STAT 101, Module 3: Numerical Summaries for



STAT 101, Module 5: Logarithms and Regression

Factor Changes versus Percentage Changes

Example: CEO S.O.Rich of the company O.S.O.Poor has his total compensation increased by 7%. It is rising from $1,000,000 to $1,070,000.

Q: By what factor is total compensation changing?

A: By a factor 1.07 = 1 + 7/100.

Q: If compensation is lowered by 3%, what’s the factor change?

A: The factor is 0.97 = 1 – 3/100. (New compensation: $970,000)

Comment: When people calculate percentage changes, they calculate hundreths and add/subtract them from the amount:

(1,000,000 /100) · perc + 1,000,000

This, however, is the same as:

1,000,000 · (1+perc /100) = 1,000,000 · f

In what follows we will need this last form of expressing changes.

In general:

If a quantity Z changes by perc %, it changes by a factor

f = 1 + perc /100 ,

Conversely, if the quantity changes by a factor f, it changes by this many percentages:

perc = ( f – 1) ·100

The change is commonly computed as

Znew = Zold + (Zold / 100) · perc ,

which is the same as

Znew = Zold · ( 1 + perc/100) = Zold · f .

As we said earlier, it is this second form that we will need.

Q: Why percentage changes? Why not always factor changes?

A: In business, changes are often small, in the order of a few hundredths. So one keeps track of the hundredths only.

Example: Of a change by a factor 1.07, one retains the 7, which is the percentage. It’s a sort of universal laziness. Of course it comes to haunt us because we have to become adept at going back and forth between 7 and 1.07, and –3 and 0.97.

Practice:

• What is the factor change for a 10%, 50%, 0.2%, –80%, 1000%, 100%, –200%, –50%, –.1% change?

• What is the percentage change for a change by a factor 1.7, 2.0, 1.001, .92, –1.5, 1.000078, .999935, .5?

Factor/Percent Changes and Logarithms

Basics of logarithms:

• Two types of commonly used logarithms: base 10 and base e=2.718282 (e is here the transcendental number, not a residual variable). The base-10 logarithm is written log10(Z) or log10(Z), the natural base-e logarithm ln(Z).

(JMP, however, uses log(Z) for the natural logarithm).

• Logarithms are the inverses of exponentiations:

ln(ea) = a , log10(10a) = a .

For example, log10(1,000,000) = 6.

• ln(1) = 0, log10(1) = 0, because e0 = 100 = 1.

[pic]

The reason for using logarithms:

Logarithms transport multiplicative changes to additive changes:

ln( Z · f ) = ln( Z ) + ln( f ) .

In other words: If a quantity Z changes by a factor f, its logarithm ln(Z) changes by an amount ln( f ). In yet different words: Logarithms transport factor changes to amount changes.

(In English, “amount” is something that gets added, whereas “factor” is something that gets multiplied.)

JMP: You can use JMP as a logarithm table or logarithm calculator by forming two columns, one containing the input values, the other being a formula with the logarithm of the input values.

1. Create a blank spreadsheet by clicking the left-most icon in the second tool bar.

2. Create a column with name ‘ f ‘, say: Cols > New Column… > (Column Name) > f ,

3. Create a column with name ‘ln(f)’: Cols > New Column… > (Column Name) > ln(f) , (Column Properties) > Formula > (Functions grouped) > Transcendental > Log (=ln in JMP)

4. Enter numbers in the column ‘f’. Their natural logarithms will appear instantly in the column ‘ln(f)’.

Small factor approximation for natural logarithms:

ln( 1 + perc / 100 ) ≈ perc / 100

This approximation is quite good for –10 ≤ perc ≤ 10.

It can also be written as follows:

ln( 1 + h ) ≈ h ,

which is quite good for –0.1 ≤ h ≤ 0.1 .

(Try a few values between f = 0.9 and f = 1.1 on the above JMP calculator.)

Important: This approximation holds only for the natural logarithm, not the 10-based logarithm.

Check the graph of the natural logarithm: the approximation means that the 45 degree line and the curve of the logarithm hug each other at 1. Technically, the line is the first order Taylor approximation to the natural logarithm at 1. It uses the fact that the derivative of the natural logarithm at 1 is 1, that is, d ln(1+h) / dh = 1 at h=0. This implies (ln(1+h)–ln(1))/h ≈ 1 for small h, that is, ln(1+h)/h ≈ 1, or ln(1+h) ≈ h.

