Question

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Yaqeel Services conduct an annual audit of its financial records. A total of 25,056 sales invoices were issued and on the basis of past experience, the standard deviation of the amount of invoices is RM30. 2/10 a) Apakah saiz sampel yang patut dipilih oleh juruaudit jika dia mahu 95% keyakinan dalam menganggarkan min populasi untuk amaun invois pada ralat persampelan tRM5? What sample size should the auditor select if he wants to be 95% confident of estimating the population mean of the invoice amount to within a sampling error of tRMS? (2 markah/marks) Dengan menggunakan saiz sampel yang dipilih di (a), satu audit dijalankan dengan keputusan berikut Using the sample size selected in (a), an audit is conducted with the following results Amaun invois jualan AMOUNT OF SALES INVOICES S RM34.55 b) Jika juruaudit mahu menggangarkan min populasi,patutkah dia menggunakan sishan piawai untuk populasi atau sampel? Terangkan If the auditor wants to estimate the population mean of the invoice amount should he use the population or sample standard deviation? Explain (2 markah/marks) 12 amaun invois yang tidak tepat telah dimasukkan ke dalam sistem jualan. Perbezaan (dalam RM) di antara amaun invois yang tepat dan tidak tepat adalah seperti berikut: 12 incorrect amount of invoices were entered into the sales information system. The differences (in RM) between the correct and incorrect amount of invoices were as follows 17 25 14 -10 20 40 35 30 28 5 c) Bina selang keyakinan 95% bagi jumlah perbezaan antara amaun invois jualan yang tepat dan tidak tepat Construct a 95% confidence interval estimate for the total difference between the correct and incorrect amount of the sales invoices (6 markah/marks)

Answer 1

a)

Sample size, n = (z $\sigma$ / E)²

where z is z score for 95% confidence interval.

$\sigma$ is the population standard deviation

E is the margin of error.

z score for 95% confidence interval is 1.96

$\sigma$ = 30

E = 5

Sample size, n = (1.96 * 30 / 5)²

= 138 (Rounding to nearest integer)

(b)

To estimate the population mean, we should use the population standard deviation as it will provide smaller margin of error for small sample size.
If we use sample standard deviation, it will give large margin of error for small sample sizes.

(c)

Sample mean difference, d = (17 + 25 + 14 - 10 + 20 + 40 + 35 + 30 + 28 + 22 + 15 + 5)/12 = 20.08

Sample Std deviation, s = SQRT[(17 - 20.08)^2 + (25- 20.08)^2 + (14- 20.08)^2 + ( - 10- 20.08)^2 + (20- 20.08)^2 + ( 40- 20.08)^2 + (35 - 20.08)^2 + (30- 20.08)^2 + (28 - 20.08)^2 + (22 - 20.08)^2 + (15 - 20.08)^2 + (5- 20.08)^2 )/11]

= 13.53

Standard error of mean = s / $\sqrt{n}$ = 13.53 / $\sqrt{12}$ = 3.905775

Degree of freedom = n - 1 = 12 - 1 = 11

Critical value of t at 95% confidence interval and df = 11 is 2.2

Margin of error = t * Std Error = 2.2 * 3.905775 = 8.59

95% confidence interval is,

(20.08 - 8.59,  20.08 + 8.59)

(11.49,  28.67)