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