Question

A credit bureau analysis of undergraduate students credit records found that the average number of credit cards in an undergraduate's wallet was \(4.09\) ("Undergraduate Students and Credit Cards in 2004," Nellie Mae, May 2005). It was also reported that in a random sample of 132 undergraduates, the sample mean number of credit cards carried was 2.6. The sample standard deviation was not reported, but for purposes of this exercise, suppose that it was \(1.2\). Is there convincing evidence that the mean number of credit cards that undergraduates report carrying is less than the credit bureau's figure of \(4.09\) ?

Short answer

Based on the hypothesis testing via the one sample z-test, it can be determined if there is convincing evidence that the mean number of credit cards that undergraduates report carrying is less than the credit bureau's figure of \(4.09\). The exact numerical determination requires calculation based on the steps provided.

Step-by-step solution

- Stating the hypotheses

In hypothesis testing, we start by stating the Null hypothesis and the Alternative hypothesis. For this problem, the Null hypothesis \(H_0\) is that the mean number of credit cards is equal to 4.09, i.e. \(H_0: \mu = 4.09\). The Alternative hypothesis \(H_1\) reflects the claim we're testing, that the mean number of credit cards is less than 4.09, i.e. \(H_1: \mu < 4.09\)

- Calculating the test statistic

Standardize the sample mean to see how unusual it is, measuring in standard deviations. Here, Z-statistic (or test statistic) can be calculated using the formula: \(Z = \frac{(\overline{x}-\mu)}{(\sigma/\sqrt{n})}\) where, \(\overline{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the standard deviation and \(n\) is the sample size. Substituting the given values in the formula, we get \(Z = \frac{(2.6-4.09)}{(1.2/\sqrt{132})}\)

- Comparing to the critical value

After computing the value of the Z statistic, we will compare it with the critical z value for a given confidence level, generally \( Z_{\alpha}\). If the absolute value of the Z statistic is greater than \( Z_{\alpha}\), we can reject the null hypothesis. In this case, the critical z-value we compare with is for \(\alpha = 0.05\) (standard for 95% confidence), which is -1.645 (for a one-sided test).

- Making the decision

Based on the comparison in Step 3, if our calculated z-value is less than the critical z-value, we can reject the null hypothesis and say that there is enough evidence to support the claim that the mean number of credit cards that undergraduates report carrying is less than the credit bureau's figure of 4.09. If not, we do not reject the null hypothesis.