Question

5.8. Consider the following regression output: Î; = 0.2033 + 0.6560X, se = (0.0976) (0.1961) r2 = 0.397 RSS = 0.0544 ESS = 0.0358 where Y= labor force participation rate (LFPR) of women in 1972 and X = LFPR of women in 1968. The regression results were obtained from a sample of 19 cities in the United States. a. How do you interpret this regression? b. Test the hypothesis: Ho: B2 = 1 against H: B2 > 1. Which test do you use? And why? What are the underlying assumptions of the test(s) you use? c. Suppose that the LFPR in 1968 was 0.58 (or 58 percent). On the basis of the regres- sion results given above, what is the mean LFPR in 1972? Establish a 95 percent con- fidence interval for the mean prediction. d. How would you test the hypothesis that the error term in the population regression is normally distributed? Show the necessary calculations.

Answer 1

Solution: ñ = 0.8033 +0.6560X+ se = (0.0976) (0.1961) r² = 0.397 RSS = 0.05uy ESS = 0.0358. Y = labor fore participation rate (LFPR) in 1972 X = LFPR Of women in 1968. Sample = 19 cities. (a) How do you interpret this regression, The formula used is : +(Bk) = Bk - E (BK) tvalue = The interval estimaty is calculated as: bk & tcra/2, N-k) & se (bk)-

bk = estimated sample value t(1-42, Nok) = critical value of t. se C6k) = Standard error of bks E(bk) = expected value of bk The relationship between LF PR of women in 1968 is positively related with 2FPR of women in 1972. This means that there is lof. increase in the LFPR of women in 1968 and also there is an increase of LFPR of women in 1972 by 65064. 6) Hypothel's testing: Noll hypother is to : B2=1 Alternative hypothesis Hl: R31

using equation ) the calwlated tvale is: +(f2) = ² - E (B2) Se (ße) - 0.656-1 0.1961 = -1,4842 . 6 (0.025, 23) 51.73961) since calculated absolute tvake is higher than the critical vake, it rejects the null hypothesis. © The expected rake of LFPR in 1972: Since the given LFPR in 1968 was 0.58, in ege) substituting weget, yî =0.2033 +0.6560X; Y = 0.2033 +0.6560*0.58 = 10.58378)

Interval estimate: using evation (3), the required interval & timate can be calculated or follows: Requried interval estimate = battutto 27 (0.978 N-2=17) sec62) = 0.58378£1,73965ec62! Since, se chal is not given the integral estimate is impossible to derive. (d) to test hypothesis of normal distribution of error term, use the normal probability plot or calculate Jarque Bera test. Since data is not available, the calculation can not be shown.