -3.91 0.001
Humidity 0.0001141 0.0006266 0.18 0.858
Temperature 0.001768 0.001920 0.92 0.371
Pressure 0.4223 0.1013 4.17 0.001
S = 0.0501715 R-Sq = 62.6% R-Sq(adj) = 55.5%
Analysis of Variance
Source DF SS MS F P
Regression 3 0.067300 0.022433 8.91 0.001
Residual Error 16 0.040275 0.002517
Total 19 0.107575
????????2, and ????????3 are 0.
5) An experiment was conducted to compare the effects of inpatient and outpatient protocols on the inlaboratory measurements of resting metabolic rate (RMR) in humans. A previous study had indicated measurements of resting metabolic rate on elderly individuals to be 8% higher using an outpatient protocol than with an inpatient protocol. If the measurements depended on the protocol, then comparison of the results of studies conducted by different laboratories using different protocols would be difficult.
The experimental treatments consisted of three protocols: (1) an inpatient protocol in which meals were controlled – the patient was fed the evening mean and spent the night in the laboratory, then RMR was measured in the morning; (2) an outpatient protocol in which meals were controlled – the patient was fed the same evening mean at the laboratory but spent the night at home, then RMR was measured in the morning; and (3) an outpatient protocol in which meals were not strictly controlled – the patient was instructed to fast for 12 hours prior to measurement or RMR in the morning.
Since subjects tend to differ substantially from each other, the subjects were treated as block in the experiment. In this experiment there were nine subjects (healthy, adult males of similar age). Every subject was measured under all three treatments.
A 2 way ANOVA model was fitted to this data. The following is a graph of the residuals versus the fitted values.
a) Comment on this plot, does the plot suggest any problems with the 2 way ANOVA model? Why or why not?
The output from the 2 way ANOVA model is given below.
Source DF SS MS F P
Subject 8 23117462 2889683 37.42 0.000
Protocol 2 35949 17974 0.23 0.795
Error 16 1235483 77218
Total 26 24388894
S = 277.9 R-Sq = 94.93% R-Sq(adj) = 91.77%
b) Based on this output test the null hypothesis that the means for the three different protocols are all the same versus the alternative hypothesis that the means for the three different protocols are not all the same.
Based on this experiment what conclusion can be made about the use of the three different protocols? Justify your answer.
Source DF SS MS F P
Passes 4 243.16 60.79 29.48 0.000
Error 15 30.93 2.06
Total 19 274.09
S = 1.436 R-Sq = 88.72% R-Sq(adj) = 85.71%
Individual 95% CIs For Mean Based on
Pooled StDev
Level N Mean StDev --+---------+---------+---------+-------
0 4 18.000 1.992 (---*----)
25 4 12.000 2.022 (---*----)
75 4 11.975 0.506 (---*----)
200 4 9.000 1.003 (----*---)
500 4 8.000 0.997 (----*---)
--+---------+---------+---------+-------
7.0 10.5 14.0 17.5
Pooled StDev = 1.436
Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons among Levels of Passes
Individual confidence level = 99.25%
Passes = 0 subtracted from:
Passes Lower Center Upper --+---------+---------+---------+-------
25 -9.137 -6.000 -2.863 (----*----)
75 -9.162 -6.025 -2.888 (----*----)
200 -12.137 -9.000 -5.863 (----*----)
500 -13.137 -10.000 -6.863 (----*-----)
--+---------+---------+---------+-------
-12.0 -6.0 0.0 6.0
Passes = 25 subtracted from:
Passes Lower Center Upper --+---------+---------+---------+-------
75 -3.162 -0.025 3.112 (----*----)
200 -6.137 -3.000 0.137 (----*----)
500 -7.137 -4.000 -0.863 (----*-----)
--+---------+---------+---------+-------
-12.0 -6.0 0.0 6.0
Passes = 75 subtracted from:
Passes Lower Center Upper --+---------+---------+---------+-------
200 -6.112 -2.975 0.162 (----*----)
500 -7.112 -3.975 -0.838 (----*-----)
--+---------+---------+---------+-------
-12.0 -6.0 0.0 6.0
Passes = 200 subtracted from:
Passes Lower Center Upper --+---------+---------+---------+-------
500 -4.137 -1.000 2.137 (----*-----)
--+---------+---------+---------+-------
-12.0 -6.0 0.0 6.0
Source DF SS MS F P group 13 5063.1 389.5 11.66 0.000
Error 42 1402.8 33.4
Total 55 6465.8
S = 5.779 R-Sq = 78.31% R-Sq(adj) = 71.59%
The following plot is a histogram of the residuals from this model. |
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Comment on this plot. Does this plot suggest any problems with this model? Justify your answer.
9) An experiment compared the relative strengths of two similarly priced brands of paper towels under varying levels of moisture saturation and liquid type. The treatment factors were “amount of liquid” 5 drops and 10 drops), “brand of towel” (brand A and brand B), and “type of liquid” (beer and water). The data from the experiment are in the following table. This experiment was replicated 3 times.
