Question

QUESTION 7

Use computer software to find the multiple regression equation. Can the equation be used for prediction?

A wildlife analyst gathered the data in the table to develop an equation to predict the weights of bears. He used WEIGHT as the dependent variable and CHEST, LENGTH, and SEX as the independent variables. For SEX, he used male=1 and female=2.

Weight	Chest	Lenght	Sex
344	45.0	67.5	1
416	54.0	72.0	1
420	41.0	70.0	2
360	49.0	68.5	1
332	44.0	73.0	1
140	32.0	63.0	2
436	48.0	72.0	1
132	33.0	61.0	2
356	48.0	64.0	2
150	35.0	59.0	1
202	40.0	63.0	2
365	50.0	70.5	1

		WEIGHT = -442.6 + 12.1CHEST + 3.6LENGTH - 23.8SEX; Yes, because the adjusted R2 is high
		WEIGHT = 196 + 2.35CHEST + 3.40LENGTH + 25SEX; Yes, because the R2 is high
		WEIGHT = -320 + 10.6CHEST + 7.3LENGTH - 10.7SEX; Yes, because the P-value is high
		WEIGHT = 442.6 + 12.1CHEST + 4.2LENGTH - 21SEX; Yes, because the P-value is low

QUESTION 8

Use computer software to obtain the multiple regression equation and identify R², adjusted R², and the P-value.

A visitor to Yellowstone National Park sat down one day and observed Old Faithful, which faithfully spurts throughout the day, day in and day out. He surmised that the height of a given spurt was caused by the pressure build-up during the interval between spurts and by the momentum build-up during the duration of the spurt. He wrote down the data to test his hypothesis, but he didn't know what to do with his data. Can you help him out with his theory?

HEIGHT	INTERVAL	DURATION
150	86	240
154	86	237
140	62	122
140	104	267
160	62	113
140	95	258
150	79	232
150	62	105
160	94	276
155	97	248
125	86	243
136	85	241
140	86	214
155	58	114
130	89	272
125	79	227
125	83	237
139	82	238
125	84	203
140	82	270
140	82	270
140	78	218
135	87	270
140	70	241
100	56	102
105	81	271

		0.089, 0.032, 0.634
		0.025, -0.060, 0.750
		0.049, -0.021, 0.123
		0.213, 0.182, 0.213

Answer 1

Solution-

A.

After running the regression analysis in the excel, regression output is-

SUMMARY OUTPUT Regression Statistics MultipleR R Square Adjusted R Square Standard Error Observations 0.937009 0.877986 0.83223 47.18321 12 ANOVA MS Significance F Regression Residual Total 128156.8744 42718.96 19.18871 0.0005181 17810.04232 2226.255 145966.9167 Coefficients Standard Error tstat P-value Lower 95% Uper 95% Intercept Chest Length 987.192 262.8933748-3.75511 0.0055831593.425607-380.9591886 9.231013 3.148796833 2.9316 0.018952 1.969874686 16.49215172 12.55495 4.556794642 2.755215 0.024858 2.046961101 23.06293568 36.5007 33.64607923 1.084843 0.309598 -41.08729828 114.0886974

So multiple regression equation is-

Weight = -987.192 +9.231 * chest + 12.555 * length + 36.5 * Sex

value of R² is high so model is a good fit.

B.

After running the regression analysis in the excel, regression output is-

SUMMARY OUTPUT Regression Statistics MultipleR R Square 0.275908488 0.076125494 Adjusted R Square 0.00421142 Standard Error14.92878102 Observations 26 ANOVA MS Significance F Regression Residual Total 2 23 25 422.3705912 211.1852956 0.947578 0.402299327 5125.975563 222.8685027 5548.346154 Coefficients Standard Error 115.5414573 22.54138417 5.125748111 3.42E-05 68.9110514162.1718632 0.643326196 0.476458885 1.350223948 0.1900820.342304102 1.628956494 0.131409717 0.1012774111.297522481 0.207315 0.340918004 0.078098569 t Stat P-value Lower 95% Upper 95% Intercept X Variable 1 X Variable 2

R² = 0.076 and adjusted R² = -0.0042

and p-value is given by -

0.4023 ( from the table)

Answers

TY!