Question

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

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

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.

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).

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.

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.

Key concepts

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.