Parametric Models for Regression _ Coursera

10/22/2016

Par am etr i c M odel s for Regr essi on | C our ser a



Parametric Models for Regression



4/12 points earned (33%) (33%) You haven't passed yet. You need at least 70% to pass. Review the material and try again! You have 3 attempts every 8 hours.

Review Related Lesson



1/1 points

1. A manufacturer has developed a specialized metal alloy for use in jet engines. In its pure f orm, orm, the alloy starts to soften at 1500 F. However, small amounts of impurities in production cause the actual temperature at which the alloy starts to lose strength to vary around that mean, in a Gaussian distribution with standard deviation = 10.5 degrees F. If the manufacturer wants to ensure that no more than 1 in 10,000 of its commercial products will suer from softening, what should it set as the maximum temperature to which the alloy can be exposed? Hint: Refer to the Excel NormSFunctions Spreadsheet.

Excel NormS Functions Spreadsheet.xlsx

39.0497 F 1496.281 1460.9503 F Correct Response https://ww w.cour ser a.or g/ g/l ear n/ n/anal yti cs- ex excel /exam /c4Qel /par am am etr ic ic- mo model s- fo for -r -r eg egr es essi on

1/12

10/22/2016


The manufacturer needs to nd the z-score on the Gaussian distribution such that .0001 of the probability will be to the left of that temperature, and .9999 to the right. Using Excel NormSInv, the z-score for .0001 is z = -3.7190. The maximum temperature that achieves the desired reliability is = mean – ((z-score)*(standard ((z-score)*(standard deviation) ) = 1500 – (3.7190*10.5) = 1460.9503 F

1539.0497



0/1 points

2. A carefully machined wire comes o an assembly line within a certain tolerance. Its diameter is 100 microns, and all the wires produced have a uniform distribution of error, between -11 microns and +29 microns. A testing machine repeatedly draws samples of 180 wires and measures the sample mean. What is the distribution of sample means? Hint: Use the CLT and Excel Rand() Spreadsheet. Spreadsheet.

CLT and Excel Rand.xlsx

A Gaussian Distribution that, in Phi notation, is written ϕ(109, .7407). A Gaussian distribution that, in Phi notation, is written, ϕ(109, 133.33). Incorrect Response

133.33 is the variance of the uniform distribution of wire diameters, not the variance of sample means. Although the probability distribution of wire diameters is Uniform, the Central Limit Theorem explains that the distribution of sample means is a

https://ww w.cour ser a.or g/ g/l ear n/ n/anal yti cs- ex excel /exam /c4Qel /par am am etr ic ic- mo model s- fo for -r -r eg egr es essi on

2/12

10/22/2016


auss an. emem er a n no a on e secon num er s the variance (standard deviation squared) not the standard deviation.

A Uniform Distribution with mean = 109 microns and standard deviation = .8607 microns. A Uniform Distribution with mean = 109 microns and standard deviation = 11.54 microns.



1/1 points

3. A population of people suering from Tachycardia (occasional rapid heart rate), agrees to test a new medicine that is supposed to lower heart rate. In the population being studied, before taking any medicine the mean heart rate was 120 beats per minute, with standard deviation = 15 beats per minute. After being given the medicine, a sample of 45 people had an average heart rate of 112 beats per minute. What is the probability that this much variation from the mean could have occurred by chance alone? Hint: Use the Typical Problem with NormSDist Spreadsheet.

Typical Problem_ NormSDist .xlsx

99.9827% 1.73% .0173% Correct Response

The distribution of sample means has an expected mean of 120 and standard deviation of 15/sqrt(45) = 2.236. The z-score of the sample average of 112 is (112-120)/2.236 = -3.577. A z-Score this small or smaller has a probability of occurring by chance of only ( 1 –NormSDist(3.577)) = .0173% .

https://ww w.cour ser a.or g/ g/l ear n/ n/anal yti cs- ex excel /exam /c4Qel /par am am etr ic ic- mo model s- fo for -r -r eg egr es essi on

3/12

10/22/2016

.




0/1 points

4. Two stocks have the following expected annual returns: Oil stock – expected return = 9% with standard deviation = 13% IT stock – expected return = 14% with standard deviation = 25% The Stocks prices have a small negative correlation: R = -.22. What is the Covariance of the two stocks? Hint: Use the Algebra with Gaussians Spreadsheet.

