Respuesta :
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:
[tex]H_{0}: p = 0.05[/tex]
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:
[tex]H_{a}: p < 0.05[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
0.05 is tested at the null hypothesis:
This means that
[tex]\mu = 0.05[/tex]
[tex]\sigma = \sqrt{0.05*0.95}[/tex]
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 [tex]n = 1124, X = \frac{45}{1124} = 0.04[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.04 - 0.05}{\frac{\sqrt{0.05*0.95}}{\sqrt{1124}}}[/tex]
[tex]z = -1.54[/tex]
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.
