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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 .

Short Answer

Expert verified
To determine whether we can provide convincing evidence that the mean minimum purchase amount for Canadians considering the use of a debit card is less than \$10, we need to carry out a t-test as outlined in the steps above. The decision will depend on the calculated t-value and p-value.

Step by step solution

01

State the Hypotheses

The first step in hypothesis testing is to clearly state the null hypothesis and the alternative hypothesis. In this case, we have:Null hypothesis (H0): µ = \$10 Alternative hypothesis (H1): µ < \$10 where µ is the population mean minimum purchase amount considered acceptable for using a debit card.
02

Calculate the Test Statistic

Given the sample size (n=2000), the sample mean (x̄=\$9.15), and the sample standard deviation (s=\$7.60), we can calculate the t statistic using the formula:t = (x̄ - µ₀) / (s / √n)where:µ₀ = \$10 (value in null hypothesis),x̄ = \$9.15 (sample mean),s = \$7.60 (sample standard deviation),n = 2000 (sample size).
03

Find the Critical Value and P-Value

With a significance level of 0.01 and being a left-tailed test, we look up the critical value for t in the t-distribution table based on n-1 = 1999 degrees of freedom. Because the degrees of freedom is quite large, this value will be approximately equal to the z-value, which is -2.33 for a one-tailed test at the 0.01 significance level. Next, we find the P-value associated with the observed value of the test statistic. The P-value is the probability that a t statistic falls in the extreme (that is, more negative) tail of the distribution if the null hypothesis is true.
04

Make the Decision

We reject the null hypothesis if our calculated t-value is less than our critical t-value or if the P-value is less than the significance level.

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Key Concepts

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

Null Hypothesis
Every hypothesis test begins with the formation of a null hypothesis, often denoted as \(H_0\). It serves as the starting assumption about a population parameter. In the context of our problem, the null hypothesis is \(H_0: \mu = \\(10\). This means that, according to the null hypothesis, the average minimum purchase amount believed to be appropriate for using a debit card by Canadians is exactly \(\\)10\).
This hypothesis reflects a state of no effect or no difference, and it is tested against the collected sample data.
The null hypothesis is applied to determine if there is sufficient statistical evidence to consider the alternative view.
It's crucial to remember that the null hypothesis can never be proven; it can only be rejected or fail to be rejected based on the data.
Alternative Hypothesis
While the null hypothesis assumes no difference or effect, the alternative hypothesis proposes what we are trying to support. It's symbolized by \(H_1\) or \(H_a\). For the exercise we are examining, the alternative hypothesis is \(H_1: \mu < \\(10\).
This suggests that the sample data implies the true mean minimum purchase amount is less than \(\\)10\).
Picking an alternative hypothesis depends heavily on the research question and determines whether the test will be one-tailed or two-tailed.
  • One-tailed tests, like ours, make predictions in one direction, i.e., "less than" or "greater than."
  • Two-tailed tests contemplate both possible directions.

Acceptance of the alternative hypothesis leads to the conclusion that a significant difference exists compared to the null hypothesis.
Significance Level
The significance level, denoted by \(\alpha\), is a critical part of hypothesis testing. It represents the probability of rejecting the null hypothesis when it is true. In our problem, the significance level is set at 0.01.
This means we're willing to accept a 1% chance of incorrectly rejecting the null hypothesis. Why choose 0.01? Because a lower \(\alpha\) level like 0.01 indicates stronger evidence against the null hypothesis is needed before it can be rejected.
Common significance levels include 0.05, 0.01, and 0.10. These values are not rigid but should be decided upon based on the context of the test.
T-Distribution
T-distribution plays a crucial role when dealing with sample sizes and unknown population variances. It resembles a normal distribution but has heavier tails, meaning more variability.
This feature makes it suitable for smaller sample sizes or when population variance is unknown.
In our example, we rely on the t-distribution because the sample standard deviation is used in place of the population standard deviation, and we have a large sample size (n = 2000).
  • The test statistic falls under the t-distribution category whenever the sample standard deviation is utilized.
  • The critical value derived from t-distribution aids in concluding the hypothesis test.

It's crucial to use the appropriate degrees of freedom, which is usually \(n-1\), when referencing t-distribution tables.
P-Value
The P-value quantifies the probability of obtaining test results at least as extreme as the observed results, assuming that the null hypothesis is true.
In simpler terms, it provides evidence against \(H_0\) by showing how likely the observed data would occur under the null hypothesis.
P-values are compared against the chosen significance level \(\alpha\) to help decide whether to reject \(H_0\).
  • If \(P < \alpha\), there is strong evidence against the null hypothesis, leading to its rejection.
  • If \(P \geq \alpha\), we won't reject the null hypothesis as the evidence is not strong enough.

In our case, a calculated P-value gives us the probability that the observed mean difference (or more extreme) could occur, assuming the null hypothesis (\(\mu = \$10\)) is true, and helps us decide the next action in hypothesis testing.

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Most popular questions from this chapter

An automobile manufacturer is considering using robots for part of its assembly process. Converting to robots is an expensive process, so it will be undertaken only if there is strong evidence that the proportion of defective installations is lower for the robots than for human assemblers. Let \(p\) denote the proportion of defective installations for the robots. It is known that human assemblers have a defect proportion of .02 . a. Which of the following pairs of hypotheses should the manufacturer test: \(H_{0}: p=.02\) versus \(H_{a}: p<.02\) or \(H_{0}: p=.02\) versus \(H_{a}: p>.02\) Explain your answer. b. In the context of this exercise, describe Type I and Type II errors. c. Would you prefer a test with \(\alpha=.01\) or \(\alpha=.1 ?\) Explain your reasoning.

Occasionally, warning flares of the type contained in most automobile emergency kits fail to ignite. A consumer advocacy group wants to investigate a claim against a manufacturer of flares brought by a person who claims that the proportion of defective flares is much higher than the value of .1 claimed by the manufacturer. A large number of flares will be tested, and the results will be used to decide between \(H_{0}: p=.1\) and \(H_{a}: p>.1,\) where \(p\) represents the proportion of defective flares made by this manufacturer. If \(H_{0}\) is rejected, charges of false advertising will be filed against the manufacturer. a. Explain why the alternative hypothesis was chosen to be \(H_{a}: p>.1 .\) b. In this context, describe Type I and Type II errors, and discuss the consequences of each.

In a survey of 1005 adult Americans, \(46 \%\) indicated that they were somewhat interested or very interested in having web access in their cars (USA Today, May I. 2009 ). Suppose that the marketing manager of a car manufacturer claims that the \(46 \%\) is based only on a sample and that \(46 \%\) is close to half, so there is no reason to believe that the proportion of all adult Americans who want car web access is less than \(.50 .\) Is the marketing manager correct in his claim? Provide statistical evidence to support your answer. For purposes of this exercise, assume that the sample can be considered as representative of adult Americans.

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.

Explain why the statement \(\bar{x}=50\) is not a legitimate hypothesis.

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