Algebra with Gaussians.xlsx

-.00573 -.00219 -.00715 -.0286 Incorrect Response

Covariance is dened as R*Std.Dev(1)*Std.Dev(2). R*Std.Dev(1)*Std.Dev(2).



0/1 points

5.

Two stoc stocks ks have have the the follow followin in ex ected ected annua annuall return returns: s: https://ww w.cour ser a.or g/ g/l ear n/ n/anal yti cs- ex excel /exam /c4Qel /par am am etr ic ic- mo model s- fo for -r -r eg egr es essi on

4/12

10/22/2016


Oil stock – expected return = 9% with standard deviation = 13% IT stock – expected return = 14% with standard deviation = 25% The Stocks prices have a small negative correlation: R = -.22. Assume return data for the two stocks is standardized so that each is represented as having mean 0 and standard deviation 1. Oil is plotted against IT on the (x,y) axis. What is the covariance? Hint: Use the Standardization Spreadsheet.

Standardization Spreadsheet.xlsx

-.22 -.00573 Incorrect Response

See the Standardization Spreadsheet. Standardization Spreadsheet.

0 -1



0/1 points

6.

Two stocks have the following expected annual returns: https://ww w.cour ser a.or g/ g/l ear n/ n/anal yti cs- ex excel /exam /c4Qel /par am am etr ic ic- mo model s- fo for -r -r eg egr es essi on

5/12

10/22/2016


Oil stock – expected return = 9% with standard deviation = 13% IT stock – expected return = 14% with standard deviation = 25% The Stocks prices have a small negative correlation: R = -.22. What is the standard deviation of a portfolio consisting of 70% Oil and 30% IT? Hint: Use either the Algebra with Gaussians or the Markowitz Portfolio Optimization Spreadsheet.

Algebra with Gaussians.xlsx

Markowitz Portfolio Optimization.xlsx

10.44% 12.68% 11.79% Incorrect Response

Use the Markowitz Portfolio Optimization Spreadsheet. Enter Oil Expected return in Cell C6, Oil standard deviation in cell C7, IT expected return in Cell D6, IT standard deviation in Cell D7, Oil weighting in Cell C9 and IT weighting in Cell D9. Results in Cell E12.

17.93%



0/1 points

7.

Two stocks have the following expected annual returns: https://ww w.cour ser a.or g/ g/l ear n/ n/anal yti cs- ex excel /exam /c4Qel /par am am etr ic ic- mo model s- fo for -r -r eg egr es essi on

6/12

10/22/2016


Oil stock – expected return = 9% with standard deviation = 13% IT stock – expected return = 14% with standard deviation = 25% The Stocks prices have a small negative correlation: R = -.22. Use MS Solver and the Markowitz Portfolio Optimization Spreadsheet to Find the weighted portfolio of the two stocks with lowest volatility.

Solver Add-In.xlsx

Markowitz Portfolio Optimization.xlsx What is the minimum volatility? 11.58% 10.43% 10.36% 9.5% Incorrect Response



1/1 points

8.

An automobile parts manufacturer uses a linear regression model to https://ww w.cour ser a.or g/ g/l ear n/ n/anal yti cs- ex excel /exam /c4Qel /par am am etr ic ic- mo model s- fo for -r -r eg egr es essi on

7/12

10/22/2016


orec orecas astt t e o ar va ue o t e next next year years’ s’ or ers ers rom rom curr curren entt cust custom omer erss as a function of a weighted sum of their past-years’ orders. The model error is assumed Gaussian with standard deviation of $130,00 0. To the nearest dollar, what is the range above and below each Point Forecast required to have 90% conတdence that the dollar value of next years’ orders will fall within that range? Hint: you can answer this question by making small modications to the Correlation and Model Error Spreadsheet. Spreadsheet.

Correlation and Model Error.xlsx

The 90% condence interval is from $213,831 below to $213,831 above the point forecast. Correct Response

A 90% condence interval means 5% above and 5% below the interval. Using p = .95, Excel "Normsinv(.95)" = 1.64485. This is the z-Score. Therefore the interval ranges from (1.64485)*(130,00) or $213,831 below the point forecast, to (1.64485)*($130,000) = $213,831 above the forecast.

The 90% condence interval is from $83,831 below to $83,831 above the Point Forecast The 90% condence interval is from $316,831 below to $316,831 above the Point Forecast. The 90% condence interval is from $164,831 below to $164,831 above the Point Forecast.



1/1 points

9.

