This assignment asks you to carry out a difference in means test using a small number of actual observations from the 2015 American Community Survey. As you will see from examining the small Excel file, I slightly modified only one value to make the problem below easier to calculate by hand. You should be able to complete this assignment with a pen and paper only. You can check your work with a computer or calculator, but you will learn the most if you try it by hand. The six households below contain one mom, one dad, and either one or two children. The table below indicates the Census-assigned household number (for reference purposes; it is not used in this assignment), the number of children in the household, a two-child household indicator variable, and the number of hours the mom works in a typical week.householdNumber kidsTwo kidsMom’s work hours1676121024438214541729210428021035635281040148848100Here is your task: Carry out a difference in means test of the null hypothesis that moms with one child work the same number of hours as moms with two kids; in other words, the null hypothesis is that average work hours is the same in both types of households. To do this, you’ll have to answer the following questions:What is the average female work hours among one-child households? What is the average female work hours among two-child households? What is the difference in means?What is the estimated standard error of the difference in means? (Hint: you have to calculate the variance of work hours for each group and plug these values, along with the number of observations in each group, in the formula in MM Chapter 1, footnote 17. You then have to take the square root of a large number; if you are doing this by hand, an approximate value is fine, and you can use the following hint to help you: the square root of 361 is 19 and the square root of 400 is 20. )What is the value of the test statistic?Is the difference in means statistically different from zero? In other words, do you reject the null hypothesis of equal work hours? (Hint: if the absolute value of the test statistic is greater than 1.96 we reject the null and say the difference in means is statistically different from zero, or “statistically significant”. Angrist and Pischke describe the critical value as “about two” but the number 1.96 is the precise value at which the cumulative standard normal distribution (this is a “bell shaped curve”) has 2.5% of the area under the curve and to the right. Thus, 95% of the area is under the curve between -1.96 and 1.96, and so 1.96 is known as the critical value for a test at the 5% significance level.)
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