Question

Homework assistance for accounting students. Refer to the Journal of Accounting Education (Vol. 25, 2007) study of providing homework assistance to accounting students, Exercise 8.18 (p. 468). Recall that one group of students was given a completed homework solution and another group was given only check figures at various steps of the solution. The researchers wanted to compare the average test score improvement of the two groups. How many students should be sampled in each group to estimate the difference in the averages to within .5 point with 99% confidence? Assume that the standard deviations of the test score improvements for the two groups are approximately equal to 1

Short answer

The required sample size is 54.

Step-by-step solution

Given Information

The standard deviation of two groups are given below

${\sigma }_1={\sigma }_2=1$

The sampling error is

$SE=.5$

Z-value

For $\alpha =0.01$

The z-value is given by

$$\begin{gathered} z_{\frac{\alpha }{2}}=z_{0.005} \\ =2.58 \end{gathered}$$

Compute the sample

For, $z=2.58,{\sigma }_1={\sigma }_2=1 andSE=.5$

The sample is calculated as

$\begin{array}{rcl}{n}_1& =& n_2\\ & =& \frac{{\left(z_{0.005}\right)}^2\times \left[{\sigma }_1^2+{\sigma }_2^2\right]}{{\left(SE\right)}^2}\\ & =& \frac{{2.58}^2\times \left[1^2+1^2\right]}{{.5}^2}\\ & =& \frac{6.6564\times 2}{.25}\\ & =& \frac{13.3128}{.25}\\ & =& 53.2512\\ & \approx & 54\end{array}$

Therefore, the required sample size is 54.