Chapter 8: Q65E (page 452)

Last name and acquisition timing. Refer to the Journal of Consumer Research (August 2011) study of the last name effect in acquisition timing, Exercise 8.13 (p. 466). Recall that the mean response times (in minutes) to acquire free tickets were compared for two groups of MBA students— those students with last names beginning with one of the first nine letters of the alphabet and those with last names beginning with one of the last nine letters of the alphabet. How many MBA students from each group would need to be selected to estimate the difference in mean times to within 2 minutes of its true value with 95% confidence? (Assume equal sample sizes were selected for each group and that the response time standard deviation for both groups is $σ$ ≈ 9 minutes.)

Short Answer

Expert verified

The required sample size is 156.

Step by step solution

Given Information

The standard deviation of two groups are given below

$σ_{1} = σ_{2} = 9$

The sampling error is

$S E = 2$

Z-value

For

The z-value is given by

$\begin{array}{rcl} z_{\frac{α}{2}} & = & z_{0.025} \\ = & 1.96 \end{array}$

Compute the sample

For, $z = 1.96, σ_{1} = σ_{2} = 9 a n d S E = 2$

The sample is calculated as

$\begin{array}{rcl} n_{1} & = & n_{2} \\ = & \frac{{(z_{0.005})}^{2} \times [σ_{1}^{2} + σ_{2}^{2}]}{{(S E)}^{2}} \\ = & \frac{{1.96}^{2} \times [9^{2} + 9^{2}]}{2^{2}} \\ = & \frac{3.8416 \times 162}{4} \\ = & \frac{622.3392}{4} \\ = & 155.58 \\ \approx & 156 \end{array}$