Question

1) (7 points) An investigator hypothesizes that educating parents on proper nutrition and exercise will reduce the total cholesterol level of children. A sample of 20 children between the ages of ten to fifteen have agreed to have their total cholesterol level measured for this study. Suppose the investigator wishes to run a hypothesis test at a 5% level of significance. a) (1 point) What is the outcome variable? Is this outcome variable considered to be continuous or dichotomous? b) (2 points) Set up the appropriate null and alternative hypotheses. The researcher assumes that the true (null) value for the mean total cholesterol in all children is 191 mg/dL. c) (1 point) Select the appropriate test statistic. Should a z ort test statistic be used for this hypothesis test? Explain your reasoning and/or show your work. d) (1 point) Write out the decision rule for this hypothesis test. Make sure to include the appropriate z or t critical value. e) (1 point) The researcher determined the sample mean cholesterol level to be 175 mg/dL with a standard deviation of 13.77 mg/dL. Use this information to calculate the appropriate test statistic. f) (1 point) Make a decision using the decision rule in (d) and the calculated test statistic in (e). Write out an appropriate conclusion.

Answer 1

a The & outcome of the 290oble the agriable is choles for level is consideo to be continous blHypothesis: - Hoi- l = 191 mg/dL diffrouse 1.e. There is no grifronlea between the educating ponents os pappa supotos & excessibl_will_oedule the total cholesterol. Karel of childoen. 5 H:- 191 mg/dl je. These is Onliche differedie bem -the educating paserys on poopis putos tios escensze will Lcoeduce the 'otay cholesterol level of childbes . Here we will 2 test for hypothesis test Deigion Rules Accept Ho.- az < Taba (ritical value] Regeef Ho:- (az > Tab(z) (ie Coitia Nalues/ Costiccer Yalye = 241 = 70.05% =_96 Cobtejo tooro Standaod Woony Toble } epfonie gehog pagina rejection region - 200/2 Z805/ logo 1.96

e we have LDE 2016=) 3:77 = 1755, 43191. 2. gref 6 150 175 - 191_-_-16 13.77 lino 3:07906 Cal z= -5.1963. Cal 12) = 5.1962] a ta z of Henie Here because the we take bypothesis absolute value is two tailed. a 2= 5:1968 Cal z 7 Tab 2 , Heole we seject Ho of 57. Level of Gryficance & conclude that There is agnificant effect of educating parents on -porpes surton & excesuse will redure the cholespol level of children