Question

[5] Q2. Problem 3.14. Refer to Plastic hardness Problem 1.22. Use the data modified for this problem from Datasets for Assignment_3. (a) Perform the F test to determine whether or not there is lack of fit of a linear regression function; use a = .01. State the alternatives, decision rule, and conclusion. (b) Is there any advantage of having an equal number of replications at each of the X levels? Is there any disadvantage? (c) Does the test in part (a) indicate what regression function is appro- priate when it leads to the conclusion that the regression function is not linear? How would you proceed?

USE THE DATA LISTED IN THE EXCEL BELOW PLEASE

THANK YOU !!

Answer 1

Sol:

Q2).

In order to calculate the F statistic, we need to calculate the regression coefficients and the sum of squares and the sum of products.

For this, we form the table below with squares and cross products.

	X	Y	X^2	Y^2	X*Y
	16	199	256	39601	3184
	16	205	256	42025	3280
	16	196	256	38416	3136
	16	200	256	40000	3200
	24	218	576	47524	5232
	24	220	576	48400	5280
	24	215	576	46225	5160
	24	223	576	49729	5352
	32	237	1024	56169	7584
	32	234	1024	54756	7488
	32	235	1024	55225	7520
	32	230	1024	52900	7360
	40	250	1600	62500	10000
	40	248	1600	61504	9920
	40	250	1600	62500	10000
	40	240	1600	57600	9600
Total	448	3600	13824	815074	103296

We calculate the following quantities:

Now

Quantities for ANOVA:

Total SS=

SSreg=

SSerror=Totalss-SSreg=5074-4867.2=206.8

The total df=16-1=15

Regression SS=2-1=1

Error df=15-1=14.

The null hypothesis for an F-test:

ie there no linear relationship between X and Y

We shall form the ANOVA table

	df	SS	MS	F	Significance F
Regression	1	4867.2	4867.2	329.501	3.98E-11
Residual	14	206.8	14.77143
Total	15	5074

Conclusion:

From the table above, we can see that the p-value is 0.0000<0.05, and we reject the null hypothesis. Hence we conclude that there is a linear relationship between X and Y.

(b).

Having equal number observations are always advantageous in that we can estimate the pure error term which helps us to get a more reliable estimate of the error. There are disadvantage that it takes time and efforts as well.

(c).

No, the test in part (a) indicates whether a liner model is a good fit or not. If not, it doesn't give any indication of which model fits the data. If a linear model doesn't fit then one has to procced by transformation, or to a non linear model.

1.22).

(a)

The regression output is:

	r²	0.973	n	16
	r	0.986	k	1
	Std. Error	3.234	Dep. Var.	Y
ANOVA table
Source	SS	df	MS	F	p-value
Regression	5,297.5125	1	5,297.5125	506.51	2.16E-12
Residual	146.4250	14	10.4589
Total	5,443.9375	15
Regression output				confidence interval
variables	coefficients	std. error	t (df=14)	p-value	95% lower	95% upper
Intercept	168.6000
X	2.0344	0.0904	22.506	2.16E-12	1.8405	2.2283
Predicted values for: Y
		95% Confidence Interval	95% Prediction Interval
X	Predicted	lower	upper	lower	upper	Leverage
40	249.975	247.073	252.877	242.456	257.494	0.175

The regression function is:

Hardness in Brinell = 168.6 + 2.0344*Elapsed time

The plot of the data is:

Since the coefficient of determination value is very high, we can say that the regression function appear to give a good fit here.

(b)

The point estimate will be:

Hardness in Brinell = 168.6 + 2.0344*Elapsed time

Elapsed time = 40 hours

Hardness in Brinell = 168.6 + 2.0344*40

Hardness in Brinell = 249.975

(c)

When the elapsed time is increased by 1 hour, then the mean hardness will increase by 2.0344.

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