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 that the students said they 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

If the calculated z-score in Step 3 is found to be less than the critical z-value (which corresponds to the chosen significance level) or the p-value is found to be less than the chosen significance level, there is convincing evidence to reject the null hypothesis and conclude that the mean number of credit cards that students report carrying is less than the credit bureau's figure of 4.09.

Step-by-step solution

Hypothesize

First start by formulating the null hypothesis $H_0$ and the alternative hypothesis $H_A$. For this case, the null hypothesis is that the mean number of credit cards that students report carrying is the same as the credit bureau's figure, i.e., $H_0:\mu =4.09$. The alternative hypothesis is that the mean number of credit cards that students report carrying is less than the reported average of 4.09, i.e., $H_A:\mu <4.09$.

Calculate Standard Error

Next, calculate the standard error of the sample mean. The formula for standard error is $SE=\frac{s}{\sqrt{n}}$, where $s$ is the sample standard deviation and $n$ is the sample size. Substitute the given numbers: $SE=\frac{1.2}{\sqrt{132}}$.

Calculate Z-score

Now calculate the z-score, which shows how far and in what direction the sample mean deviates from the population mean. Use the formula $z=\frac{(\overline{x}-\mu )}{SE}$, where $\mu$ is the population mean, $\overline{x}$ is the sample mean, and $SE$ is the standard error. Substitute in the given values: $z=\frac{(2.6-4.09)}{SE}$.

Make a Decision

Finally, compare the calculated z-score with the critical z-value for the given significance level (typically 0.05). If the calculated z-score is less than the critical value, reject the null hypothesis. Alternatively, compute the p-value associated with the observed z-score. If the p-value is less than the chosen significance level, reject the null hypothesis.

Key concepts

Null Hypothesis

The null hypothesis, often denoted as $H_0$, is a foundational concept in hypothesis testing. It represents a statement of no effect or no difference and serves as the default assumption that there is nothing unusual happening. In the exercise, the null hypothesis posits that the mean number of credit cards carried by students is equal to the credit bureau's reported average, specifically $H_0:\mu =4.09$. Essentially, it's saying, "Let's assume that nothing has changed from the known average."
The null hypothesis is crucial because it sets the stage for the testing process. By assuming $H_0$ to be true until evidence suggests otherwise, hypothesis testing maintains an objective standard. For students, it's important to remember: the null hypothesis provides a baseline and is always what you test against. You only reject it if strong evidence emerges that an observed effect or difference truly exists.

Alternative Hypothesis

In any hypothesis test, the alternative hypothesis, symbolized as $H_A$, challenges the null hypothesis. It suggests that there is an effect or a difference. For our exercise, the alternative hypothesis is that the mean number of credit cards that students carry is less than the reported average of $4.09$. Mathematically, this is represented as $H_A:\mu <4.09$.
Understanding the alternative hypothesis is critical because it represents what you want to test or prove. It is the claim you are interested in establishing over the null hypothesis. When the test results provide enough statistical evidence, the alternative hypothesis is accepted. In our scenario, if the evidence supports it, it means students on average carry fewer credit cards than reported by the credit bureau.

Standard Error

The standard error (SE) measures the accuracy with which a sample mean approximates the population mean. It's essentially a measure of variability and is crucial for determining how spread out the sample data points are around the mean.
The formula for standard error is $SE=\frac{s}{\sqrt{n}}$, where $s$ is the sample standard deviation and $n$ is the sample size. In our exercise, to calculate the standard error, we use the sample standard deviation of 1.2 and a sample size of 132. Thus, the calculation would be $SE=\frac{1.2}{\sqrt{132}}$.
Standard error is important in hypothesis testing because it forms the basis for determining the margin of error. It helps us understand how much the sample mean from a study deviates from the actual population mean, giving us an indication of the reliability of the sample as a representation of the greater population.

Z-score

A z-score quantifies the number of standard deviations a data point (or sample mean) is from the population mean. It is a crucial step when performing hypothesis testing, as it helps us understand the position of the sample mean relative to the hypothesized population mean. In our context, the z-score formula used is $z=\frac{(\overline{x}-\mu )}{SE}$. Here, $\overline{x}$ represents the sample mean, $\mu$ the population mean, and $SE$ the standard error.
Using this formula, we calculate the z-score to be $z=\frac{(2.6-4.09)}{SE}$. The z-score helps determine whether the observed sample mean is a likely outcome, assuming the null hypothesis is true. If it falls beyond a certain threshold in standard deviation units (the critical value), it suggests that the sample mean is unlikely under the null hypothesis, offering grounds to reject it. Understanding the z-score is key for students as it objectively indicates whether evidence supports maintaining or rejecting the null hypothesis.