Fox Module 11: Statistical inference for simple linear regression HW


Fox Module 11: Statistical inference for simple linear regression HW

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NEAS
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Module 11: Statistical inference for simple linear regression: homework assignment

 

(The attached PDF file has better formatting.)

 

Homework assignment: Estimating regression parameters

 

Some final exam problems give a set of points and ask to compute ordinary least squares estimators, sums of squares, t values, confidence intervals, and other regression statistics.

The final exam may give one set of points and ask about several statistics and estimates, or separate points for each statistic. 

 

This homework assignment continues the exercise from Module 8.

 

An actuary fits a two-variable regression model ( )

to the relation between the incurred loss ratio (x) and the retrospective ratio (y):

 

Policy Year

(x)

(y)

(x – 0)

(x – 0)2

(y– )

(y– )2

(x–0)(y– )

20X1

66.00%

22.50%

0.00%

0.00%

0.60%

0.0036%

0.0000%

20X2

67.00%

19.50%

1.00%

0.01%

-2.40%

0.0576%

-0.024%

20X3

68.00%

21.00%

2.00%

0.04%

-0.90%

0.0081%

-0.018%

20X4

65.00%

22.50%

-1.00%

0.01%

0.60%

0.0036%

-0.006%

20X5

64.00%

24.00%

-2.00%

0.04%

2.10%

0.0441%

-0.042%

Average

66.00%

21.90%

0.00%

0.02%

0.00%

0.0234%

-0.0180%

 

~      The column captions in the table use lower case x and y for the variables; the deviations are shown explicitly as (x – 0) and (y– ).

~      The last line has averages, not totals.  Some formulas in the textbook use totals.

 

We regress retrospective premium ratios on reported loss ratios to estimate premium assets for retrospectively rated business. Actuarial issues of retrospectively rated business are not important; the homework deals with the regression analysis only.

 

The Module 8 homework assignment solves for the least squares estimators of á and â.

 


A.       What is the variance of the ordinary least squares estimator of â?

B.       What is the t statistic for testing the null hypothesis that â = 0?

C.       What is the 95% confidence interval for the true value of â?

D.       What is the p value for testing the null hypothesis that â = 0?

E.       What is the variance of the ordinary least squares estimator of á?

F.        What is the t statistic for testing the null hypothesis that á = 21.90%?

G.       What is the 95% confidence interval for the true value of á?

H.       What is the p value for testing the null hypothesis that á = 21.90%?

 

Show the computations for the homework assignment, not just the solution. You can check your solutions with Excel or other statistical software.

 

The null hypotheses for á and â depend on the scenario.

 


 

            A null hypothesis of â = 0 means that retrospective rating has no effect on the premium.

            A null hypothesis of â = 1 means that a dollar of loss causes a dollar of premium.


 

 

The null hypothesis for the value of á depends on the situation.

 


 

            A null hypothesis of á = 0 means that retrospective rating is the standard premium.

            This hypothesis is not realistic, since large insureds get premium discounts.

            A null hypothesis of á = 21.90% means retrospective rating has no effect on premium.

            The average observed discount is the expected discount regardless of losses.

 

These retrospective rating issues are not part of the regression analysis course.


 

 

The confidence intervals require a t distribution for the appropriate degrees of freedom. Use Excel to find the t values for a 95% confidence interval.

 


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Matt Feipel
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For homework items D and H, how are we suppose to calculate the p values? Is there a table? In the textbook, the readings for this section don't have a formula or even talk about p values. Also, is there a t table somewhere? How do I get the t value for the confidence interval? Also, how does the hypothesis testing work?

[NEAS: The p-value is the flip side of the critical t-value. If the critical t-value for a significance level of 5% is 1.96, then the p-value for a t-value of 1.96 is 5%. Years ago, we used tables; now we get the significance from Excel of other software. Read the discussion of Type 1 errors, Type 2 errors, and the null hypothesis. Classical regression analysis tests whether a null hypothesis should be rejected at a given significance level. We can never say that a specific hypothesis should be accepted. There is always a range of hypotheses that can not be rejected.

Several student comments below discuss getting t values and p values from Excel. Questions on final exam issues also have responses from NEAS. Excel, R, SAS, and other statistical software have many ways of getting t values and p values. Feel free to comment on the statistical software in this discussion thread.]
Edited 10 Years Ago by NEAS
wangxy
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Which excel function do I use to get the P value from T value?

I got V(B) = 198, S^2 = 0.0198( from Module 8 assignment), sum of (Xi-X bar)^2 = 0.02*5. Is it correct? 198 seems too high. My B from module 8 equals -0.9, therefore I got my T0= 0.06396.
Edited 10 Years Ago by NEAS
CalLadyQED
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wangxy, from module 8, I got SE^2 = .00012 = .012%.

In Excel, I think we want TINV(probability, degrees_of_freedom). Probability = a and degrees_of_freedom = n - 2. I was able to get the t-value in the example on page 105.
CalLadyQED
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For the p-value, I use TDIST(x, degrees_of_freedom, tails). x = t statistic; degrees_of_freedom = n - 2; and tails = 1 for strictly > or strictly < tests and 2 for either < or < tests.
Edited 10 Years Ago by NEAS
noturbizniss
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I am getting -1.5 (or-150%) for part B and 205.26% for part F...does this make sense?

I got SE(B) is .6 and SE(A) is 15.69%...is that correct?


Michelle2010
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Is there anyway to get the p value in part D by hand?  (without using Excel)

I don't see any examples of this in the book or notes so far.

[NEAS: The p value is the CDF of a standard normal distribution. Before the advent of microcomputers with Excel, statisticians used CDF tables. Textbooks on statistics had the CDF table in the back, and students learned to look up values. The CDF of a normal distribution has no closed form solution, so one can't work out the p-value by pencil and paper.]


Gautham
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I am getting -2.59808 for Part B. Does a negative t statistic make sense? Because of a negative t statistics I cannot find a p-value for it.

[NEAS: t statisics are positive or negative, depending if the estimated parameter is positive or negatiave. Use the absolute value of the t statistic for the p-value.]
Edited 10 Years Ago by NEAS
Briggs
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I am getting -2.59808 for Part B. Does a negative t statistic make sense? Because of a negative t statistics I cannot find a p-value for it.

[NEAS: p-value is the same for a negative t as a positive t.]

I got the same answer... it matched the result in Excel. I'm still working on the p-value.


atkinsmt
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For part C, is anyone getting CI = -0.9 +/- 3.1825(0.3464)= -0.9 +/- 1.1025 ?

The +/- 1.1025 seems awfully wide to me. Are my units messed up when calculating SE(B)?

Thanks for the help!
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