Question

An auditor for a hardware store chain wished to compare the efficiency of two different auditing techniques. To do this he selected a sample of nine store accounts and applied auditing techniques A and B to each of the nine accounts selected. The number of errors found in each of techniques A and B is listed in the table below:

Errors in A	Errors in B
25	11
28	17
26	19
28	17
32	34
30	25
29	29
20	21
25	30

Select a 90% confidence interval for the true mean difference in the two techniques.

a) [0.261, 8.627]

b) [-4.183, 4.183]

c) [2.195, 6.693]

d) [3.050, 5.838]

e) [2.584, 6.304]

f) None of the above

Answer 1

For calculating 90% confidence interval for the true mean difference, we need to find mean and standard deviation for the difference between errors in A and B.

first, we need to find the difference between the errors in A and B

Now, we need to find the mean and standard deviation for the difference column

we know that mean =(sum of all the data values)/(total number of data values)

setting the values, we get

Mean= (14+11+...+(-1)+(-5))/9 = 40/9 = 4.44

Formula for the standard deviation is given as

where di are all difference values, d(bar) is mean difference and n is the sample size equal to 9

setting the values, we get

We will use t critical value in this case because population variance is unknown.

Degree of freedom = n-1 = 9-1 = 8

alpha level = 1-0.9 = 0.10

Using the t distribution table for two tailed hypothesis (for confidence interval we always use two tailed test)

we get , t critical =

Using the confidence interval for the mean difference, we can write

CI=

setting calculated values, we get

CI =

this gives us

CI = (0.262,8.627)

So, option A is the correct answer