According to a survey conducted in October 2001, consumers were trying to reduce their credit card debt (Extracted from M. Price, “Credit Debts Get Cut Down to Size." Newsday, November 25, 2001, p. F3). Based on a sample of 1000 consumers in October 2001 and in October 2000. the mean credit card debt was $2411 in October 2001as compared to $2814 in October 2000. Suppose that the sample standard deviation was $847.43 in October 2001 and $976.93 in October 2000. Assuming the population variances from both years are equal, is there evidence that the mean credit card debt was lower in October 2001 than in October 2000? (Use the 0.05 level of significance)

Question

According to a survey conducted in October 2001, consumers were trying to reduce their credit card debt (Extracted from M. Price, “Credit Debts Get Cut Down to Size." Newsday, November 25, 2001, p. F3). Based on a sample of 1000 consumers in October 2001 and in October 2000. the mean credit card debt was $2411 in October 2001as compared to $2814 in October 2000. Suppose that the sample standard deviation was $847.43 in October 2001 and $976.93 in October 2000. Assuming the population variances from both years are equal, is there evidence that the mean credit card debt was lower in October 2001 than in October 2000? (Use the 0.05 level of significance)

Answer 1

Hypotheses:

Ho: µ2001 = µ2000
Ha: µ2001 < µ2000

You will need to use a 2-sample formula for this type of test. You have all the data needed for both samples. This is a one-tailed test because the alternative hypothesis is showing a specific direction. If the null is rejected in favor of the alternative hypothesis, then there is a difference (in other words, µ2001 < µ2000).

Answer 2

To determine if there is evidence that the mean credit card debt was lower in October 2001 compared to October 2000, we can conduct a hypothesis test.

Let's define the null and alternative hypotheses as follows:

Null hypothesis (H0): The mean credit card debt in October 2001 is greater than or equal to the mean credit card debt in October 2000.
Alternative hypothesis (H1): The mean credit card debt in October 2001 is lower than the mean credit card debt in October 2000.

We can conduct a two-sample t-test to compare the means of the two samples. The test statistic is calculated as:

t = (x1 - x2) / sqrt((s1^2 / n1) + (s2^2 / n2))

Where:
x1 = mean credit card debt in October 2001
x2 = mean credit card debt in October 2000
s1 = standard deviation of credit card debt in October 2001
s2 = standard deviation of credit card debt in October 2000
n1 = sample size in October 2001
n2 = sample size in October 2000

Based on the information provided, we have the following values:

x1 = $2411
x2 = $2814
s1 = $847.43
s2 = $976.93
n1 = n2 = 1000

Using these values, we can calculate the t-value.

t = (2411 - 2814) / sqrt((847.43^2 / 1000) + (976.93^2 / 1000))

After performing the calculations, the t-value is obtained. We can then compare the t-value with the critical t-value from the t-distribution table with (n1 + n2 - 2) degrees of freedom at a 0.05 significance level.

If the calculated t-value is less than the critical t-value, we can reject the null hypothesis and conclude that there is evidence that the mean credit card debt was lower in October 2001 than in October 2000. Otherwise, if the calculated t-value is greater than the critical t-value, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest a significant difference in the mean credit card debt between the two years.

Please note that in this explanation, the assumption of equal population variances (based on the information provided) was made. If the population variances are assumed to be different, a modified test statistic and different degrees of freedom would be used.