Amount |
Brand |
Type |
Strength R1 |
Strength R2 |
Strength R3 |
-1 |
-1 |
-1 |
3279.0 |
4330.7 |
3843.7 |
-1 |
-1 |
1 |
3260.8 |
3134.2 |
3206.7 |
-1 |
1 |
-1 |
2889.6 |
3019.5 |
2451.5 |
-1 |
1 |
1 |
2323.0 |
2603.6 |
2893.8 |
1 |
-1 |
-1 |
2964.5 |
4067.3 |
3327.0 |
1 |
-1 |
1 |
3114.2 |
3009.3 |
3242.0 |
1 |
1 |
-1 |
2883.4 |
2581.4 |
2385.9 |
1 |
1 |
1 |
2142.3 |
2364.9 |
2189.9 |
The following is the output from the analysis of this data.
Estimated Effects and Coefficients for Strength (coded units)
Term Effect Coef SE Coef T P
Constant 2979.5 66.82 44.59 0.000
Amount -247.0 -123.5 66.82 -1.85 0.083
Brand -837.6 -418.8 66.82 -6.27 0.000
Type -378.2 -189.1 66.82 -2.83 0.012
Amount*Brand -25.2 -12.6 66.82 -0.19 0.853
Amount*Type 20.4 10.2 66.82 0.15 0.880
Brand*Type 95.9 48.0 66.82 0.72 0.483
Amount*Brand*Type -122.7 -61.3 66.82 -0.92 0.372
S = 327.332 PRESS = 3857257
R-Sq = 76.51% R-Sq(pred) = 47.16% R-Sq(adj) = 66.24%
Amount of liquid:
Brand of towel:
Type of liquid:
consistency of the measurement on the film.
Industry Type
Primary Reason Manufacturing Retail Tourism
Emerging Technology 53 25 10
Tax Credits 67 36 20
Labor Force 30 40 33
The following output is for a chi-squared test that the proportions for the three different primary reasons for locating in Florida are the same for all the different industry types.
Expected counts are printed below observed counts
Chi-Square contributions are printed below expected counts
Manufacturing Retail Tourism Total
Emerging Tech 53 25 10 88
42.04 28.31 17.66
2.858 0.386 3.320
Tax Credits 67 36 20 123
58.76 39.56 24.68 1.156 0.321 0.887
Labor Force 30 40 33 103
49.20 33.13 20.67 7.495 1.424 7.362
Total 150 101 63 314
Chi-Sq = 25.210, DF = 4, P-Value = 0.000
Based on this output what conclusion can be made about the reasons for locating in Florida? Are the proportions the same for all three industry groups? Justify your answer.
14) A subcontracting company processed sheet metal panels for an electrical machine maker. One key feature was the hardness of the metal. The following plot is a histogram of the hardness of 50 metal
15) In a sample of 392 vehicles we are interested in predicting the miles per gallon (mpg) for the car as a function of the weight of the car (weight) and the engine displacement (displacement). In exploring the relationship between these two variables, the following regression models were fit to the data.
Model 1: ???????????????? = ????????0 + ????????1????????1,???????? + ???????????????? where ???????????????? is the mpg of the ith car and ????????1,???????? is the weight of the ith car.
Model 2:
where ???????????????? is the mpg of the ith car and ????????1,???????? is the weight of the ith car.
Model 3:
where ???????????????? is the mpg of the ith car, ????????1,???????? is the weight of the ith car, and ????????2,???????? is the displacement of the ith car.
a) Both model 1 and model 2 only have the mpg dependent upon the weight of the car. Write a short sentence that describes the difference between model 1 and model 2.
The regression output for all three models are provided below.
Model 1:
The regression equation is mpg = 46.2 - 0.00765 weight
Predictor Coef SE Coef T P Constant 46.2165 0.7987 57.87 0.000 weight -0.0076473 0.0002580 -29.65 0.000
S = 4.33271 R-Sq = 69.3% R-Sq(adj) = 69.2%
Model 2:
The regression equation is
mpg = 62.3 - 0.0185 weight + 0.000002 weightsq
Predictor Coef SE Coef T P Constant 62.255 2.993 20.80 0.000 weight -0.018496 0.001972 -9.38 0.000 weightsq 0.00000170 0.00000031 5.55 0.000
S = 4.17635 R-Sq = 71.5% R-Sq(adj) = 71.4%
Model 3:
The regression equation is
mpg = 64.6 - 0.0182 weight + 0.000002 weightsq - 0.0611 horsepower
Predictor Coef SE Coef T P Constant 64.596 2.908 22.22 0.000 weight -0.018189 0.001897 -9.59 0.000 weightsq 0.00000202 0.00000030 6.75 0.000 horsepower -0.06107 0.01070 -5.71 0.000
S = 4.01643 R-Sq = 73.7% R-Sq(adj) = 73.5%
16) Assume that the test scores from a college admissions test are Normally distributed, with a mean score of 476 and a standard deviation of 85. A certain college will automatically give a distinguished alumni scholarship to any student who is in the top 1% of those taking the test. What score does a student need to get on this test to qualify for this scholarship? (Hint: The score for which 1% of the students are above also has 99% of the students below.)
Please use this google drive link to download the answer file.
https://drive.google.com/file/d/1P08IfN7rP7Xnad7ahHWr4xsREWZNVT3D/view?usp=sharing
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https://helpinhomework.org/blog/how-to-obtain-answer-through-google-drive-link