HW11_skeleton

10.2

• 1. statistically significant
• 2. True
• 3. false
• 4. 2.33
• 5. -1.28
• 6. 1.645

7.)

a) Classical: Z= 2.31 which is greater than the Z for alpha of .05 which is 1.645. We reject null hypothesis at the .05 significance level
  
b) P value of .0104, which is less than alpha of .05. we can now reject null hypothesis at the .05 level of significance

11.)

a) Classical, Z value = -1.49 and the z value for alpha of .10 is -1.96. Therefore we fail to reject null hypothesis at the alpha level of .10
  
b) P value,  P value= .1360 which is greater than the alpha level of .05. Therefore we also fail to reject null hypothesis at the .05 alpha level.

 

17. Classical, Z value= .65 which is less than the alpha z value of .01 which is 2.33, therefore we fail to reject the null hypothesis at the .01 alpha level

18. Classical, Z value is .0159 which is less than alpha z value of .01 which is 2.33. Therefore we fail to reject null hypothesis.

 

10.3

1.)

a) 2.602
  
b) -1.729
  
c) + or - 2.179

12.)

a) Ho: mu= 63.7
   
   H1: mu> 63.7
  
b) The p value of .35 represents the probability that the you accepted the alternative hypothesis when it in fact the null was true
  
c) p value= .35 > .10 meaning we fail to reject the null hypothesis

13.)

a) Ho:  mu = 22

   H1: mu > 22
  
b) Sample size is larger than 30 as it equals 200, scores also independant
  
c) Classical, T=2.176 which is greater than the t value at .05 which is 1.660 so we will reject the null hypothesis.

d) We have enough evidence to conclude that students are scoring higher than 22 on their scores

14.)

a) Ho:mu= 501

   H1:mu > 501

b) The scores are indendant and the sample size is well over 30
  
c) Classical, T value is -1.38, p value is between -1.28 and 1.28

d) Seeing that the t value falls in the critical region, We reject the null hypothesis

15. Ho= 9.02 cm H1 < 9.02 cm P value = .0004 < alpha of .01 We reject the null hypothesis as there is sufficent evidence to conclude hippocamal valume is less than 9.02