DATA: 10.127 I need answer for 1b. YEARP UCOSTP PRODP CUMP LNUC LNY LNCP 1960 0.
ID: 2929512 • Letter: D
Question
DATA:
10.127
I need answer for 1b.
YEARP UCOSTP PRODP CUMP LNUC LNY LNCP 1960 0.58 140 260 -0.545 4.942 5.561 1961 0.50 200 400 -0.693 5.298 5.991 1962 0.46 400 600 -0.777 5.991 6.397 1963 0.43 700 1,000 -0.844 6.551 6.908 1964 0.39 500 1,700 -0.942 6.215 7.438 1965 0.33 1,100 2,200 -1.109 7.003 7.696 1966 0.31 1,200 3,300 -1.171 7.090 8.102 1967 0.30 1,300 4,500 -1.204 7.170 8.412 1968 0.29 1,400 5,800 -1.238 7.244 8.666 1969 0.25 2,100 7,200 -1.386 7.650 8.882 1970 0.23 10,700 9,300 -1.470 9.278 9.138 1971 0.18 5,000 20,000 -1.715 8.517 9.903 1972 0.15 10,000 25,000 -1.897 9.21010.127
I need answer for 1b.
The purpose of this exercise is to have you estimate and interpret parameters of a learning curve. In this exercise you have a choice of using one of two data files, both of which are in your data diskette subdirectory CHAP3.DAT. One file, called POLY, contains annual data on deflated unit costs (UCOSTP) current production (PRODP), and cumulative production to year t -I (CUMP) for a typical manufacturer of polyethylene over the 13-year period from 1960 to 1972. The second file, called TIO2, contains annual data for the DuPont Corporation in its production of titanium dioxide from 1955 to 1970 The variables in this data file include undeflated unit costs (UCOSTT), cur- rent production (PRODT), cumulative production to year - 1 (CUMT) and a GNP-type deflator (DEFL).9 The YEAR variable is called YEARP in the POLY file and YEART in the TIO2 file. Notice that both files contain data on costs and production, which are typically more difficult to obtain than data on prices and production Choose one of these two data files, and then perform steps (a) and (b) Important note: If you are using the TIO2 data file, you will first need to deflate the UCOSTT data by dividing UCOSTT by DEFL Take the logarithm of cumulative production up to time period t -1, and name this variable LNCP. Do the same for unit cost (LNUC) and current output (LNY). Plot unit cost against cumulative production, and then plot LNUC against LNCP. Comment on the two plots and what they might imply for the mathematical form of the learning curve relationship Using least squares, estimate the parameters of the simple learning curve model (3.8) in which LNUC is regressed on a constant and on LNCP Interpret your estimate of the learning curve elasticity. Is this estimate reasonable? What is the corresponding slope of the learning curve? Using a reasonable level of significance, construct a confidence interval for your estimated learning curve elasticity, and then test the null hypothesis that the learning curve elasticity is zero against the alterna- tive hypothesis that it is not equal to zero (a) (b)Explanation / Answer
LNUC^ = 1.003488 -0.2716*LNCP
estimate for slope = -0.2716
when LNCP increase by 1 unit , there is decrease of 0.2716 unit in LNUC
Since R^2 = 0.9693 , which is > 0.8 ,hence the estimate is reasonable
p-value for slope = 1.124*10^(-)<<0.05
hence it is signiificant
95% confidence interval of slope is (-0.30365,-0.23957)
SUMMARY OUTPUT Regression Statistics Multiple R 0.984566558 R Square 0.969371307 Adjusted R Square 0.96658688 Standard Error 0.072713877 Observations 13 ANOVA df SS MS F Significance F Regression 1 1.840725305 1.840725305 348.1403619 1.12408E-09 Residual 11 0.058160388 0.005287308 Total 12 1.898885692 Coefficients Standard Error t Stat P-value Lower 95% Intercept 1.003488554 0.117331052 8.552625532 3.44221E-06 0.745244649 LNCP -0.271614799 0.014557146 -18.65851982 1.12408E-09 -0.30365486Related Questions
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