Example 9

EXAMPLE 9 Confidence interval for $μ_{1} - μ_{2}$

Bank Loans

Find a 90% confidence interval for the difference in population mean credit scores for those approved and those denied bank loans, using the data in Table 10.

Solution

Both sample sizes are large ( $(n_{1} = n_{2} = 100 \geq 30)$ ), so we may construct the interval. For $t_{α / 2}$ , the required degrees of freedom is the smaller of $n_{1} - 1$ and $n_{2} - 1$ , which is $100 - 1 = 99$ . Because $df = 99$ is not listed in the $t$ table, we use the next lower value listed as a conservative alternative: $df = 90$ . For 90% confidence, then $t_{α / 2} = 1.662$ .

The margin of error is

$E = t_{α / 2} \cdot \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}} \approx (1.662) \cdot \sqrt{\frac{{(34)}^{2}}{100} + \frac{{(70)}^{2}}{100}} \approx 12.934$

The 90% confidence interval is then

$({\bar{x}}_{1} - {\bar{x}}_{2}) \pm E = (708 - 635) \pm 12.934 = (60.066, 85.934)$

We are 90% confident that the difference in population mean credit scores $μ_{1} - μ_{2}$ lies between 60.066 and 85.934. Because 0 is not contained in this interval, we may conclude that $μ_{1} \neq μ_{2}$ , just as we did in Example 8.

NOW YOU CAN DO

Exercises 11–16.