1.06 1.04 To check the accuracy of the prediction method, the linear model y= Bo+B1x+ε is fit. If the prediction method is accurate, the value of Bo will be 0 and the value of ẞ1 will be 1. Note: This problem has a reduced data set for ease of performing the calculations required. This differs from the data set given for this problem in the text.
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- Consider a model with an interaction term between being female and being married. The dependent variable is the log of the hourly wage: log(wage) = 0.151 - 0.038 female + 0.1 married - 0.301 female* married + 0.079 educ + 0.027 exper+0.029 tenure (0.072) (0.056) (0.055) (0.007) (0.005) (0.007) n = 536, R2 = 0.461 Numbers in parantheses are standard errors of coefficients. Given the estimation result and the observation number fill in the blanks below which aim at discussing the statistical significance of variables. The test statistic of the interaction term is The critical value at 1% significance level is Then the interaction term statistically significant at 1% significance level. (Hint: to fill the blank make a choice between "is" and "is not".)EXAMPLE 8.5 | Alloy Adhesion An article in the Journal of Materials Engineering ["Instrumented Tensile Adhesion Tests on Plasma Sprayed Thermal Barrier Coatings" (1989, Vol. 11(4), pp. 275-282)] describes the results of tensile adhesion tests on 22 U-700 alloy specimens. The load at specimen failure is as follows (in megapascals): 19.8 15.4 11.4 19.5 10.1 18.5 14.1 8.8 14.9 7.9 17.6 13.6 O 0.01 O 0.025 O 0.05 O 0.95 O 0.975 7.5 12.7 16.7 11.9 15.4 11.9 15.8 11.4 The sample mean is x = 13.71, and the sample standard deviation is s = 3.55. Figures 8.6 and 8.7 show a box plot and a normal probability plot of the tensile adhesion test data, respectively. These displays provide good support for the assumption that the population is normally distributed. We want to find a 95% CI on μ. Since n = 22, we have n - 1 = 21 degrees of freedom for t, so to.025,21 = 2.080. The resulting CI is X-1/2-1/√x+1a/2n-1³/√n 13.71-2.080 (3.55)/√/22 ≤ ≤ 13.71 +2.080 (3.55)/√22 13.711.57 ≤ ≤ 13.71 +1.57 12.14 ≤…Lecture(8.8): The amount of time people engage in physical activity mat be related to health outcome. Those who report that they spend more than 15 hours are put into one while those who spend less 10 were put into another group. (Those who fall in between 10 and 15 were left out of the study); The reaearcher then ask the participants to wear a monitor for one month. The average time in minutes is recorded and shown below . Is there any evidence that on average people who watch less than 10 hours watching televsion spend more time on physical activity.?. Test the hypotheses at (alpha=0.05) using the 5 step procedure <10 hours 75 63 118 35 82 >15 hours 62 6 78 43 22 33
- 6 A study was conducted that measured the total brain volume (TBV) (in mm3) of patients that had schizophrenia and patients that are considered normal. Table #1 contains the TBV of the normal patients and Table #2 contains the TBV of schizophrenia patients ("SOCR data Oct2009," 2013). Table #1: Total Brain Volume (in mm3) of Normal Patients 1663407 1583940 1299470 1535137 1431890 1578698 1453510 1650348 1288971 1366346 1326402 1503005 1474790 1317156 1441045 1463498 1650207 1523045 1441636 1432033 1420416 1480171 1360810 1410213 1574808 1502702 1203344 1319737 1688990 1292641 1512571 1635918 Table #2: Total Brain Volume (in mm3) of Schizophrenia Patients 1331777 1487886 1066075 1297327 1499983 1861991 1368378 1476891 1443775 1337827 1658258 1588132 1690182 1569413 1177002 1387893 1483763 1688950 1563593 1317885 1420249 1363859 1238979…2. Do children diagnosed with attention deficit/hyperactivity disorder (ADHD) have smaller brains than children without this condition? This question was the topic of a research study described in the paper "Developmental Trajectories of Brain Volume Abnormalities in Children and Adolescents with Attention Deficit/Hyperactivity Disorder" (Journal of the American Medical Association [2002]: 1740–1747). Brain scans were completed for 152 children with ADHD and 139 children of similar age without ADHD. Summary values for total cerebral volume (in milliliters) are given in the following table: Children with ADHD 152 1059.4 117.5 Children without ADHD 139 1104.5 111.3 Do these data provide evidence that the mean brain volume of children with ADHD is smaller than the mean for children without ADHD? Let's test the relevant hypoth- eses using a .05 level of significance.CONSTRUCT MODIFIED BOXPLOT To understand better the effects of exercise and aging on various cireulatory functions, the article "Cardiac Output in Male Middle-Aged Runners" (Journal of Sports Medicine [1982]: 17–22) presented data from a study of 21 middle-aged male runners. The following data set gives values of oxygen capacity values (in milliliters per kilo- gram per minute) while the participants pedaled at a speci- fied rate on a bicycle ergometer: 12.81 14.95 15.83 15.97 17.90 18.27 18.34 19.82 19.94 20.62 20.88 20.93 20.98 20.99 21.15 22.16 22.24 23.16 23.56 35.78 36.73 a. Compute the median and the quartiles for this data set. b. What is the value of the interquartile range? Are there outliers in this data set? c. Draw a modified boxplot, and comment on the interest- ing features of the plot.
