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19 ERA 20 Win% 21 0.021 0.028 0.028 0.530 0.038 0.432 0.03 0.022 0.026 0.554 0.028 0.542 0.570 0.620 0.740 0.504 Compute the IQR for ERA. 23 24 25 26 27 28 29 30 31 32 Question 1A Question 1B Compute the covariance between ERA and win%. Question 1C Compute the correlation between ERA and win96. 34 36 Professor Smith took a sample of 211 students from her morning class and a sample of 160 students from her evening class. On Exam 1, all 37 those sampled from the morning class got a 64 while those sampled from the evening class averaged a 63 with a variance of 179. If 38 Professor Smith combines these two samples 39 40 42 43 45 46 47 48 49 Question 2 What is the combined Exam 1 average? Question 3A What is the combined Exam 1 median?Question 3B What is the shape of the combined Exam 1 grade distribution? Question4 What is the combined Exam 1 standard deviation? Note: Answer Questions 5-8 by Using the Excel Output The U.S. Department of Energys Fuel Economy Guide provides fuel efficiency data for cars and trucks. (See exercise 35 on page 77 in your textbook for a similar problem.) Use the data in the FuelData sheet in this workbook to generate excel output to answer the following questions. Please note that units are liters for engine size and miles per gallon for highway fuel efficiency. Put the output in given space on FuelData sheet. Each orange numerical answer cell below MUST reference generated Excel output cells in the FuelData sheet. Question SA What is range for engine size?Question SA What is range for engine size? Question 5B What is the variance for highway fuel efficiency? Question SC What is IQR for engine size? What is the shape of the highway fuel efficiency distribution? Why? NOTE 1: Skewed right must select>, skewed left must select<, etc.) NOTE 2: The shape of the distribution is based on the formatted 4-decimal numbers showing in your output.) Question 6 because Question 7 rule predicts that percent of the engine size data should lie between 0.9253 and 4.8536 liters, which are +/ standard deviation(s) away from the mean. Question 8What varied the most in this study, engine size or highway fuel efficiency? Prove using the best measure of variability! varied the most.Question 7nale predicts tht percent of the engine size data should lie 0.9253 4.8536 liters,which are + standard deviation(s) away from the mean. Question 8 What varied the most in this study, engine size or highway fuel efficiency? Prove using the best measure of variability! varied the most. Proof Engine Size Hwy Fuel EfficiencyEngine Size Hwy Fuei Efficiency 3.72 3.85 2.70 7.A3 1.93 2.05 2.31 2.57 2.83 2.95 2.95 44.00 28.00 19.00 25.00 32.DD 27.DD 43.00 Place Englne Size output (starting with label) In E2.> Place Hwy Fuel Eff output (starting with label) in H2 Mcan -> Mean> 22.00 19.00 8.00 45.DD 36.DD 43.00 42.00 37.00 42.00 .21 3.34 1.11 4.37 3.47 3.60 3.34 3.47 1.67 1.80 2.31 2.57 3.60 .72 2.95 3.08 Count - Count - 01- 01-> 21.00 39.DD 44.00 43.00 21.DU 23.00 40.00 45.DU305762714158298528052961293216 244424423443432232442233332444 2584735733581817317 7903490588902915824 3 2 3 3-3-1-2-2 2-2-2-3-33|4 2 2 3 3 1 1 2 2 1 2 2233 1 031 73535700 89 359089468 78901234567890123 22233333333334444 0123456 455 5 5 5 5 556 57 58 59 600 61 62 65.3 64 65 1.80 3.47 2.31 2.57 2.95 3.08 2.83 1.93 3.85 2.83 3.47 2.57 46.00 47.00 21.00 23.00 38.00 28.00 45.00 35.00 47.00 22.00 28.00 20.00 24.00 27.00 31.00 67 69 70 71 72 73 74 75 76 3.21 2.83 2.31

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Answer #1

Solution

Let for convenience of representation, x denote ERA and y represent Win%

[NOTE: Formulas and Answers are given below. Detailed Working follow at the end.]

Q1 Part (a)

First quartile, Q1 = average of the second and the third observation in the ordered set

= (0.022 + 0.026)/2

= 0.024

Third quartile, Q3 = average of the sixth and the seventh observation in the ordered set

= (0.028 + 0.030)/2

= 0.029

IQR = Q3 – Q1

= 0.005 Answer

Q1 Part (b)

Covariance = (1/n)Σ(i = 1, n){(xi – Mean X)(yi – Mean Y)}

= - 0.00027   Answer

Q1 Part (c)

V(X) = (1/n)Σ(i = 1, n)(xi – Mean X)2

V(Y) = (1/n)Σ(i = 1, n)(yi – Mean Y)2

Correlation coefficient = Covariance/sqrt{V(X).V(Y)}

= - 0.00027/(0.0049 x 0.0843)

= - 0.6653    Answer

Details of Working

Let for convenience of representation, dxi = (xi – Mean X) and dyi = (yi – Mean Y)

i

x (ERA)

y(Win%)

dxi^2

dyi^2

(dxi)(dyi)

1

0.028

0.53

1.40625E-07

0.00099225

-1.2E-05

2

0.021

0.542

4.38906E-05

0.00038025

0.000129

3

0.038

0.432

0.000107641

0.01677025

-0.00134

4

0.03

0.57

5.64063E-06

7.225E-05

2.02E-05

5

0.028

0.62

1.40625E-07

0.00342225

2.19E-05

6

0.022

0.74

3.16406E-05

0.03186225

-0.001

7

0.026

0.554

2.64062E-06

5.625E-05

1.22E-05

8

0.028

0.504

1.40625E-07

0.00330625

-2.2E-05

Total

0.221

4.492

0.000191875

0.056862

-0.0022

Mean

0.0276

0.5615

2.39844E-05

0.00710775

-0.00027

Square Root of Variance

0.0049

0.0843

Ordered Set

0.021
0.022
0.026
0.028
0.028
0.028
0.03
0.038

DONE

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