The convexity of the logarithm yields not only an approximation but an inequality as well: ln( 1 + h ) ≤ h, with equality for h = 0.

Changes by small percentages

If a quantity Z changes by fewer than 10%, we can use the small factor approximation:

ln( Z · (1 + perc / 100) ) ≈ ln(Z ) + perc / 100 ,

which is quite good for –10 ≤ perc ≤ 10.

In words: If a quantity Z changes by perc%, then ln(Z) changes by an approximate amount perc / 100.

Examples:

• If Z changes by +1%, then ln(Z) gets added 0.01.

• If Z changes by +5%, then ln(Z) gets added ….

• If Z changes by +10%, then ln(Z) gets added ….

• If Z changes by –1%, then ln(Z) gets added ….

• If Z changes by –5%, then ln(Z) gets added ….

• If Z changes by –10%, then ln(Z) gets added ….

For percent changes greater than +10% or less than –10%, one has to calculate ln(1+perc/100) and add to ln(Z).

Examples:

• If Z changes by +15%, then ln(Z) gets added +0.140.

• If Z changes by +50%, then ln(Z) gets added +….

• If Z changes by –15%, then ln(Z) gets added –….

• If Z changes by –50%, then ln(Z) gets added –….

Get the added values to three decimals from the above logarithm calculator in JMP, or from your pocket calculator.

[xxx Add reverse list of examples: ln(Z) was added 0.01 => Z was multiplied by 1.01 => Z was added 1%]

Factor/Percent Changes and Exponentials

Basics of exponentials:

• Two types of commonly used exponentials: base 10 and base e=2.718282 (again, e is here the transcendental number, not a residual variable). The base-10 exponential is written 10a, the base-e exponential is written ea or exp(a).

• Exponentiations are the inverses of logarithms:

exp(ln(Z)) = Z , 10log10(Z) = Z .

• exp(0) = 1, 100 = 1.

Exponential bases:

The natural exponential function exp(x) can be used to represent exponential functions to any other basis. In particular, f x can be written

f x = exp(bx) , where b = ln(f) .

Expressions such as f x occur naturally in business contexts, for example, when a company grows by a factor f = 1+perc/100 every year. It also occurs in fixed-income investments where the interest gets compounded monthly or quarterly or yearly. Lastly, it occurs in depreciation when some object loses a fixed percentage in value every year.

Small factor approximation for exponentials:

The natural exponential has a slope (derivative) of 1 at 0 and exp(0)=1. Therefore, the first order Taylor approximation is

exp(h) ≈ 1 + h ,

which is a satisfactory for –0.1 ≤ h ≤ +0.1. As the following graph shows, the exponential is convex, hence the approximation is actually an inequality: exp(h) ≥ 1 + h , with equality for h=0.

[pic]

Logarithms and Regression

The purpose of logarithms in regression is to extend the reach of linear regression so it can be used to describe certain types of non-linear association. To be exact, logarithms can be used to reduce the following three types of non-linear association to linear associations:

1. Exponential growth/decay/depreciation: This is an association of the form

y = exp( b0 + b1 x ) .

Taking a logarithm on both sides reduces it to a linear assocation between ln(y) and x:

ln( y ) = b0 + b1 x .

[pic]

Practical use: In many business applications x is related to time. We then speak of exponential growth or decay, depending on whether b1 is positive or negative. Examples:

• Exponential growth: Revenue of a company might grow by a constant 7% every year.



• Exponential decay: The customer base of a company might shrink every year by a constant 8%.



• Exponential depreciation: This term is used when x = age, y = value of assets, and b1 < 0. The value of assets may decrease, for example, by a fixed 15% every year.

[We are concerned here with empirical/economic depreciation, that is, loss in actual market value; this is different from stylized accounting rules that use, for example, fixed amount depreciation over 5 years, which may have very little to do with actual values of aging assets but are used to satisfy reporting and tax requirements.



We think of the descending exponential as a potentially realistic approximation to actual values as a function of age. Whether this is realistic or not needs to be decided by residual analysis after fitting an exponential to age-value data, or, as we will actually do, after fitting a straight line to age-ln(value) data.)

Math: The exponential formula above describes constant factor changes as follows.

A unit difference in x is associated with a difference in y by a factor exp(b1).

