HW11_skeleton

10.2

• 1. statistically significant
• 2. T
• 3. F
• 4. 2.33
• 5. -1.28
• 6. ???1.96

7.) np0(1-p0)=42>10 a) Z0=2.315>z0.95=1.645, reject the null hypothesis

b) p(z>2.315)=0.0104<a=0.05, we reject the null hypothesis, there is sufficient evidence at the a=005 level of significance to reject the null hypothesis

11.) np0(1-p0)=45>10 a) z0=-1.493, it is between the zone -z0.025=-1.96 and z0.025=1.96, do not reject the null hypothesis

b) 2p(z<-1.493)=2x0.0681=0.1362>a=0.05, do not reject the null hypothesis,there is not sufficient evidence at the a=005 level of significance to reject the null hypothesis

 

17. H0:p=0.019, H1:p>0.019 a=0.01 np0(1-p0)=16.1>10, n<=0.05N phat=x/n=0.022 z0=0.645 p(z>0.645)=0.2578>a=0.01, do not reject the null hypothesis, there is no sufficient evidence at the a=0.01 level of significance to conclude more than 1.9% of Lipitor users experience flulike symptoms as a side effect

18. h0:p=0.94,h1:p>0.94 a=0.01 np0(1-p0)=12.6>10 phat=213/224=0.951 z0=0.692 p(z>0.692)=0.2451>a=0.01, do not reject the null hypothesis, there is no sufficient evidence at the a=0.01 level of significance to conclude more than 94% of patients taking Nexium are healed within 8 weeks

 

10.3

1.)

a) t0.01=2.602
  
b) -t0.05=-1.729
  
c) +-t0.025=+-2.179

12.)

a) Ho:mu=6.37
   
   H1:mu>6.37
  
b) the P-value of 0.35 means that, if the null hypothesis that mu=6.37 is true, we expect abhou 35 out of 100 samples to result in a sample mean as extreme or more extreme than one obtained.
  
c) do not reject the null hypothesis

13.)

a) Ho:mu=22

   H1:mu>22
  
b) the sample is random, n=200>30, we can assume that the sample is small relative to the population, so the scores are independent
  
c)  a=0.05
    t0=2.174
    200-1=199degrees of freedom
    t0.05=1.66
    t0=2.174>t0.05=1.66
    we reject the null hypothesis
d) there is sufficient evidence to conclude that students who complete the core curriculum are scoring above 22 on the math portion of the ACT

14.)

a) Ho:mu=501

   H1:mu<501

b) the sample is random, n=100>30,we can assume that the sample is small relative to the population, so the scores are independent
  
c)  a=0.1
    t0=-1.379
    100-1=99 degrees of freedom
    -t0.1=-1.29
    t0=-1.379<-t0.1=-1.29
    we reject the null hypothesis
d) there is sufficient evidence to conclude that students who learn English as well as another language simultaneously score worse on the SAT Critical Reading exam than the general population of test takers

15. H0:mu=9.02 H1:mu<9.02 a=0.01 t0=-4.555 12-1=11 degrees of freedom -t0.01=-2.718 t0=-4.555<-t0.01=-2.718 we reject the null hypothesis, there is sufficient evidence to conclude that the mean hippocampal volume in alcoholic adolescents is less than the normal mean volume of 9.02cm^3