## Related questions with answers

A large discount chain compares the performance of its credit managers in Ohio and Illinois by comparing the mean dollar amounts owed by customers with delinquent charge accounts in these two states. Here a small mean dollar amount owed is desirable because it indicates that bad credit risks are not being extended large amounts of credit. Two independent, random samples of delinquent accounts are selected from the populations of delinquent accounts in Ohio and Illinois, respectively. The first sample, which consists of $10$ randomly selected delinquent accounts in Ohio, gives a mean dollar amount of $\$ 524$ with a standard deviation of $\$ 68$. The second sample, which consists of $20$ randomly selected delinquent accounts in Illinois, gives a mean dollar amount of $\$ 473$ with a standard deviation of $\$ 22$.

Set up the null and alternative hypotheses needed to test whether there is a difference between the population mean dollar amounts owed by customers with delinquent charge accounts in Ohio and Illinois.

Solutions

VerifiedThe goal of the exercise is to set up the null and alternative hypotheses needed to test whether there is a difference between the population mean dollar amounts owed by customers with delinquent charge accounts in Ohio and Illinois.

The given information is:

$\text {Sample }$ | $\mathrm{N}$ | $\text { Mean }$ | $\text{ StDev }$ | $\text { SE Mean }$ |
---|---|---|---|---|

$\text { Ohio }$ | 10 | 524.0 | 68.0 | 22 |

$\text { Illinois }$ | 20 | 473.0 | 22.0 | 4.9 |

The formula we have to take a look at is a formula for a $t$-test about the difference between two population means and it consists two equations:

$H_0:\mu_1- \mu_2=D_0,$

which represents null hypothesis and

$t=\frac{(\=x_1-\=x_2)-D_0}{\sqrt{s_p^2 \left(\frac{1}{n_1}+\frac{1}{n_2}\right)}},$

which represents test statistic.

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