Question

PLEASE SHOW IN EXCEL

A professor wants to know if his lecture had an impact on student learning in his class. He gives his students a pre-test at the beginning of the lecture. After he gives the lecture he gives a post-test. The post-test has similar questions in order to gauge learning. The results of the tests are:

Pre-test	Post-test
78	80
67	69
56	70
78	79
96	96
82	84
84	88
90	92
87	92

Did the students score significantly higher on the post-test than they did on the pre-test?

State your hypothesis and your null hypothesis:

Decide which t-test you will use (Dependent, Independent – equal variance, Independent – unequal variance):

Test your null hypothesis at the .05 level (decide if this is one-tailed or two-tailed). Which tailed test did you use and what is your observed t value?

What is the critical value for this test?

Do you accept or reject your null hypothesis?

State your conclusion:

Answer 1

given that, null, ho: ud = 0; alternate, hi: ud < 0 level of significance,alpha = 0.05 from standard normal table,left tailed talpha/2 = 1.86 since our test is left-tailed reject ho, if to <-1.86 we use test statistic to= d/ (s/Vn) where value of s^2 = [ di^2-( di)^2 /n]/(n-1)). d = (xi-yi)/n) = -3.5556 we have d = -3.5556 pooled standard deviation = calculate value of sd= Vs^2 = sort [ 254-(-32^2/9] / 8 = 4.1866 => to = d/ (s/Vn) = -2.5478 critical value the value of talpha with n-1 = 8 d.f is 1.86 we got It op = 2.5478 & talpha| =1.86 make decision; hence value of to | > | talpha and here we reject ho p-value :left tail - ha :(p < -2.5478 ) = 0.01715 hence value of p0.05 > 0.01715, here we reject ho

null, ho: ud = 0; alternate, hi: ud < 0 t-test dependent test statistic: -2.5478, p-value: 0.01715 critical value: reject ho, if to < -1.86