Kinh tế lượngTrắc nghiệm

500 câu trắc nghiệm Kinh tế lượng – 9A

14) Find the explained variation for the paired data. The equation of the regression line for the paired data below is \(\hat y\) = 3x. Find the explained variation.

x 2 4 5 6
y 7 11 13 20

○ 10.00
○ 88.75
● 78.75
○ 72.45

15) Construct the indicated prediction interval for an individual y. The equation of the regression line for the paired data below is \(\hat y\) = 3x and the standard error of estimate is se = 2.2361. Find the 90% prediction interval of y for x = 3.

x 2 4 5 6
y 7 11 13 20

○ 7.1 < y < 10.9
○ 6.8 < y < 11.2
○ 4.5 < y < 13.5
● 1.2 < y < 16.8

16) Construct a scatterplot and identify the mathematical model that best fits the data. Assume that the model is to be used only for the scope of the given data and consider only linear, quadratic, logarithmic, exponential, and power models. Use a calculator or computer to obtain the regression equation of the model that best fits the data. You may need to fit several models and compare the values of R2.

x 1 2 3 4 5 6
y 9 13 25 27 31 46

○ y = 3.14 + 6.59x
○ y = 4.87 + 18.5lnx
○ y = 8.34 + 0.88x
● y = 1.07 + 6.89x

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

17) FPEA, the Farm Production Enhancement Agency, regressed corn output against acreage, rainfall, and a trend line. The trend line is proxy for technological advancement in farming from improved pest control, fertilization, land management, and farming implements.

CORNPROD ACRES RAINFALL TREND
456 9896 29.1 1
421 9680 42.3 2
653 10449 29.8 3
573 10811 26.0 4
546 10014 34.3 5
499 10293 22.7 6
504 9413 24.2 7
611 9860 31.6 8
646 9782 25.6 9
789 12139 37.9 10
773 12166 33.9 11
753 9976 37.4 12
852 10645 27.0 13
755 9738 31.5 14
815 9933 39.9 15
902 10132 25.3 16
986 11145 30.4 17
909 9775 32.7 18
945 9549 35.0 19
866 10077 33.8 20
1178 11550 29.4 21
1230 10600 37.1 22
1207 11280 42.9 23
968 12100 32.2 24
1118 12420 30.5 25

○ CORNPROD = -21.1 + .036ACRES + 2.62RAINFALL + 27.6TREND;
No, because the P-value is low
● CORNPROD = -21.1 + .036ACRES + 2.62RAINFALL + 27.6TREND;
Yes, because the R2 is high
○ CORNPROD = -.9 + 1.68ACRES + .79RAINFALL + 10.2TREND;
Yes, because the adjusted R2 is high
○ CORNPROD = -16.3 + 2.6ACRES + 3.9RAINFALL + 21.3TREND;
Yes, because the the R2 is high

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

Use computer software to obtain the regression and identify R2, adjusted R2, and the P-value.
18) 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? Interpret the statistics.

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 79 248
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

In the order requested by the question, the answers are: .025, -.060, and .750. The negative adjusted coefficient of determination and the high P-value indicate that the variation in height cannot be explained by pressure build-up during the intervals and duration of the spurt.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Use computer software to obtain the regression equation. Use the estimated equation to find the predicted value.
19) 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. He took his equation “to the forest” and found a male bear whose chest measured 70.3 inches and who was 64.0 inches long.

WEIGHT CHEST LENGTH SEX
344 45.0 67.5 1
416 54.0 72.0 1
220 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

○ 635.72 pounds
○ 601.83 pounds
● 615.18 pounds
○ 674.30 pounds

Use computer software to find the best regression equation to explain the variation in the dependent variable, Y, in terms of the independent variables, X1 and X2.

20)

Y X1 X2
15 1.2 16
15 1.2 16
17 1.0 16
6 0.8 9
1 0.1 1
8 0.8 8
10 0.8 10
17 1.0 16
15 1.2 15
11 0.7 9
18 1.4 18
16 1.0 15
10 0.8 9
7 0.5 5
18 1.1 16

CORRELATION COEFFICIENT: Y/X1= .886; Y/X2= .965

COEFFICIENTS OF DETERMINATION: Y/X2= .932; Y/X2,X1= .977

○ \(\hat y\) = 1.3 – 1.3X2

○ \(\hat y\) = 1.25 – 1.55X1 + 5.79X2

○ \(\hat y\) = 1.37 – 5.50X1

● \(\hat y\) = 1.37 – 5.53X1 + 1.33X2

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