ECON 3810 HW #4.docx - Christian Ball HW#4 1 n=105 xbar=78 sample standard deviation=12 H 0 u>=80 H1 u<80 a t*=78-80(12\/sqrt(105 t*=-1.71 b Standard

# ECON 3810 HW #4.docx - Christian Ball HW#4 1 n=105 xbar=78...

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Christian Ball HW #4 1. n=105, xbar=78, sample standard deviation=12, H 0 : u>=80, H 1 : u<80 a. t*=78-80/(12/sqrt(105)), t*=-1.71 b. Standard normal distribution because n is larger than 100 c. 1 critical value on the left, 5%=level of significance (LS)=0.05, in R qnorm(0.05)=-1.6448, conclusion: you reject the null hypothesis because the test statistic (TS) is in the rejection region(RR) at 0.05 d. p-value=p(TS<-1.71), in R pnorm(-1.71)=0.044, conclusion: could reject H 0 if LS>0.044, the null is not rejected at LS<0.044 2. N=51, xbar=14, sample standard deviation=3, H 0 : u=15, H 1 : u does not equal 15 a. t*=14-15/(3/sqrt(51)), t*=-2.38 b. Student T with a 50 degrees of freedom because (n-1) -> (51-1)=50 c. 2 critical values, LS=0.005, in R qt(0.025, df=50)=-2.01 on left and 2.01 on right, conclusion: reject the null because TS is in RR at 0.05 d. p-value=2*p(|TS|>2.38) -> in R 2*pt(2.38, df=50, lower.tail=FALSE) -> (0.011)2=0.022, conclusion: could reject H 0 if LS>0.022 3. n=66, pbar=0.8, H 0 : p=0.88, H 1 does not equal 0.88 a. z=(0.8-0.88)/(sqrt(0.88(1-0.88))/66)=-2 b. The sampling distribution will be standard normal distribution c. 2 critical values, LS= 0.05, in R qnorm(0.025)=-1.96, conclusion: you reject the null

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