Proof: y = exp( b0 + b1 x )

= exp(b0) exp(b1)x

= Z f x where Z = exp(b0) and f = exp(b1) .

Note: Amount difference in x → Factor difference in y

Approximation: If b1 is small, –0.1 ≤ b1 ≤ +0.1, then:

Unit difference in x

→ difference in y by a factor f = exp(b1) ≈ 1+ b1

→ difference in y is approximately by 100 · b1 %

Mechanics: Exponential associations are linear associations between x and ln(y). Hence the only “difficulty” is in creating a ln(y)-variable and interpreting the meaning of the coefficients appropriately.

Restriction: Exponential associations can exist only when y takes on positive values, y > 0.

2. Logarithmically diminishing returns: These are associations of the form

y = b0 + b1 ln(x) with b1 > 0.

[pic]

General concept of diminishing returns: In general, diminishing returns are situations where constant increases in x are associated with lesser increases in y for larger x. Mathematically speaking, diminishing returns are present whenever the association between x and y is described by a monotone increasing but concave function, as in the following three examples.



[pic]

The third is the worst in that y is bounded above, whereas the others at least aren’t bounded. We are limiting ourselves to the middle case of logarithmically diminishing returns.

Practical use: Diminishing returns frequently arise in business contexts when x measures effort to raise y, such as

• x = advertizing expenditure, or

• x = display space allocation for a product in a store, and

the response y is an outcome such as sales of, or profits from, the advertized or displayed product.

It is a common experience in such situations that initial effort has considerable effect, but raising the effort encounters “diminishing returns”, that is, the same increase in effort brings a lesser increase in outcomes (sales) when the effort is large.

Math: Logarithmically diminishing returns describe the association between x and y as follows:

A difference in x by a factor f is associated with a difference in y in the amount b1 ln( f ).

Proof: ynew =  b0 + b1 ln( f x)

= ( b0 + b1 ln(x) ) + b1 ln( f )

= yold + b1 ln( f )

Note: Factor difference in x → Amount difference in y

Approximations: It is natural to use a 1% or 10% difference in x for simple interpretation of b1. (We’ll throw in a 100% difference also, which is not small.)

• 1% difference in x → f = 1.01 → ln( f ) ≈ 0.01

→ difference in y in the amount ~ b1/100

• 10% difference in x → f = 1.1 → ln( f ) ≈ 0.1

→ difference in y in the amount ~ b1/10

• 100% difference in x → f = 2 → ln( f ) ≈ 0.7

→ difference in y in the amount ~ 0.7 b1 [ln(2)=0.693]

Illustration: Assume advertizing has a logarithmically diminishing returns relation with sales. If one gets an increase in sales of $1,000,000 by increasing advertizing from $100,000 to $200,000, one gets also an increase in sales of $1,000,000 by increasing advertizing from $500,000 to $1,000,000. In both cases, effort is doubled, resulting in the same amount increase in sales. To make it even more drastic, increasing advertizing from $10,000,000 to $20,000,000 would also result in a sales increase of $1,000,000.

[In this hypothetical illustration, what is b1?

From the given information, can you infer b0 ?]

Mechanics: Logarithmically diminishing returns are linear associations between ln(x) and y. Hence the only “difficulty” is in creating a ln(x)-variable and interpreting the meaning of the coefficients appropriately.

Restriction: Logarithmically diminishing returns can only hold when x takes on only positive values, x > 0. Otherwise the logarithm is not defined.

3. Constant elasticity: This type of association is really a power law,

y = a xb.

The power or “elasticity” b can be positive or negative. Which of the power law curves below have negative elasticity? The plot shows powers –2, –1, –1/2, 1/2, 1, 2. Annotate it with the correct power.

[pic]

Power laws can be reduced to a linear assocation by taking logarithms on both sides:

ln(y) = ln(a) + b ln(x) ,

which can be rewritten as

ln(y) = b0 + b1 ln(x)

with b0 = ln(a) and b1 = b. Thus a power law becomes a linear association between ln(x) and ln(y) and can therefore be estimated with a straight line fit.

Math: Constant elasticity (= a power law) describes the association between x and y as follows:

A difference in x by a factor f is associated with a difference in y by a factor f b.

Proof: ynew =  a ( f x ) b

= ( a xb ) f b

= yold f b

Note: Factor difference in x → Factor difference in y

Percent difference in x → Percent difference in y

Terminology: Associations described as factor-to-factor or percent-to-percent mappings are called “elasticities”.

