Question

A check cashing service found that approximately 5% of all checks submitted to the service were fraudulent. After instituting a check-verification system to reduce its losses, the service found that only 45 checks out of a random sample of 1124 were fraudulent. Does this sample provide sufficient evidence at the .01 significance level that the check verification system is effective

Answer 1

Answer:

Pvalue of the test is 0.0618 > 0.01, which means that this sample does not provide sufficient evidence at the .01 significance level that the check verification system is effective.

Step-by-step explanation:

A check cashing service found that approximately 5% of all checks submitted to the service were fraudulent.

This means that the null hypothesis is:

$H_{0}: p = 0.05$

Test if the check verification system is effective:

If it is effective, the proportion will decrease, so we are going to test if after the change the proportion is lower than 0.05, that is:

$H_{a}: p < 0.05$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.05 is tested at the null hypothesis:

This means that

$\mu = 0.05$

$\sigma = \sqrt{0.05*0.95}$

After instituting a check-verification system to reduce its losses, the service found that only 45 checks out of a random sample of 1124 were fraudulent.

This means that $n = 1124, X = \frac{45}{1124} = 0.04$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.04 - 0.05}{\frac{\sqrt{0.05*0.95}}{\sqrt{1124}}}$

$z = -1.54$

Pvalue of the test and decision:

Probability of finding a sample proportion lower than 0.04, which is the pvalue of z = -1.54.

Looking at the z-table, z = -1.54 has a pvalue of 0.0618.

Pvalue of the test is 0.0618 > 0.01, which means that this sample does not provide sufficient evidence at the .01 significance level that the check verification system is effective.