Testing For Significance: Test

Testing for Significance: t Test ■

Hypotheses

H 0: β 1 = 0 H a: β 1 ≠ 0 ■

Test Statistic

b1 t= sb1

where

sb1 =

s Σ(xi − x)

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1

Testing for Significance: t Test ■

Rejection Rule Reject H0 if p-value < α or t < -tα/ 2 or t > tα/ 2 where: tα/ 2 is based on a t distribution with n - 2 degrees of freedom

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Testing for Significance: t Test 1. Determine the hypotheses. H 0: β 1 = 0

H a: β 1 ≠ 0 α = .05 2. Specify the level of significance.

b1 3. Select the test statistic.t = sb1 4. State the rejection rule. Reject H0 if p-value < .05 or |t| > 3.182 (with 3 degrees of freedom)

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Testing for Significance: t Test 5. Compute the value of the test statistic.

b1 5 t= = = 4.63 sb1 1.08 6. Determine whether to reject H0. t = 4.541 provides an area of .01 in the upper tail. Hence, the p-value is less than .02. (Also, t = 4.63 > 3.182.) We can reject H0.

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Confidence Interval for β

1

 We can use a 95% confidence interval for β the hypotheses just used in the t test.

1

 H0 is rejected if the hypothesized value of β included in the confidence interval for β 1.

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to test 1

is not

5

Confidence Interval for β ■

1

The form of a confidence interval for β 1 is: tα / 2 sb1

b1 ± tα / 2 sb1

is the margin of error

b1 is the point tα / 2 is the t value providing an area where estimat of α /2 in the upper tail of a t distribution or

with n - 2 degrees of freedom

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Confidence Interval for β ■

■

1

Rejection Rule Reject H0 if 0 is not included in the confidence interval for β 1. 95% Confidence Interval for β 1

b1 ± tα / 2=sb15 +/- 3.182(1.08) = 5 +/- 3.44 or ■

1.56 to 8.44

Conclusion 0 is not included in the confidence interval. Reject H0

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Testing for Significance: F Test ■

Hypotheses

H 0: β 1 = 0 H a: β 1 ≠ 0 ■

Test Statistic F = MSR/MSE

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Testing for Significance: F Test ■

Rejection Rule Reject H0 if p-value < α or F > Fα

where: Fα is based on an F distribution with 1 degree of freedom in the numerator and n - 2 degrees of freedom in the denominator

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Testing for Significance: F Test 1. Determine the hypotheses. H 0: β 1 = 0

H a: β 1 ≠ 0

α 2. Specify the level of significance.

= .05

3. Select the test statistic.F = MSR/MSE 4. State the rejection rule. Reject H0 if p-value < .05 or F > 10.13 (with 1 d.f. in numerator and 3 d.f. in denominator)

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Testing for Significance: F Test 5. Compute the value of the test statistic. F = MSR/MSE = 100/4.667 = 21.43 6. Determine whether to reject H0. F = 17.44 provides an area of .025 in the upper tail. Thus, the p-value corresponding to F = 21.43 is less than 2(.025) = .05. Hence, we reject H0. The statistical evidence is sufficient to conclude that we have a significant relationship between the number of TV ads aired and the number of cars sold. © 2009 Thomson South-Western. All Rights Reserved

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Some Cautions about the Interpretation of Significance Tests  Rejecting H0: β 1 = 0 and concluding that the relationship between x and y is significant does not enable us to conclude that a causeand-effect  Justrelationship because weisare able to reject Hx0:and β 1= present between y. 0 and demonstrate statistical significance does not enable us to conclude that there is a linear relationship between x and y.

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Using the Estimated Regression Equation for Estimation and Prediction ■

Confidence Interval Estimate of E(yp) y p ± tα / 2sy p

■

Prediction Interval Estimate of yp

yp ± tα / 2sind where: confidence coefficient is 1 - α tα /2 is based on a t distribution with n - 2 degrees of freedom © 2009 Thomson South-Western. All Rights Reserved

and

13

Point Estimation If 3 TV ads are run prior to a sale, we expect the mean number of cars sold to be: ^ y= 10 + 5(3) = 25 cars

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Confidence Interval for E(yp) ■

yˆpof Estimate of the Standard Deviation (xp − x)2

1 syˆp = s + n ∑ (xi − x)2 (3− 2)2 1 syˆp = 2.16025 + 5 (1− 2)2 + (3− 2)2 + (2 − 2)2 + (1− 2)2 + (3− 2)2 1 1 syˆp = 2.16025 + = 1.4491 5 4

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Confidence Interval for E(yp) The 95% confidence interval estimate of the mean number of cars sold when 3 TV ads are run is: y p ± tα / 2sy p 25 + 3.1824(1.4491) 25 + 4.61 20.39 to 29.61 cars

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Prediction Interval for yp ■

Estimate of the Standard Deviation of an Individual Value of yp 2 ( x − x ) 1 p sind = s 1+ + n ∑ (xi − x)2 1 1 syˆp = 2.16025 1+ + 5 4 syˆp = 2.16025(1.20416) = 2.6013

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Prediction Interval for yp The 95% prediction interval estimate of the number of cars sold in one particular week when 3 TV ads are run is:

yp ± tα / 2sind 25 + 3.1824(2.6013) 25 + 8.28 16.72 to 33.28 cars

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Residual Analysis  If the assumptions about the error term ε appear questionable, the hypothesis tests about the significance of the regression relationship and the interval estimation results may not be valid.  The residuals provide the best information about ε .  Residual for Observation i

yi − yˆi  Much of the residual analysis is based on an examination of graphical plots.

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Residual Plot Against x ■

If the assumption that the variance of ε is the same for all values of x is valid, and the assumed regression model is an adequate representation of the relationship between the variables, then The residual plot should give an overall impression of a horizontal band of points

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Residual Plot Against x

Residual

y − yˆ

Good Pattern

0

x

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Residual Plot Against x

Residual

y − yˆ

Nonconstant Variance

0

x

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Residual Plot Against x y − yˆ

Residual

Model Form Not Adequate

0

x

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Residual Plot Against x ■

Residuals Observation

Predicted Cars Sold

Residuals

1

15

-1

2

25

-1

3

20

-2

4

15

2

5

25

2

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Residual Plot Against x TV Ads Residual Plot

3

Residuals

2 1 0 -1 -2 -3 0

1

2

TV Ads

3

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4

25

End of Chapter 12

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