2) Consider the following regression predicting birth weight using the variables listed: (10 points) variable name variable label bwght birth weight, ounces cigs cigs smked per day while preg faminc 1988 family income, $1000s motheduc mother's yrs of educ male =1 if male child Source SS df MS Number of obs = 674 Model 12066.6671 4 3016. 66677 Residual 1 305549. 726 669 456. 726048 Total | 317616.393 673 471 . 941149 bwght | Coef. Std. Err. t P>It| [95% Conf. Interval] cigs -. 4370689 . 1358532 faminc . 0812292 . 050534 motheduc . 3334542 . 3955639 male 4. 538161 1 . 661007 cons 110 .1121 4. 849262 a) Explain carefully what the intercept and each of four slope parameter estimates means, including all relevant units. b) Write the formula for R and calculate it for this regression. Explain what R tells us . c) Write the formula for a 95% confidence interval and calculate it for male only. d) Write the formula for a t-statistic that tests the null hypothesis Ho: 3,=0. Calculate the t-statistics for cigs, faminc, motheduc, and male. Which of these are statistically significant at the 18 level (two-tailed), and how do you know?

a) Here the intercept is 110.1121

And the slope parameter estimates each of four as below -

1) Cigs = -0.4371

2) Faminc = 0.0813

3) Motheduc = 0.3334

4) Male = 4.5381

 

The regression equation is given as,

Bwght = 110.1121 - 0.4371(Cigs) + 0.0813( Faminc) + 0.3334(Motheduc) + 4.5381( Male) 

 

Where, Y = Bwght

And β1,β2,β3,β4 are coefficients and X1, X2, X3, X4 are predictors

 

              β1=−0.4371 , X1 = Cigs

              β2=0.0813,     X2 = Faminc

              β3=0.3334,      X3 = Motheduc

               β4=4.5381,     X4 = Male

 

intercept= a = 110.1121

 

 

 b) The formula for 

     R2=1−TSSRSS?

 

R2=  Coefficient of determination

RSS = sum of square due to residual

TSS = Total sum of square

 

Here, our model RSS = 305549. 726

       and,          TSS = 317616. 393

 

Put this values in R square formula 

      R2=1−317616.393305549.726?

            = 0.9620

 

R square represents the how well regression model fits the observed data

 

Here, the R square value is 0.9620

96.20% of the data fit the regression model

 

 

 

   

Step-by-step explanation

c) The formula of 95% confidence interval for slope male(β4) 

 

      C.I.=β4±t2∗(1−0.095),n−2?∗SEβ4? 

        Here,     β4(Coefficient of male predictor) = 4.5382 

              SE = standard error of β4(Coefficient of male predictor) = 1.6610

 

               For finding t0.95,n−2? value from the standard t table or manually in excel

Firstly here we find 95% confidence interval 

       1 - 0.95 = 0.05

                     = 2 * 0.05

                     = 0.10

This the probability and 

  n-2 is degrees of freedom of residual sum of square = 669

      So, we use function in table is TINV(0.10, 669) 

    and enter then we get value is 1.6471

 

so, thus t0.95,n−2? = 1.6471

 

      By using TINV function in excel we find this value

            

          

        C.I.=4.5382±1.6471∗1.6610

 

                 Lower bound = 4.5382−1.6471∗1.6610

                                         =  1.8024

                  Upper bound = 4.5382+1.6471∗1.6610 

                                          = 7.2740

 

  d) The hypothesis for this test is given as,

 

    Ho - βi=0

There is evidence of no statistically significant effect 

 

   H1 - βi?=0

There is  evidence of statistically significant effect 

 

where, i = 1, 2, 3, 4 for β1,β2,β3,β4 coefficients 

 

 

β is the slope assumed in a null hypothesis 

SE is the standard error of coefficient of predictor.

 

For cigs-

t statistic as - 

 

    t = SE(β1)β1−β?

      = 0.1359−0.4371−0?

      = - 3.2163

 

For faminc coefficient - 

t statistic as - 

 

t = SE(β2)β2−β?

 

   = 0.05050.0813−0?

 

  = 1.6099

 

For motheduc coefficient- 

t statistic as - 

 

t = SE(β3)β3−β?

 

   = 0.39560.3335−0?

 

   = 0.8430

 

 

For male coefficient - 

t statistic as - 

 

t = SE(β4)β4−β?

   = 1.66104.5321−0?

   = 2.7285

 

By using excel we find for two tailed test the t critical value at 1%(0.01) level of significance

 

The formula we use as,

t1−α/2,DF(SSR)?

 

DF of SSR = DF of residual sum of square

Here two tailed test at 0.01 level of significance

 

SO, the probability is given as 

 1 - α/2 = 1 - 0.01/2 

              = 0.995 

 

 

 

t critical value = TINV(0.995,669)

                          = 0.0063

   

 

Decision criteria is as below - 

 

If the calculated t is in the critical region (smaller than -0.0063 or greater than 0.0063) then we reject null hypothesis

Thus, we can say that there is evidence of statistically significant effect of cigs coefficient at 1% level of significance.

 

Only Calculated t statistic of β1 coefficient is smaller than -0.0063 then we reject Ho

Calculated t statistic of β1 = - 3.2163 which is less than -0.0063

 

For Calculated t statistic of β2= 1.6099 is greater than -0.0063 then we do not reject Ho.

Thus, we can say that there is no evidence of statistically significant effect of faminc coefficient at 1% level of significance.

 

 

For Calculated t statistic of β3= 0.8430 is greater than -0.0063 then we do not reject Ho.

Thus, we can say that there is no evidence of statistically significant effect of motheduc coefficient at 1% level of significance.

 

For Calculated t statistic of β4= 2.7285 is greater than -0.0063 then we do not reject Ho.

   Thus, we can say that there is no evidence of statistically significant effect of male coefficient at 1% level of significance.