Question

16761 2 0 24438 2 45 41729 2 0 42802 1 0 35 63528 1 0 40 148848 1 0 0 Here 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 1. 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? 2. 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.) 3. What is the value of the test statistic? 4. 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 differenc ein 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

Answer 1

Solution: a. Obtain the values. MINITAB procedure: Step 1: Choose Stat > Basic Statistics > 2-Sampleit Step 2: Choose Samples in different columns. Step 3: In First, enter the column of l kid Step 4: In Second, enter the column of2 kid Step 5: Choose Options. Step 6: In Confidence level, enter 95 Step 7: In Alternative, select not equal. Step 8: Click OK in all the dialog boxes MINITAB output Two-Sample T-Test and Cl 1 kid, 2 kid wo-asple T for 1 kid ve 2 kid 2 kid 3 5.0 26.0 15 Difference·瓢(1 kid) -mu (2 kid) Estimate for difference 10.0 95% C; for difference ; {-44.4, 64.4) 빻- .at of differer e@ 0 4vs not s): -"Zalue Both use Pooled StDev 23.9792 0.51 p-valué a Ο.636 D『 From the MINITAB output,

Avg 1kid)-25.0, Avg (2 kid)-15.0,difference -10.0 2. Obtain the standard error From the MINITAB output, the standard error is, SE-23.9792 Obtain the test statistic From the MINITAB output, the test statistic is, t 0.51 4. Conclusion Since, / <1.960, thus, the mll hypothesis is not rejected. Therefore, it can be concluded that the mean working hours is same for both the groups.