An automobile parts manufacturer uses a linear regression model to ’ https://ww w.cour ser a.or g/ g/l ear n/ n/anal yti cs- ex excel /exam /c4Qel /par am am etr ic ic- mo model s- fo for -r -r eg egr es essi on

8/12

10/22/2016


as a function of a weighted sum of their past-years’ orders. The model error is assumed Gaussian with standard deviation of $130,00 0. If the correlation is R = .33, and the point forecast orders $5.1 million, what is the probability that the customer will order more than $5.3 million? Hint: Use the Typical Problem with NormSDist Spreadsheet.


93.8% 4.3% 6.2% Correct Response

Convert $5.3 million to a z-score. The z-Score is (5.3-5.1)/.13 = (.2/.13) = 1.538. Input the z-score in the Excel formula (1 - Normsdist(z)) to get the probability of a score at least that high. The probability is 1- 93.8% = 6.2%.

12.4%



0/1 points

10.

An automobile parts manufacturer uses a linear regression model to forecast the dollar value of the next next ears’ orders from current customers https://ww w.cour ser a.or g/ g/l ear n/ n/anal yti cs- ex excel /exam /c4Qel /par am am etr ic ic- mo model s- fo for -r -r eg egr es essi on

9/12

10/22/2016


as a function of a weighted sum of that customer’s past-years orders. The linear correlation is R = .33. After standardizing the x and y data, what portion of the uncertainty about a customer’s order size is eliminated by their historical data combined with the model? Hint: Use the Correlation and P.I.G. Spreadsheet.

Correlation and P.I.G..xlsx

4.2% 3.5% Incorrect Response

4.5% 5.2%



0/1 points

11. Customers who use online chat support can rate the help they receive from a customer support worker as a 0 (useless), a 1, 2, 3, 4, or 5 (excellent). (excellent). The mean rating is 3.935, with standard deviation = 1.01. A new support worker named Barbara has received, over her rst 100 chat sessions, an average rating of 3.7. Her boss calls her in and threatens to re her if her performance does not improve. Barbara replies “Its just bad luck - I’ve had more than my share of unhappy customers today.” Who is most likely right? Hint: Use the Typical Problem with NormSDist Spreadsheet.


The boss https://ww w. w.cour se ser a. a.or g/ g/l ea ear n/ n/anal yt yti cs cs- ex excel /e /exam /c /c4Qel /p /par am am et etr ic ic- mo model ss- fo for -r -r eg egr es essi on on

10/12

10/22/2016


Barbara Incorrect Response

Barbara’s performance has a Z score = (mean - actual) /standard deviation of the sample means = (3.7 - 3.935) / (1.01/sqrt(100)) = -2.327. If Barbara’s performance is due to chance, the probability of observing a z-score as low, or lower, than -2.327 would be less than 1%. Note that due to the Central Limit Theorem, even though the raw customer scores do not have a Gaussian distribution, the sample means will have a Gaussian distribution, so that use of the NormSDist function is appropriate. NormSDist(-2.327) = .0099898.



0/1 points

12. Your company currently has no way to predict how long visitors will spend on the Company’s web site. All it known is the average time spent is 55 seconds, with an approximately Gaussian distribution and standard deviation of 9 seconds. It would be possible, after investing some time and money in analytics tools, to gather and analyzing information about visitors and build a linear predictive model with a standard deviation of model error of 4 seconds. How much would the P.I. G. of that model be? Hint: Use the Correlation and P.I.G. Spreadsheet

How to use the AUC calculator.pdf

48.2% 61.5% Incorrect Response

With no model, the correlation R = 0 and the “model error” is equal to the standard deviation of Y = 9 seconds. Standardized, the model error when R = 0 is equal to 1. Reducing the model error to 4 seconds is equivalent to reducing the standardized model error to 4/9 = .4444 Since Sqrt(1-R^2) = .4444, R^2 = 1https://ww w. w.cour se ser a. a.or g/ g/l ea ear n/ n/anal yt yti cs cs- ex excel /e /exam /c /c4Qel /p /par am am et etr ic ic- mo model ss- fo for -r -r eg egr es essi on on

11/12

10/22/2016


.4444^2 = .8958.

53.3% 57.2%



https://ww w. w.cour se ser a. a.or g/ g/l ea ear n/ n/anal yt yti cs cs- ex excel /e /exam /c /c4Qel /p /par am am et etr ic ic- mo model ss- fo for -r -r eg egr es essi on on





12/12

Parametric Models for Regression _ Coursera

Recommend Documents