1. Suppose\tyou\tare\tdoing\ta\tstudy\ton\thousehold\tmilk\tconsumption,\tand\tusing a\tsample\tof\tn\t=\t33\thouseholds\tyou\tobtained\ta\tregression\tline\tof: y\t=\t.012\t+\t.088x1\t+\t1.76x2\t+\t1.62x3 (.005) (.079) (.792) (.56) where\tthe\tnumbers\tin\tparentheses\trepresent\tthe\tstandard\terror\tof\tthe coefficient\testimates.\tFurthermore,\teach\tvariable\tis: y\t=\tMilk\tconsumption\tin\tquarts per\tweek\tx1\t=\tweekly\tincome,\tin hundreds\tof\tdollars\tx2\t=\tfamily\tsize x3\t=\ta\tdummy\tvariable\tthat\tequals\t1\tif\tone\tof\tthe\tmembers\tof\tthe\tfamily\tis lactose\tintolerant,\tand\t0\totherwise (a) What\twould\tyou\tpredict\twould\tbe\tthe\thousehold\tmilk\tconsumption\tfor a\tfamily\twith\t4\tmembers,\ta\tweekly\tincome\tof\t$1,500,\tand\twith\ta lactoseintolerant\tfamily\tmember? (b) Test\tthe\tnull\thypothesis\t1\t=\t0\tagainst\tthe\ttwosided\talternative\tat\tthe\t =\t.05\tsignificance\tlevel\tusing\tthe\tt\tdistribution. (c) Test\tthe\tnull\thypothesis\t2\t=\t0\tagainst\tthe\ttwosided\talternative\tat\tthe\t =\t.05\tsignificance\tlevel\tusing\tthe\tt\tdistribution (d) Construct\ta\t95%\tconfidence\tinterval\tfor\t3. (e) The\t99%\tconfidence\tinterval\tfor\t1\tis\t(.1297,.3057).\tThe\t99% confidence\tinterval\tfor\t2\tis\t(.423,3.94). The\t99%\tconfidence\tinterval\tfor\t3\tis\t(3.16,.0767) Without\tdoing\tthe\thypothesis\ttest,\tanswer\tthe\tfollowing:\tfor\twhich\tof these\ti\twould\tyou\treject\tthe\tnull\tH0\t:\ti\t=\t0\tagainst\tthe\ttwosided alternative\tat\tthe\t\t=\t.01\tsignificance\tlevel