Question

The Dynamic Credit Company issues credit cards and uses software to detect fraud. After tracking the spending habits of one customer, it is found that charges over $100 constitute 35.8% of the credit transaction. Among 30 charges made this month, 18 involve totals that exceed $100. Does this constitute an unusual spending pattern that should be verified? Explain.

Answer 1

We would check for the same using Z test for one population proportion.

The following information is provided: The sample size is N=30, the number of favorable cases is X=18, and the sample proportion is , and the significance level is α=0.05

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: p=0.358

Ha: p>0.358

This corresponds to a right-tailed test, for which a z-test for one population proportion needs to be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a right-tailed test is zc​=1.64.

The rejection region for this right-tailed test is R={z:z>1.64}

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that z=2.765>zc​=1.64, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0028, and since p=0.0028<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p is greater than p0​, at the α=0.05 significance level.

So this does constitute an unusual spending pattern that should be verified.

Confidence Interval

The 95% confidence interval for p is: 0.425<p<0.775.

Graphically

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