- The least-squares regression equation is y = 689.9x + 14,803 where y is the median income and x is the percentage of 25 years and older with at least a bachelor's degree in the region. The scatter diagram indicates a linear relation between the two variables with a correlation coefficient of 0.7256. Complete parts (a) through (d). (a) Predict the median income of a region in which 20% of adults 25 years and older have at least a bachelor's degree. (Round to the nearest dollar as needed.) . TOLED dian Income Media 55000- 20000- 15 20 25 30 35 40 45 50 55 60 Bachelor's 96 QSuppose that you wish to estimate the effect of class attendance on student performance. A basic model is examscore = β0 + β1attendance + β2priorGP A + u where examscore is students’ score on the exam (from 1 to 6), attendance is the number of TA sessions attended on Zoom (from 0 to 9), and priorGPA is the average exam grade last year. (a) Let internet be the quality of internet at the student’s study place. Do you think internet satisfies the independence assumption? What about the exclusion restriction? (b) Assuming that internet satisfies the conditions above, what other condition must internet satisfy in order to be a valid IV for attendance? (c) Suppose, we add the interaction term priorGP A × attendance. Interpret the coefficient on the interaction term. (d) (Difficult) If attendance is endogenous, then, in general, so is priorGP A × attendance. What might be a good IV for priorGP A × attendance?A coffee shop owner offers two brands of coffee, Brand "A" and Brand "B". The proportion of clients that prefer Brand "A" is 60%. Recently, the price of Brand "A" went up. The proportion of clients that buy brand "A" among the first 15 customers that enter the shop after the change in price is an estimator of the current proportion of clients that prefer Brand "A". Assume that the change in price has no effect on the preference of the clients. Then the mean squared error (MSE) of the estimator is equal to: (Give an answer of the form x.xxx, i.e. with 3 significance digits) (HINT: Carefully read the definition of MSE in Chapter 10 of the text book.)
- Explain the Stationarity in the AR(1) Model?Composites are materials that are made by embedding a fiber, such as glass or carbon, inside a matrix, such as a metal or a ceramic. Composites are used in engineering structures, and their degradation when subjected to whether conditions is an important issue. In an experiment to investigate the effect of the moisture on a certain kind of composite, the weight gains of a collection of n=28 samples of composite subjected to water diffusion were obtained. The sample mean is x=0.42% with a sample std. of S=0.04% Can we 95% confident from the results of this experiment that the average weight gain for composites of this kind is smaller than 0.43%?Consider a single variable model to estimate the effect of in-person lecture attendance on university students' WAMs (weighted average marks): (C1) WAM = β0 + β1Attend + u Where: WAMis a student's weight average mark Attendis the proportion of lectures attended by a student in an academic year Using the information above, answer the following 3 questions. [i] Explain why Attend might be endogenous in Model (C1). What does this suggest about E [u ∣ Attend]? [ii] Suppose that student attendance is related to a student's conscientiousness. Does this mean that conscientiousness would be a good instrumental variable for Attend? Why or why not? Explain your reasoning. [iii] Unfortunately a major strike by railway works disrupts transport links to 3 of the 7 universities in a city for a period of 2 weeks. This temporarily means that students who are enrolled at those 3 affected universities cannot attend lectures in person. Carefully explain how you would use this information to…