Example 7.17

EXAMPLE 7.17 Planning a New Smart Shopping Cart Study

smrtcrt

As part of Example 7.10 (pages 381–382), we calculated a 95% confidence interval for the mean difference in spending when shopping with and without real-time feedback. The 95% margin of error was roughly $2.70. Suppose that a new study is being planned and the desired margin of error is $1.50. How many shoppers per group do we need?

The sample standard deviations in Example 7.10 were $6.59 and $6.85. To be a bit conservative, we’ll guess that the two population standard deviations are both $7.00. To compute an initial $n$ , we replace $t^{*}$ with $z^{*}$ . This results in

$n = {(\frac{\sqrt{2} z^{*} s^{*}}{m})}^{2} = {[\frac{\sqrt{2} (1.96) (7)}{1.5}]}^{2} = 167.3$

We round up to get $n = 168$ . The following table summarizes the margin of error for this and some larger values of $n$ .

$n$	$t * s * \sqrt{2 / n}$
168	1.502
169	1.498
170	1.493

The requirement is first satisfied when $n = 169$ . In SAS, we’d perform these calculations using the command

proc power;

twosamplemeans CI=diff stddev=7 halfwidth=1.5

probwidth=0.50 npergroup=.;

run;

Page 402

This sample size is almost 3.5 times the sample size used in Example 7.10. The researcher may not be able to recruit this large a sample. If so, we should consider a larger desired margin of error.