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![Page 1: Time Series Analysis – Chapter 4 Hypothesis Testing Hypothesis testing is basic to the scientific method and statistical theory gives us a way of conducting.](https://reader035.fdocuments.in/reader035/viewer/2022062421/56649e555503460f94b4cb04/html5/thumbnails/1.jpg)
Time Series Analysis – Chapter 4Hypothesis Testing
Hypothesis testing is basic to the scientific method and statistical theory gives us a way of conducting tests of scientific hypotheses. Scientific philosophy today rests on the idea of falsification: For a theory to be a valid scientific theory it must be possible, at least in principle, to make observations that would prove the theory false. For example, here is a simple theory:
All swans are white
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Time Series Analysis – Chapter 4Hypothesis Testing
All swans are white This is a valid scientific theory because there is a way to falsify it: I can observe one black swan and the theory would fall. For more information on the history and philosophy of falsification I suggest reading Karl Popper.
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Time Series Analysis – Chapter 4Hypothesis Testing
Besides the idea of falsification, we must keep in mind the other basic tenant of the scientific method: All evidence that supports a theory or falsifies it must be empirically based and reproducible.
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All evidence that supports a theory or falsifies it must be empirically based and reproducible.
In other words, data! Just holding a belief (no matter how firm) that a theory is true or false is not a justifiable stance. This chapter gives us the most basic statistical tools for taking data or empirical evidence and using it to substantiate or nullify (show to be false) a hypothesis.
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All evidence that supports a theory or falsifies it must be empirically based and reproducible.
I have just used the word hypothesis and this chapter is concerned with hypothesis testing, not theory testing. This is because theories are composed of many hypotheses and, usually, a theory is not directly supported or attacked but one or more of it’s supporting hypotheses are scrutinized.
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Discrimination or Not Activity
Null Hypothesis Ho: No Discrimination
Alternative Hypothesis Ha: Discrimination
How do we choose which hypothesis to support?
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Discrimination or Not Activity
Null Hypothesis Ho: No Discrimination
Alternative Hypothesis Ha: Discrimination
How do we choose which hypothesis to support?
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The p-value
• p-value measures amount of support for alternative hypothesis.
• The smaller the p-value the more support for the alternative hypothesis.
• Typical level of support is 5% or 0.05
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Fourth Graders Feet Data Set
Predictor variable (x): Childs AgeResponse variable (y): Foot LengthModel: Test: Ho: -> x has no effect on y
H1: -> x has an effect on y
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Fourth Graders Feet Data Set – Minitab Output
Predictor variable (x): Childs AgeResponse variable (y): Foot Length
The regression equation isFoot Length = 18.1 + 0.0358 Childs Age
Predictor Coef SE Coef T PConstant 18.138 3.753 4.83 0.000Childs Age 0.03575 0.02922 1.22 0.229
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Fourth Graders Feet Data Set – Minitab Output
Predictor variable (x): Childs AgeResponse variable (y): Foot Length
Ho: -> x has no effect on y
H1: -> x has an effect on y
P-value = 0.229 -> x has no STATISTICAL effect on y given the model we used!
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Fourth Graders Feet Data Set – One Tailed Alternative
Predictor variable (x): AgeResponse variable (y): foot length
Ho: -> x has no effect on y
H1: -> x has a positive effect on y
P-value = (0.229)/2 = 0.1145 -> x has no statistical positive effect on y given the model we used.
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Statistical vs. Practical Significance
401K data setPredictor variables x1: mrate
x2: age
x3: totemp
Response variable (y): prate
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Statistical vs. Practical Significance
The regression equation is:prate = 80.3 + 5.44 mrate + 0.269 age - 0.000130 totemp
Predictor Coef SE Coef T PConstant 80.2943 0.7777 103.25 0.000mrate 5.4414 0.5244 10.38 0.000age 0.26941 0.04515 5.97 0.000totemp -0.00012978 0.00003672 -3.53 0.000
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Statistical vs. Practical Significance
The regression equation is:prate = 80.3 + 5.44 mrate + 0.269 age - 0.000130 totemp
All predictors are statistically significant.
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Statistical vs. Practical Significance
The regression equation is:prate = 80.3 + 5.44 mrate + 0.269 age - 0.000130 totemp
If total number of employees increases by ten thousand then participation rate decreases by -0.000130*10,000 = 1.3% (other predictors held constant)
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Boeing 747 Jet
What does an empty Boeing 747 jet weigh?
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Boeing 747 Jet
What does an empty Boeing 747 jet weigh?
My point estimate: 250,000 lbs
Answer: 358,000 lbs
I am wrong! A point estimate is almost always wrong!
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Boeing 747 Jet
What does an empty Boeing 747 jet weigh?
My confidence interval estimate: (0, ∞)
Answer: 358,000 lbs
I am right! But, my interval is not useful!
