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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

Expert verified
The short answer should be determined by completing the calculations in step 3 and comparing the resulting P-value to the chosen significance level in step 5. Without these exact calculations, a short answer cannot be specified.

Step by step solution

01

State the hypotheses

The null hypothesis is that the mean number of credit cards undergraduates reporting carrying is equal to the credit bureau's figure, \(4.09\), or \(H_0: \mu = 4.09\). The alternative hypothesis is that the mean number is less than \(4.09\), or \(H_1: \mu < 4.09\).
02

Calculate the sample mean and standard deviation

We know that in a sample of \(132\) undergraduates, the sample mean number of credit cards the students reported carrying was \(2.6\) and the sample standard deviation was \(1.2\). This information will be used in subsequent calculations.
03

Calculation of Test Statistic

The next step is to calculate the z-test statistic using the formula \[Z = \frac{{\bar{X} - \mu}}{{s / \sqrt{n}}}\] where \(\bar{X}\) is the sample mean, \(\mu\) is the population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. Substituting the given values into the formula gives: \[Z = \frac{{2.6 - 4.09}}{{1.2 / \sqrt{132}}}\] After performing the necessary calculations, the Z value is then determined.
04

Find the P-value

Next, the student will look up the P-value associated with the computed Z value in a standard normal (Z) table. The P-value is the probability that you have falsely rejected the null hypothesis.
05

Make a decision

If the P-value is less than the chosen significance level (typically \(0.05\), but it may vary depending on the problem), we reject the null hypothesis in favor of the alternative hypothesis. If the P-value is larger, we do not reject the null hypothesis. It means that there is not strong evidence that undergraduates on an average carry less than \(4.09\) credit cards.
06

Interpretation

The result of hypothesis test is to be interpreted in the context of the problem. If we chose a significance level of \(0.05\) and our P-value is less than this, we would conclude that we have strong evidence that the mean number of credit cards undergraduates carry is less than \(4.09\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null and Alternative Hypothesis
When we talk about hypothesis testing in statistics, the first step is always laying out the hypotheses. The null hypothesis (\(H_0\)) represents the status quo or a statement of no effect, which we presume true until evidence suggests otherwise. In the context of our credit card example, the null hypothesis posits that the average number of credit cards carried by undergraduates is equal to the reported figure of 4.09. Formally, this is stated as \(H_0: \mu = 4.09\).

Conversely, the alternative hypothesis (\(H_1\)) represents what the researcher is trying to establish, which is a statement of an effect or difference. In this example, the alternative hypothesis asserts that the average number of credit cards undergraduates carry is less than 4.09, formally stated as \(H_1: \mu < 4.09\). This sets the stage for the investigation: we are essentially trying to find strong evidence against the null hypothesis and in favor of the alternative hypothesis.
Test Statistic Calculation
Once the hypotheses are specified, we proceed with the calculation of the test statistic. The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It's crucial because it allows us to determine how far the sample mean deviates from the null hypothesis' claim.

In this scenario, the test statistic is determined using the formula \[Z = \frac{{\bar{X} - \mu}}{{s / \sqrt{n}}}\], where \(\bar{X}\) is the sample mean, \(\mu\) is the population mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size. Substituting the provided values into this formula, we arrive at a Z value that quantifies the difference in terms of the standard error of the mean. This Z value will then guide us in finding out how likely or unlikely our sample result is under the assumption that the null hypothesis is true.
P-value Interpretation
The P-value plays a pivotal role in hypothesis testing. It provides a way of quantifying the strength of the evidence against the null hypothesis. The P-value is the probability of observing a test statistic as extreme as the one calculated, or more so, assuming the null hypothesis is true. It is not the probability that the null hypothesis is true.

If the P-value is small—typically below a threshold known as the significance level (\(\alpha\)), which is often set at 0.05—it suggests that our sample result is unusual under the assumption of the null hypothesis. In such a case, we reject the null hypothesis, suggesting that there is evidence to support the alternative hypothesis. If the P-value is above the threshold, we do not reject the null hypothesis as there's insufficient evidence to support the alternative claim. The crucial part here is interpreting the P-value correctly: it is not about the probability of hypotheses being true or false, but rather about how consistent the data is with each hypothesis.
Sample Mean and Standard Deviation
The sample mean (\(\bar{X}\)) is simply the average of all the measurements in a sample. In our example, the sample mean represents the average number of credit cards reported by the sample of undergraduates. It is a central estimate of a population parameter — the population mean (\(\mu\)).

The sample standard deviation (\(s\)) measures the variability or dispersion of the sample. A larger standard deviation indicates that the data points are spread out over a wider range of values. In hypothesis testing, we use the sample standard deviation to estimate the standard error of the mean (\(s / \sqrt{n}\)), which in turn is used in the calculation of the test statistic. It is essential to note that while these sample statistics provide important insights, they are estimates derived from sample data and are subject to sampling variability. Therefore, it's necessary to be cautious when inferring about the population based on these statistics.

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Most popular questions from this chapter

The paper titled "Music for Pain Relief" (The Cochrane Database of Systematic Reviews, April \(19 .\) 2006) concluded, based on a review of 51 studies of the effect of music on pain intensity, that "Listening to music reduces pain intensity levels .... However, the magnitude of these positive effects is small, the clinical relevance of music for pain relief in clinical practice is unclear." Are the authors of this paper claiming that the pain reduction attributable to listening to music is not statistically significant, not practically significant, or neither statistically nor practically significant? Explain.

The mean length of long-distance telephone calls placed with a particular phone company was known to be \(7.3\) minutes under an old rate structure. In an attempt to be more competitive with other long-distance carriers, the phone company lowered long-distance rates, thinking that its customers would be encouraged to make longer calls and thus that there would not be a big loss in revenue. Let \(\mu\) denote the mean length of long-distance calls after the rate reduction. What hypotheses should the phone company test to determine whether the mean length of long-distance calls increased with the lower rates?

The international polling organization Ipsos reported data from a survey of 2000 randomly selected Canadians who carry debit cards (Canadian Account Habits Survey, July 24, 2006). Participants in this survey were asked what they considered the minimum purchase amount for which it would be acceptable to use a debit card. Suppose that the sample mean and standard deviation were \(\$ 9.15\) and \(\$ 7.60\), respectively. (These values are consistent with a histogram of the sample data that appears in the report.) Do these data provide convincing evidence that the mean minimum purchase amount for which Canadians consider the use of a debit card to be appropriate is less than \(\$ 10\) ? Carry out a hypothesis test with a significance level of \(.01\).

For the following pairs, indicate which do not comply with the rules for setting up hypotheses, and explain why: a. \(H_{0}: \mu=15, H_{a}: \mu=15\) b. \(H_{0}: p=.4, H_{a}: p>.6\) c. \(H_{0}: \mu=123, H_{a}: \mu<123\) d. \(H_{0}: \mu=123, H_{a}: \mu=125\) e. \(H_{0}: \hat{p}=.1, H_{a}: \hat{p} \neq .1\)

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