Keep in mind that we are only dealing in constant elasticity. Whether this is a reasonable assumption or not must be checked. For background on elasticity, see:



Approximation: It is natural to use a 1% difference in x for simple interpretation of b (= b1).

• 1% difference in x

→ difference in x by a factor f = 1.01

→ difference in y by a actor f b

→ difference in ln(y) by amount ln( f b ) = b ln(f ) ≈ b/100

→ difference in y by a factor ≈ 1+b/100

→ difference in y by ~ b%

… if: –10 ≤ b ≤ +10.

1% difference in x → ~ b % difference in y

assuming |b|≤ 10

Practical uses: Elasticities occur primarily in the relation between x = price and y = quantity sold, with negative b: If the price is raised by 1%, quantity sold changes by b% (assuming | b | ≤ 10). In business and economics, elasticities are almost always negative, and the minus sign is usually not mentioned but implied. Hence an elasticity of 0.8 is meant to mean b = –0.8.

• In a homework we will encounter very different types of elasticities, where b > 0: in biology, the association between brain weight and body weight across species.

• It should also be mentioned that for 0 < b < 1 (and a > 0), the association is increasing and concave, hence could be used to describe certain types of diminishing returns: a 1% increase in effort (advertizing) results in a b% increase in outcomes (sales).

Mechanics: Constant elasticities are linear associations between ln(x) and ln(y). Hence the only “difficulty” is in creating a ln(x)-variable, a ln(y)-variable, and interpreting the meaning of the coefficients appropriately.

Restriction: Since we take logarithms of both x and y, both must take on only positive values, as is the case for prices and quantities sold.

Note on causality: In the contexts of depreciation, diminishing returns and elasticity, one assumes that there is a causal relationship between x and y based on economic theory. Hence a change in x is assumed to be responsible for a change in y.

Simplified Summary:

1. Exponential association:

If x → x+c, then y → y · f

amount change in x → factor change in y

2. Logarithmically diminishing returns:

If x → x · f, then y → y + c ( f > 1)

factor change in x → amount change in y

3. Elasticity, or elastic association:

If x → x · 1.01, then y → y · f

factor change in x → factor change in y

Exponential Growth: The Internet 1997-2000

The dataset Web Servers 1997-2000.JMP  contains the number of web servers for the internet for the booming years of 1997 through 2000. The following plots show the time series twice, on two differently labeled time axes.

[pic]

[pic]

This is clearly a candidate for exponential growth.

JMP has two ways of fitting an exponential, none replacing the other. Hence we will typically run both ways:

1. Analyze > Fit y by x > … select x and y >

click little red icon, top left of scatterplot > Fit Special…

> Y Transformation: (click) ○Natural Logarithm: log(y)

> OK

[pic]

Log(Web.Servers) = 13.408146 + 0.9001777 Yrs.since.Jan.97

Summary of Fit

| | |

|RSquare |0.995876 |

|Root Mean Square Error |0.06414 |

|Mean of Response |15.35853 |

|Observations (or Sum Wgts) |45 |

Before analyzing the results, we create the second type of JMP output.

2. Create one new column with name ‘ln(y)’ with the log function in the formula (here: ‘ln(Web.Servers)’), then:

Analyze > Fit y by x > … select x and ln(y) >

click little red icon, top left of scatterplot > Fit Line > OK

[pic]

ln(Web.Servers) = 13.408146 + 0.9001777 Yrs.since.Jan.97

Summary of Fit

| | |

|RSquare |0.995876 |

|Root Mean Square Error |0.06414 |

|Mean of Response |15.35853 |

|Observations (or Sum Wgts) |45 |

The two outputs agree in the numbers even though they differ in the plots. The agreement in numbers stems from the fact that the Fit Special version (1.) also fits a straight line to the x-ln(y) data and reports those numbers, even though it shows an x-y plot.

Interpretation of numeric output:

• Equation, slightly rounded:

ln(Web.Servers) = 13.4 + 0.90 Yrs.since.Jan.97

Exponentiated, that is, exp() applied to both sides:

Web.Servers = exp(13.4) exp(0.90) Yrs.since.Jan.97

where exp(13.4) = 660,003 and exp(0.90) = 2.46 .

This means that the estimated number of web servers at the beginning of 1997 is 660,000, and it grew every year on average by a factor 2.46 or 146%.

