Question

Entrepreneurs developing an accounting review program for persons preparing to take the Certified Public Accountant (CPA) examination are considering two possible formats for conducting the review sessions. A random sample of 10 students are trained using format 1, and then their number of errors is recorded for a prototype examination. Another random sample of 12 individuals are trained according to format 2, and their errors are similarly recorded for the same examination. For the 10 students trained with format 1, the individual performances are 11, 8, 8, 3, 7, 5, 9, 5, 1, and 3 errors. For the 12 students trained with format 2, the individual performances are 10, 11, 9, 7, 2, 11, 12, 3, 6, 7, 8, and 12 errors. In comparing the performances of the two groups, the 0.10 level of significance will be used. Based on these data, the 10 members of group 1 made an average of 6.000 errors, with a sample standard deviation of 3.127. The 12 students trained with format 2 made an average of 8.167 errors, with a standard deviation of 3.326. The sample standard deviations do not appear to be very different, and we will assume that the population standard deviations could be equal. Test to see if the two training formats are equally effective or not.

Answer 1

given that

for format 1

sample size =n1=10

sample mean =m1=6

sample SD=S1=3.127

for format 2

sample size =n2=12

sample mean =m2=8.167

sample SD=S2=3.326

we have assumed that both populations standard deviations are equal hence we will use pooled standard deviation which is given by

=3.238

we have to test that

now test statistics is given by

here t is t distribution with df =n1+n2-2 =10+12-2=20

and the test is two-tailed hence

P-Value =2*P(t20>0.6692) =2*0.2555=0.511

Since P Value is more than level of significance hence we failed to reject H0

so

we dont have sufficient evidence to support the claim that there is difference in mean of two formats.