/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 A credit bureau analysis of unde... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Expert verified
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

01

- 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. \(H_1: \mu < 4.09\)
02

- 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 \(Z = \frac{(2.6-4.09)}{(1.2/\sqrt{132})}\)
03

- 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).
04

- 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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Null Hypothesis
When performing hypothesis testing, the null hypothesis is our starting point. You can think of it as a statement that there is no effect or no difference in the specific situation. In this exercise, the null hypothesis, denoted as \(H_0\), asserts that the mean number of credit cards that undergraduates typically carry is equal to 4.09. It's a claim of no change or no deviation from the established mean.
When testing hypotheses, we assume that the null hypothesis is true until we have enough evidence to prove otherwise. The burden is on us to show evidence against \(H_0\). This process requires careful calculation and comparison with statistical benchmarks. The null hypothesis is often written in the format:
  • \(H_0: \mu = 4.09\)
This indicates that under the null hypothesis, the population mean is thought to be 4.09.
Alternative Hypothesis
The alternative hypothesis, on the other hand, represents what we want to test or the outcome that indicates change. It's important because it contrasts with the null hypothesis. In this exercise, our alternative hypothesis is that the mean number of credit cards undergraduates carry is less than 4.09. We denote this hypothesis as \(H_1\).
While the null hypothesis assumes no change, \(H_1\) suggests a decline in the mean number of credit cards among students. Formally, the alternative hypothesis is expressed as:
  • \(H_1: \mu < 4.09\)
The goal of hypothesis testing is to determine if there's enough statistical evidence to support the alternative over the null hypothesis. Essentially, we try to find reasons compelling enough to convince us that \(H_1\) could be more plausible than \(H_0\). This process involves statistical calculations and significance testing.
Z-Statistic
The Z-statistic is a critical part of hypothesis testing, especially when dealing with sample means. It helps us understand how far the sample mean is from the population mean — measured in standard deviation units. The Z-statistic provides a way to decide if the evidence is strong enough to reject the null hypothesis.
The formula to compute the Z-statistic is:
  • \(Z = \frac{(\overline{x} - \mu)}{(\sigma/\sqrt{n})}\)
Where:
  • \(\overline{x}\) is the sample mean
  • \(\mu\) is the population mean under \(H_0\)
  • \(\sigma\) is the sample standard deviation
  • \(n\) is the sample size

In our case, substituting the given values like sample mean 2.6, population mean 4.09, sample standard deviation 1.2, and sample size 132, enables us to calculate the Z-statistic and see if it indicates a significant difference.
Confidence Level
The confidence level in hypothesis testing reflects the degree of certainty we have in our results. It helps us understand the margin of error we're willing to accept when making a decision about the null hypothesis. The confidence level is usually expressed as a percentage, such as 95%.
A 95% confidence level means we are 95% confident our decision to reject or not reject the null hypothesis is correct. It leaves a 5% risk (known as alpha, \(\alpha\)) that we might be wrong. In this exercise, we use a one-sided test with \(\alpha = 0.05\) which relates directly to our critical Z value of -1.645.
The concept of confidence level ties directly into the decision-making process:
  • If the computed Z-statistic is beyond the critical Z value derived from our confidence level, we can reject the null hypothesis.
  • If not, the evidence isn't strong enough to support \(H_1\).
Confidence levels provide a statistical benchmark for testing hypotheses, ensuring decisions have an acceptable level of risk.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The report "2005 Electronic Monitoring \& Surveillance Survey: Many Companies Monitoring, Recording, Videotaping-and Firing-Employees" (American Management Association, 2005) summarized the results of a survey of 526 U.S. businesses. Four hundred of these companies indicated that they monitor employees' web site visits. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of businesses in the United States. a. Is there sufficient evidence to conclude that more than \(75 \%\) of U.S. businesses monitor employees' web site visits? Test the appropriate hypotheses using a significance level of \(.01\). b. Is there sufficient evidence to conclude that a majority of U.S. businesses monitor employees' web site visits? Test the appropriate hypotheses using a significance level of \(.01\).

A manufacturer of hand-held calculators receives large shipments of printed circuits from a supplier. It is too costly and time-consuming to inspect all incoming circuits, so when each shipment arrives, a sample is selected for inspection. Information from the sample is then used to test \(H_{0}=\pi=.05\) versus \(H_{a}: \pi>.05\), where \(\pi\) is the true proportion of defective circuits in the shipment. If the null hypothesis is not rejected, the shipment is accepted, and the circuits are used in the production of calculators. If the null hypothesis is rejected, the entire shipment is returned to the supplier because of inferior quality. (A shipment is defined to be of inferior quality if it contains more than \(5 \%\) defective circuits.) a. In this context, define Type I and Type II errors. b. From the calculator manufacturer's point of view, which type of error is considered more serious? c. From the printed circuit supplier's point of view, which type of error is considered more serious?

Use the definition of the \(P\) -value to explain the following: a. Why \(H_{0}\) would certainly be rejected if \(P\) -value \(=.0003\) b. Why \(H_{0}\) would definitely not be rejected if \(P\) -value \(=\) \(.350\)

Typically, only very brave students are willing to speak out in a college classroom. Student participation may be especially difficult if the individual is from a different culture or country. The article "An Assessment of Class Participation by International Graduate Students" \((\) Journal of College Student Development \([1995]: 132-\) 140) considered a numerical "speaking-up" scale, with possible values from 3 to 15 (a low value means that a student rarely speaks). For a random sample of 64 males from Asian countries where English is not the official language, the sample mean and sample standard deviation were \(8.75\) and \(2.57\), respectively. Suppose that the mean for the population of all males having English as their native language is \(10.0\) (suggested by data in the article). Does it appear that the population mean for males from non-English-speaking Asian countries is smaller than \(10.0 ?\)

Many older homes have electrical systems that use fuses rather than circuit breakers. A manufacturer of 40 -amp fuses wants to make sure that the mean amperage at which its fuses burn out is in fact 40 . If the mean amperage is lower than 40 , customers will complain because the fuses require replacement too often. If the mean amperage is higher than 40 , the manufacturer might be liable for damage to an electrical system as a result of fuse malfunction. To verify the mean amperage of the fuses, a sample of fuses is selected and tested. If a hypothesis test is performed using the resulting data, what null and alternative hypotheses would be of interest to the manufacturer?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.