In rounder numbers: beginning of 1997 ~ 2/3 million servers, growing by almost 150% per year.

(Note that there are no small change approximations possible here: the growth rate of 146% is anything but small.)

• Quality of fit:

o R Square is 0.996, which is extremely high.

Recall that this is the value for the x-ln(y) data from the version 2. output.

o The RMSE is 0.064, which needs careful attention: It means the residuals on the ln(y)-scale are in the amount ±0.064, meaning on the y-scale they are in the order of a factor exp(±0.064) ≈ 1±0.064, or ±6.4%.

▪ This is something to put your mind around: the RMSE starts out as a ±amount on the ln(y)-scale and translates to a ±percentage on the y-scale.

▪ The second thing to think through is that the simple translation to ±6.4% is due to the small change approximation. It would be messier if the RMSE were something larger, like 0.3: exp(±0.3) = 1.35 and 0.74, resp. This means the residuals on the y scales are within +35% and –26%. Note the asymmetry!

Residuals: The two versions of JMP outputs result in two different residual plots.

[pic]

[pic]

1. Fit Special shows the residuals off the curve in the x-y data:

ei = yi – exp( b0 + b1 xi )

2. Fit Line shows the residuals off the line in the x-ln(y) data:

ei = ln(yi) – ( b0 + b1 xi )

Which one is preferable? Surprisingly, it is the second. One should always show the residuals where the line was fitted, here: the x-ln(y) data. They translate to ‘factor residuals’ on the x-y scale:

exp(ei) = yi / exp( b0 + b1 xi )

It says that yi is off of exp( b0 + b1 xi ) by a factor exp(ei).

Residual analysis:

1. The left plot shows that in absolute terms the exponential has very small residuals early on and very large ones later.

2. The right plot shows that in percentages, the residuals at the beginning are as large as the later ones.

Both plots, the second better than the first, show that there is a systematic pattern in that early and late residuals are positive while middle residuals are negative. We know that this indicates curvature, hence an unsatisfactory residual structure: knowing x, we can know something about the residuals. Further comments:

o It is surprising that the residuals are unsatisfactory even though the R Square value is huge.

o It seems that the data have even more convexity than the exponential because the residuals have still largely convex curvature.

o The convex curvature of the residuals ends on the very right hand side where the residuals bend back down. This bending back down makes a lot of sense because the end of the internet boom started with the economic downturn in March 2000.

Logarithmically diminishing returns of display space

The dataset DisplaySpace.JMP contains data for a chain of liquor stores where a new wine product was launched. The various stores allocated different amounts of display space to promote the new product, and so management of the chain wanted to investigate after the campaign whether there was an association between the amount of display space allocated by the store and the sales figure for the new wine at the store.

[pic]

The initial plot of sales versus display space (in feet of width) seems to indicate an association of diminishing returns: a one-foot increase in display space seems to produce progressively smaller increases in sales. We try our model of logarithmically diminishing returns.

JMP: Create a new column ‘ln(Display)’ with formula ‘log(Display.Feet)’. Then fit a straight line to Sales versus ln(Display):

[pic]

Sales = 673.934 + 1505.2329 ln(Display)

Summary of Fit

| | |

|RSquare |0.859946 |

|Root Mean Square Error |380.3813 |

|Mean of Response |2678.109 |

|Observations (or Sum Wgts) |47 |

We are not showing the alternative analysis with Fit Special, where one would choose ‘Natural Logarithm: log(x)’ and leave y untransformed.

Practice: Do the analysis with Fit Special on your own and compare.

Interpretation of numeric output:

• Equation, slightly rounded:

Sales = 674 + 1505 ln(Display)

o Slope: For a 10% increase in Display space, ln(Display) will increase by ln(1.1) ≈ 0.1, hence Sales will increase on average by about 0.1 · 1505 ≈ $150.

o Intercept: ln(Display) = 0 when Display = 1ft, in which case the average sales are estimated to be $674.

• R Square is 0.86, which is quite strong.

• The RMSE is $380, which means that the residuals are in the order of about $400.

(No problem of interpretation as we are working on the raw y-scale.)

Residual analysis: The residual plot should be taken from the ln(x)-y data, that is, it should be a plot of e versus ln(x).

[pic]

The residual do not look problematic. It would be difficult to argue that x gives us information for predicting residuals.

