1.  the random variable Y has a Normal distribution with mean 60 and standard deviation 4 find P( 61 < Y < 70)

 

2. In a random sample of 100 taken from over 55,000 fatal traffic accident reports, alcohol is the apparent cause in 39 reports.  Find a 99% confidence interval for the proportion of all accidents in which alcohol was the cause.
3. It is estimated that 55% of the freshmen entering a particular college will graduate from that college in four years.  For a random sample of 5 entering freshmen, what is the probability that 3 will graduate in four years?
4. In this problem data are taken from an experiment designed to investigate the effects of three environmental variables on exhaust emissions of light duty diesel trucks.  The variable of interest in the emissions of nitrogen oxide (NOX).  The environmental factors are the humidity (%), the temperature (degrees F) and the barometric pressure (inches Hg).  The following is the regression output for this data.

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

 

1. Test the null hypothesis that ????????₁ = ????????₂ = ????????₃ = 0 versus the alternative hypothesis that not all of  ????????₁,

????????₂, and ????????₃ are 0.  

 

 

 

2. Which of the three variables humidity, temperature, and pressure are useful in predicting the emissions of nitrogen oxide?  Justify your answer.

 

 

 

 

3. Find a 95% confidence interval for the parameter ????????₃, the coefficient for the variable pressure.  (The t value you need is t = 2.12)

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.

Two-way ANOVA: RMR versus Subject, Protocol 

 

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.

 

 

 

 

6. The shelf life of a type of battery has a Normal distribution with mean 525 days and a standard deviation of 50 days.  If a battery has been in storage for 647 days, what is the probability of it being dead?

 

 

 

 

 

 

 

 

 

 

7. Park managers need to know how resistant different vegetative types are to trampling so that the number of visitors can be controlled in sensitive areas.  The experiment deals with alpine meadows in the White Mountains of New Hampshire.  Twenty lanes were established, each .5 meters wide and 1.5 meters long.  These 20 lanes were randomly assigned to five treatments:  0, 25, 75, 200 or 500 walking passes.  Each pass consists of a 70 kg individual wearing lug-soled boots walking in a natural gait down the lane.  The response measured is the average height (in centimeters) of the vegetation along the lane one year after trampling.  Output from a one way ANOVA follows.

One-way ANOVA: Height versus Passes 

 

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

  

 

1. Based on this output is there a difference in the mean height of the vegetation for the 5 different treatments?  Justify your answer.

 

 

 

 

2. Based on this output, is there a difference in the mean height of the vegetation for the case in which there were 75 passes compared to the case in which there were 500 passes?  Justify your answer. 8)  Weeds reduce crop yields so farmers are always looking for better ways to control weeks.  14 weed control treatments were randomized in 56 experimental plots that were planted in soybeans.  The plots were later visually assessed for weed control, the fraction of the plot without weeds.  The output from a one way ANOVA is given below.

One-way ANOVA: Percent weed control versus group 

 

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.

Factorial Fit: Strength versus Amount, Brand, Type 

 

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%

 

 

 

 

1. Based on this output, which main effects and interactions affected the strength?  Justify your answer

 

 

 

2. If you wish to maximize the strength, what levels would you set each of the three varibles?

Amount of  liquid:

Brand of towel:

Type of liquid:

 

10. The Heinz frozen food corporation in Ontario, Oregon was selected to produce a new bag for Ore-Ida frozen fries.  When the bags were first produced there were problems with the consistency.  It was unclear whether the problem was the machinery or the “film” (the material used in the bags).  One of the key measurements was the distance from the UPC (universal product code) and a black mark on the bag.  A number of rolls of film were selected and this measurement was recorded.  The following is an X bar  control chart for this measurement.

 

 

 

consistency of the measurement on the film.

 

11. A credit card company wants to find out if the average time that its customers take to pay their accounts will change depending upon when they are billed.  They take a random sample of 500 customers and sent their bills on the 15^th of the month.  These customers paid their bills in an average of 18.9 days with a standard deviation of 5.3 days.  They took a second random sample of 500 customers and sent their bills on the 24^th of the month.  These customers paid their bills in an average of 13.3 days with a standard deviation of 3.7 days. 

1. Find a 95% confidence interval for the difference in the mean payment time when customers are billed on the 15^th and when customers are billed on the 24^th.

 

 

 

2. Write a sentence interpreting the result in part a).

 

 

 

 

12. Data was collected on the types of defects that were made by two workers doing the same job.  There were three types of defects:  surface scratches, cracks, and incomplete assembly.  The follow are Pareto

 

 

 

 

 

13. In a study about the reasons that business locate to Florida, the following table gives the primary reason broken down by the industry type.  The data is given in the following table.

                                                                                          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.

Chi-Square Test: Manufacturing, Retail, Tourism 

 

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

 

 

1. Comment on this histogram, what is the most important feature?

 

 

 

 

 

2. What is an appropriate follow up action based on this histogram?

 

 

 

 

 

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:

Regression Analysis: mpg versus weight 

 

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:

Regression Analysis: mpg versus weight, weightsq 

 

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:  

Regression Analysis: mpg versus weight, weightsq, horsepower 

 

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%

 

 

 

 

2. Based on the output for these three models, which model fits the data the best.  Justify your answer.
3. Suppose you have a car with weight = 2500 pounds and displacement = 127 cubic inches use the model you believe is most appropriate to predict the mpg of this car.

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.)