In this part, you are going to show some statistics of the variable net F. A thistle variable indicates the Net total financial assets in thousands of dollars. This variable has a mean of 19 up seven. The standard deviation is 63 Boyne Nice six. The minimum level is minus 502.3 and the maximum level is 1000 536 18 Remember, these numbers are in 1000. In the second part, you are going to test whether the average net total asset differs by 401 k eligibility studies and you can use a two sided alternative. The null hypothesis of this question is the average of net S A for their group that is eligible for for one cave equal that f A of the group that is not Elizabeth. For 41 K, we would do a T test and the T test. We will get a value of 13 11 The P value off the test is very small is way smaller than 0.1 So we are able to reject the non hypotheses that two groups have equal net total financial asset. The difference between the two groups in that total financial asset is well intact. The value of Met Toto A set the average one for the group that is eligible for 401 K, which is 30 points 54 These numbers are generated by their statistical software. Then we subtract from it their average net F A of the group that is not eligible for for a one K, which is 11 going 68 What we get is roughly 18 0.86 So the group that is eligible for for one K has a larger average net total financial asset, and it is larger by the average of the groups ineligible for 41 K by 18.8 $6000. We will estimate a mentally near regression model for net total financial assets. That includes income age eligibility for 401 K, and we also have age and income included in the regression as quadratic form. Okay, we will look at the coefficient of the variable E for a one K. As an indicator of the estimated dollar effect of the eligibility status, the coefficient of E 41 k is nine point 705 It means that if the family is illegible 4401 k, their net total financial asset when increased by $9705 we will add to the model estimated in Part three, the interaction terms of a legibility status and age minus 41. And another term is the interaction of eligibility status and age minus 41 square. This is the regression result. When we look at the interaction terms, we see that only this term it's significant. The term with Asian minus 41 the term with Asia minus 41 square, is not significant. In future models, we can safely drop the last term. Women compare the estimated effect of the allege ability status between model in Part three and model in Part four. The left panel show the regression results from the model. In part, we and the right panel show the results from the model. In Part four, you can tell the difference of the coefficient of E 401 K between two models. In monetary, it is 9.75 A model for it is 9.960 You should note that the meaning of the coefficient of E 401 k, the first between two models in model three. It is the effect for all ages. A model for it is the effect when age equals 40. 1 way will include family size dummies In the regression equation, you can see that in our equation we now have F s two s, three s four and fs five. These are their families side dummies F s two takes a value of one if the family has to people access to equal zero otherwise, after three takes a value of one. If the family has three people and takes a value of zero otherwise ever as far is for family of four. And Verse five is for family with more than five people. Mhm. So you don't cfs one in the equation. That is because Fs one or a family of one is the base group We have to base group So all the coefficient of the dummies should be interpreted as the difference in net total asset compared to the base group. So among the four dummies, you can see that Fs two is not significant. F s 32 Fs five are significant and the level of significance of these dummies increases with their size of the families. So have a sweet It's significant at the 5% level. But s four and s wide are significant at the 1% level. Even the P value you can say that compared to the base group, a family of a single person having three or more people in the family associate with a greater average value of net total financial assets. We can also conduct a F test to see the joint significance of their family size dummies. The null hypothesis is the coefficient of the family size dummy equals each other and equals zero. We will get an F statistic with two degrees of freedom four and NYT housing 265. The statistic is 5.44 and the P value it's very, very small is way smaller than 0.1 So we are able to reject the null hypothesis. In other words, the family size dummies are jointly significant. At the 1% level, we will do a child test forward coefficient equality across five family sized categories For the test statistic. We we need some of square residual from the unrestricted models and re restricted model. The restricted model comes from Part six and we use the phone sample to estimated The sum of square residual for the respective model is 30 million yeah, 213,000 and 115 for the and restricted models. They are five of them. They are estimated separately for each category of family size. The sum of square residual for and restricted models is the sum of the sum of square residual of all five models. Adding them up. You wouldn't get a total of 29 million, 980,000 and 959 now. Given that we have there some of square residual for the restricted model and unrestricted models, we can calculate the F statistic or the child test statistic the child has is based on the F distribution. That's why we have an F statistic. This is their formula for the F statistic. We will take the sum of square residual of the restricted model minus the sum of square of residual of the unrestricted model divided by their sum of square of residual of the unrestricted model altogether multiply with the ratio of the second degree of freedom divided by the first degree of freedom. These degrees of freedom are given from their problem. We have D F one. The first degree of freedom is the number of constraints were going to test, which is 20 and the second degrees of freedom is given as NYT housing and 245. The F statistic would be three point 57 and we can back up the P value using your statistical software. It is a very small number 0.0 03 So we are able to reject the null hypotheses at the 1% level, which means the coefficients are not equal across five family size categories.