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

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.

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 = (X - \mu)/((\sigma)/(√(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 = √(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 = (45)/(1124) = 0.04`

Value of the test statistic:

`z = (X - \mu)/((\sigma)/(√(n)))`

`z = (0.04 - 0.05)/((√(0.05*0.95))/(√(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.