Example 7.14

EXAMPLE 7.14 Does the Cash Flow Margin Differ?

cmps

CASE 7.2 Take Group 1 to be the firms that were active and Group 2 to be those that failed. The question of interest is whether or not the mean cash flow margin is different for the two groups. We therefore test

$\begin{array}{l} H_{0} : μ_{1} = μ_{2} \\ H_{a} : μ_{1} \neq μ_{2} \end{array}$

Here are the summary statistics:

Group	Firms	$n$	$\bar{x}$	$s$
1	Active	74	5.42	18.80
2	Failed	27	−7.14	21.67

The sample standard deviations are fairly close. A difference this large is not particularly unusual even in samples this large. We are willing to assume equal population standard deviations. The pooled sample variance is

$\begin{array}{l} s_{p}^{2} & = \frac{(n_{1} - 1) s_{1}^{2} + (n_{2} - 1) s_{2}^{2}}{n_{1} + n_{2} - 2} \\ = \frac{(73) {(18.80)}^{2} + (26) {(21.67)}^{2}}{74 + 27 - 2} = 383.94 \end{array}$

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so that

$s_{p} = \sqrt{383.94} = 19.59$

The pooled two-sample $t$ statistic is

$\begin{array}{l} t & = \frac{{\bar{x}}_{1} - {\bar{x}}_{2}}{s_{p} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}} \\ = \frac{5.42 - (- 7.14)}{19.59 \sqrt{\frac{1}{74} + \frac{1}{27}}} = 2.85 \end{array}$

The $P$ -value is $P (T \geq 2.85)$ , where $T$ has the $t (99)$ distribution.

In Table D, we have entries for 80 and 100 degrees of freedom. We will use the entries for 100 because $k = 99$ is so close. Our calculated value of $t$ is between the $p = 0.005$ and $p = 0.0025$ entries in the table. Doubling these, we conclude that the two-sided $P$ -value is between 0.005 and 0.01. Statistical software gives the result $p = 0.005$ . There is strong evidence that the average cash flow margins are different.

$df = 100$
$p$	0.005	0.0025
$t^{*}$	2.626	2.871