91Ó°ÊÓ

Using data from the U.S. Census Bureau and other sources, www.nerdwallet.com estimated that considering only the households with credit card debts, the average credit card debt for U.S. households was \(\$ 15,523\) in 2014 and \(\$ 15,242\) in 2013 . Suppose that these estimates were based on random samples of 600 households with credit card debts in 2014 and 700 households with credit card debts in 2013 . Suppose that the sample standard deviations for these two samples were \(\$ 3870\) and \(\$ 3764\), respectively. Assume that the standard deviations for the two populations are unknown and unequal. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the average credit card debts for all such households for the years 2014 and 2013 , respectively. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(98 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Using a \(1 \%\) significance level, can you conclude that the average credit card debt for such households was higher in 2014 than in 2013 ? Use both the \(p\) -value and the critical-value approaches to make this test.

Step by step solution

01

Calculate Point Estimate

The point estimate of \(\mu_{1} - \mu_{2}\) is simply the difference between the sample means, which are given. So, we have \(\mu_{1} - \mu_{2} = $15,523 - $15,242 = $281\)

02

Get Degree of Freedom and t-Critical

For the 98% confidence interval, the significance level is \(1 - 0.98 = 0.02\). Considering it's a two-tailed test, the significance level becomes \(0.02/2 = 0.01\). With degree of freedom \(n_1 + n_2 - 2 = 600 + 700 -2 = 1298\), looking at the t-distribution table, the t-critical value is approximately 2.576.

03

Calculate Standard Error

Next, calculate the standard error with the formula \(\sqrt{(s_{1}^{2}/N_{1}) + (s_{2}^{2}/N_{2})}\) which equals to \(\sqrt{((3870)^{2}/600) + ((3764)^{2}/700)}\) and we get SE ≈ 202.79

04

Construct a 98% Confidence Interval for \(\mu_{1} - \mu_{2}\)

The 98% confidence interval can be calculated using the formula \((\mu_{1} - \mu_{2}) \pm (t_{critical} * SE)\). Thus, the confidence interval is \(281 \pm (2.576 * 202.79)\), which equals to \((-238.10, 800.10)\)

05

Use p-value and Critical-value Approaches

For the p-value approach, we calculate the t-score with the formula \((\mu_{1} - \mu_{2}) / SE = 281 / 202.79\) which gives us the t-score ≈ 1.385. The p-value for this t-score with 1298 degrees of freedom is 0.166 which is greater than the significance level 0.01. For the critical-value approach, the calculated t-score (1.385) is less than the critical t-value (2.576). So, we can't reject the null hypothesis that the debt in 2014 is the same as in 2013 in both the p-value and critical-value approaches.

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.

Confidence Interval

A confidence interval gives us a range of values in which we expect the true value of a population parameter to fall. In this exercise, we're looking at the difference between the average credit card debts in 2014 and 2013. A 98% confidence interval suggests that we are 98% sure the true difference in average debts falls within this range.
To calculate this, we use:

• The point estimate, which is the observed difference between the sample means (\[\bar{x_1} - \bar{x_2} = \\(15,523 - \\)15,242 = \$281\]).
• The standard error of this difference.
• The t-critical value for 98%, which adjusts for the level of confidence and sample size.

By applying these, our confidence interval is calculated as \[281 \pm (2.576 \times 202.79)\]which results in a range of approximately (-238.10, 800.10). This means we are 98% confident that the true difference in mean credit card debt between the two years falls within this range.

Hypothesis Testing

Hypothesis testing helps us determine if there is enough evidence to support a claim about a population. In our case, we test if the average credit card debt in 2014 was higher than in 2013. We set up two hypotheses:

• Null hypothesis (\(H_0\)): No difference in average debt (\(\mu_1 - \mu_2 = 0\)).
• Alternative hypothesis (\(H_a\)): Debt in 2014 is higher (\(\mu_1 > \mu_2\)).

With hypothesis testing, we use a significance level (\(\alpha = 0.01\) here for 1%) to decide if we should reject the null hypothesis. A significance level represents the probability of rejecting the null hypothesis when it is actually true, and lower values indicate a stricter testing criterion. In this exercise, we use both a p-value and a critical-value approach to complete the test.

p-value

The p-value measures the strength of evidence against the null hypothesis. It tells us the probability of observing a test statistic as extreme as, or more extreme than, the observed one if the null hypothesis were true. In this exercise, our calculated t-score is around 1.385.
Depending on this t-score and the degree of freedom, we found the p-value to be approximately 0.166. Since this p-value is greater than our significance level of 0.01, we do not have sufficient evidence to reject the null hypothesis. Simply put, our data do not provide enough evidence to conclude that the average credit card debt was significantly higher in 2014 than in 2013.

Standard Error

Standard error (SE) is an essential component in statistical analyses, quantifying the variability in the sample estimate of a parameter. It gives us a measure of how much our sample mean is expected to vary from the true population mean. In our example, the standard error is crucial in calculating the confidence interval and the t-score.We compute it using the formula:\[SE = \sqrt{\left(\frac{s_1^2}{N_1}\right) + \left(\frac{s_2^2}{N_2}\right)}\]Substituting the given values, we find:\[SE \approx 202.79\]This value describes the amount of variation we might expect in our estimate of the difference between the average debts over the years 2014 and 2013. A smaller standard error would indicate more reliable estimates in practice.