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Point and Interval Estimates – Minitab will compute both
401K data setPredictor variables x1: age
Response variable (y): prate
In Minitab go to Regression -> General Regression and select the correct model variables then click on the Results box and make sure the “Display confidence intervals” box is selected.
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Point and Interval Estimates – Minitab will compute both
401K data setPredictor variables x1: age
Response variable (y): prateRegression Equation
prate = 83.4231 + 0.298893 age
Coefficients
Term Coef SE Coef T P 95% CIConstant 83.4231 0.737593 113.102 0.000 (81.9763, 84.8699)age 0.2989 0.045938 6.506 0.000 ( 0.2088, 0.3890)
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Confidence Intervals
General structure of all confidence intervals:
The standard error is an estimate of the standard deviation of the point estimator.
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Confidence Intervals
General structure of all confidence intervals:
Term Coef SE Coef T P 95% CIConstant 83.4231 0.737593 113.102 0.000 (81.9763, 84.8699)age 0.2989 0.045938 6.506 0.000 ( 0.2088, 0.3890)
0.2989 + 1.960*0.045938 = 0.38900.2989 – 1.960*0.045938 = 0.2088
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Confidence Intervals
General structure of all confidence intervals:
0.2989 + 1.960*0.045938 = 0.38900.2989 – 1.960*0.045938 = 0.2088
Where does 1.960 come from?
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Confidence Intervals
Where does 1.960 come from?
t distribution with n – k – 1 degrees of freedom where k is the number of predictors in the model.
For our model, n = 1533 and k = 1We also need to know the confidence level of the interval (typically 95%)
Then, use a t table!
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Testing Linear Combinations of Parameters
TWOYEAR data setPredictor variables x1: jc – # years attending a two-year college
x2: univ – # years attending a four-year college
x3: exper – months in workforce
Response variable (y): log(wage)
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Testing Linear Combinations of Parameters
Predictor variables x1: jc – # years attending a two-year college
x2: univ – # years attending a four-year college
x3: exper – months in workforce
Response variable (y): log(wage)
Ho:
“one year at two-year college is worth the same as one year at four-year college”
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Testing Linear Combinations of Parameters
Predictor variables x1: jc – # years attending a two-year college
x2: univ – # years attending a four-year college
x3: exper – months in workforce
Response variable (y): log(wage)
H1:
“one year at two-year college is worth less than one year at four-year college”
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Testing Linear Combinations of Parameters
Predictor variables x1: jc – # years attending a two-year college
x2: univ – # years attending a four-year college
x3: exper – months in workforce
Response variable (y): log(wage)
Ho: -> Ho:
H1: -> H1:
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Testing Linear Combinations of Parameters
Let ->
Then
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Testing Linear Combinations of Parameters
Now, after we create the new variable , we can conduct the following test:
Ho: -> Ho: -> Ho:
H1: -> H1: -> H1:< 0
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Ho: -> Ho: -> Ho:
H1: -> H1: -> H1:< 0
lwage = 1.47233 - 0.0101795 jc + 0.0768763 jc+univ + 0.00494422 exper
CoefficientsTerm Coef SE Coef T P 95% CIConstant 1.47233 0.0210602 69.9102 0.000 ( 1.43104, 1.51361)jc -0.01018 0.0069359 -1.4677 0.142 (-0.02378, 0.00342)jc+univ 0.07688 0.0023087 33.2981 0.000 ( 0.07235, 0.08140)exper 0.00494 0.0001575 31.3972 0.000 ( 0.00464, 0.00525)
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Ho: -> Ho: -> Ho:
H1: -> H1: -> H1:< 0
This is a one-tailed test so the p-value needs to be divided by 2:
0.142/2 = 0.071
Conclusion: analysis supports the null hypothesis – “one year at a junior college is worth the same as one year at a university.”
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Testing Linear Combinations of ParametersUse the TWOYEAR data set to test the following hypothesis:
Ho:
H1:
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The ANOVA F TestFor a multiple regression model:
The ANOVA F test is:
Ho:
H1: at least one is not equal to 0
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Multiple Linear Regression Assumptions
MLR Assumption 1: the model is linear in the parameters
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Multiple Linear Regression Assumptions
MLR Assumption 2: Data comes from a random sample
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Multiple Linear Regression Assumptions
MLR Assumption 3: None of the independent or predictor variables are perfectly correlated (if they were, Minitab would not run a regression analysis).
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Multiple Linear Regression Assumptions
MLR Assumption 4: The error, u, has an expected value of zero.
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Multiple Linear Regression Assumptions
MLR Assumption 5: The error, u, has the same variance given any values of the explanatory variables. This is the assumption of homoskedasticity.
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Multiple Linear Regression Assumptions
MLR Assumption 6: The error, u, is independent of the explanatory or predictor variables and is normally distributed with mean zero and variance .