Question

The international polling organization Ipsos reported data from a survey of 2,000 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 the data provide convincing evidence that the mean minimum purchase amount for which Canadians consider the use of a debit card to be acceptable is less than \(\$ 10 ?\) Carry out a hypothesis test with a significance level of 0.01 .

Short answer

The data does not provide convincing evidence that the mean minimum purchase amount for which Canadians consider the use of a debit card to be acceptable is less than $10. The null hypothesis was not rejected.

Step-by-step solution

Stating the Hypotheses

The null hypothesis \(H_0\) is that the population mean \(\mu\) is equal to $10. The alternative hypothesis \(H_1\) is that the population mean \(\mu\) is less than $10. In mathematical terms: \(H_0 : \mu = 10\) and \(H_1 : \mu < 10\)

Calculating the Test Statistic

The test statistic for a hypothesis test for a population mean when the population standard deviation is unknown is the t-statistic, which is calculated as follows: \(t = (X - \mu) / (s/ \sqrt{n})\), where \(X\) is the sample mean, \(\mu\) is the assumed population mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size. In this case, \(X = 9.15\), \(\mu = 10\), \(s = 7.60\), and \(n = 2000\). Plugging these values into the formula gives a t-statistic of approximately -1.82.

Finding the Critical Value

The critical t-value that corresponds to a significance level of 0.01 with 1999 degrees of freedom (since \(n - 1 = 1999\)) is approximately -2.33. You would normally find this value in a t-distribution table or using a calculator with a t-distribution function.

Making a decision

Since the calculated t-value is greater than the critical t-value, you don't reject the null hypothesis. This means that there is not a significant evidence to support the claim that the mean minimum purchase amount for which Canadians consider the use of a debit card to be acceptable is less than $10.

Key concepts

Null and Alternative Hypothesis

Understanding hypothesis testing in statistics begins with the concept of the null and alternative hypotheses. These are the starting assumptions used in any statistical test to determine the probability of observing the data if some presupposed condition is true.

The null hypothesis, denoted as \(H_0\), is typically a statement of no effect or no difference. It suggests that any variation observed within the data is due to mere chance or randomness. In our given exercise, \(H_0: \mu = 10\) posits that the true mean minimum purchase amount where Canadians would use a debit card is \$10.

The alternative hypothesis, represented as \(H_1\) or \(H_a\), is a statement that indicates the presence of an effect or difference that the researcher is testing for. It is considered the opposite of what is stated in the null hypothesis. In this survey example, the alternative hypothesis is \(H_1: \mu < 10\), suggesting that the true mean minimum amount for purchases where a debit card is used is less than \$10.

Hypothesis testing revolves around assessing these competing claims based on the sample data obtained.

T-Statistic Calculation

The t-statistic is a central tool in hypothesis testing, especially when dealing with small sample sizes or unknown population standard deviations. The calculation of the t-statistic aims to convert the sample data into a standardized form that allows comparing the observed results with the null hypothesis.

The formula for the t-statistic is \(t = (X - \mu) / (s/ \sqrt{n})\) where \(X\) is the sample mean, \(\mu\) the hypothesized population mean under the null hypothesis, \(s\) the sample standard deviation, and \(n\) the sample size.

In the context of our exercise, substituting the values we get \(t = (9.15 - 10) / (7.60/ \sqrt{2000})\), which simplifies to approximately -1.82. This computed t-value helps determine how far the observed sample mean is from the hypothesized mean, in units of the estimated standard error.

Significance Level

The significance level, usually denoted by \(\alpha\), is a threshold that determines the degree of certainty required to reject the null hypothesis. It represents the probability of making a type I error, which is rejecting the null hypothesis when it is, in fact, true.

A commonly used significance level is 0.05, but rigorous studies or those with a high cost of a type I error may use lower levels such as 0.01 or 0.001.

In our debit card usage example, a significance level of 0.01 (1%) is chosen. This implies that the researchers are willing to accept a 1% chance of incorrectly concluding that the true mean is less than \$10 when it is not. Using a smaller significance level makes it more difficult to reject the null hypothesis, thereby requiring stronger evidence in the data to support the alternative hypothesis.

Critical Value

The critical value in hypothesis testing is the benchmark against which the test statistic is compared. It determines the cutoff point or the boundary between the region where the null hypothesis is rejected and where it is not.

For a given significance level in a t-test, the critical value is found on the t-distribution table based on the degrees of freedom, which is one less than the sample size (\(n - 1\)).

In the survey scenario provided, with 2000 participants, the degrees of freedom is 1999. Upon consulting the t-distribution table or a statistical software at the 0.01 level of significance, we locate the critical t-value of approximately -2.33. Since the calculated t-statistic of -1.82 is not less than the critical value, we do not reject the null hypothesis, indicating insufficient evidence to support the claim that the mean is less than \$10.