Summary: The analysis is reasonably successful, and the logarithmically diminishing returns model is quite consistent with the data.

Elasticity between price and demand of parcel service

The dataset Courier.JMP has price-volume data for a type of service called ‘CourierPak’ in the early days of the Fedex company.

An anecdote: “The concept is interesting and well-formed, but in order to earn better than a "C," the idea must be feasible,” said the Yale University management professor in response to Fred Smith's paper proposing reliable overnight delivery service. Smith went on to found Federal Express Corp.

The data are more complex, namely, overlayed with exponential growth. The present version of the data is “adjusted” for this growth and shows the price-volume elasticity effect alone.

[pic]

Think of the volume as in thousands, and the price as unit price. The data is very noisy, in fact, so noisy that we can’t see a non-linearity to indicate a power law. Yet, the negative association is what one expects of a price-demand relationship. Out of curiosity we fit a straight line both to the raw x-y data and to the ln(x)-ln(y) data, the latter for a proper elasticity analysis.

[pic]

Red: Fit Line with raw x and y.

Green: Fit Special choosing both ln(x) and ln(y)

Linear Fit

Volume = 1826 - 35.680 Price

Summary of Fit

| | |

|RSquare |0.407245 |

|Root Mean Square Error |86.93111 |

|Mean of Response |1374.269 |

|Observations |26 |

Transformed Fit Log to Log

Log(Volume) = 8.05 - 0.327 Log(Price)

Summary of Fit

| | |

|RSquare |0.413282 |

|Root Mean Square Error |0.062924 |

|Mean of Response |7.222564 |

|Observations |26 |

The interpretations are:

• Linear fit (red): For a $1 increase in unit price, there is on average …

• Transformed ln-ln fit (green): For a 1% increase in unit price, there is on average …

The R-Square values are imperceptibly different, 0.407 versus 0.413, hence there is no statistical basis for distinguishing between the two models. Yet, one must report the second analysis because

1. the world speaks elasticity when it comes to price-demand, and

2. even though one refrains from extrapolating, the elasticity equation would extrapolate more sanely because it does not allow prices and volumes to go negative. (Why?)

Practice: Write the elasticity equation as a power law.

By the way, a residual analysis does not reveal any problems. Here are the residuals from Fit Special to the x-y data, and from Fit Line to the ln(x)-ln(y) data. As one should expect, the two plots are nearly identical except for the axis labeling.

[pic]

[pic]

Re-analysis of the Diamond data with a power law

Even though the Singapore diamond data do not make an elasticity situation because the association is positive, a power law would be more plausible because it extrapolates more gracefully. Here is a comparison:

[pic]

[pic]

Linear Fit

Price (Singapore dollars) =

-260 + 3721 Weight (carats)

Summary of Fit

| | |

|RSquare |0.978261 |

|RMSE |31.84052 |

|Mean |500.0833 |

|Observations |48 |

Transformed Fit Log to Log

Log(Price (Singapore dollars)) =

8.57 + 1.498 Log(Weight (carats))

Summary of Fit

| | |

|RSquare |0.970542 |

|RMSE |0.068544 |

|Mean |6.134642 |

|Observations |48 |

The interpretation of the ‘slope’ of the power law is that a 1% increase in weight is associated on average with about a 1.5% increase in price. This seems very meaningful because it expresses the increased value of large stones due to rarity.

On the other hand, the R Square does not speak in favor of the power law: while the difference between 0.978 and 0.971 seems small, re-expressed as variance unexplained, this is a more formidable difference:

1–0.978 = 2.2% versus 1–0.971=2.9%.

Therefore, on the range of the data, the straight line fit is slightly better, yet for extrapolation the power law is better.

The residuals and the RMSE of the power law have the same interpretation as for the exponential model (all that matters is the logarithm of the response, ln(y)):

e ≈ ln(y) – (8.6+1.5 x) ,

hence

exp(e) ≈ y / exp(8.6+1.5 x) ,

which means that the exponentials of residuals are the ratios of observed to predicted. Correspondingly, the RMSE of about 0.07 implies that the residuals e are in the order of ±0.07, hence the

actual prices differ from the predictions by factors around

exp(±0.07) ≈ 1 ± 0.07 ,

or in the order of ±7%.

Practice: Re-analyze the relation between Weight and MPG (Highway and City) in the CarModels2003-4.JMP dataset using a power law. The interpretation will be one of elasticity.

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