91Ó°ÊÓ

According to the credit rating firm Equifax, credit limits on newly issued credit cards decreased between 2008 and the period of January to April 2009 (USA TODAY, July 7, 2009). Suppose that random samples of 200 credit cards issued in 2008 and 200 credit cards issued during the first 4 months of 2009 had average credit limits of \(\$ 4710\) and \(\$ 4602\), respectively, which are comparable to the values given in the article. Although no information about standard deviations was provided, suppose that the sample standard deviations for the 2008 and 2009 samples were \(\$ 485\) and \(\$ 447\), respectively, and that the assumption that the population standard deviations are equal for the two time periods is reasonable. a. Construct a \(95 \%\) confidence interval for the difference in the mean credit limits for all new credit cards issued in 2008 and during the first 4 months of 2009 . b. Using the \(2.5 \%\) significance level, can you conclude that the average credit limit for all new credit cards issued in 2008 was higher than the corresponding average for the first 4 months of 2009 ?

Step by step solution

01

Calculate Sample Mean Difference

First, calculate the difference between the two sample means: \( \mu = \$4710 - \$4602 = \$108 \)

02

Calculate Pooled Standard Deviation

Next, calculate the pooled standard deviation. This is done by squaring each sample's standard deviation, dividing each by its sample size, summing these values, and taking the square root: \( s = \sqrt{ (485^2/200) + (447^2/200) } \)

03

Construct Confidence Interval

To construct a 95% confidence interval for the difference in mean credit limits, use the formula \([ \mu - (t_{\alpha/2}*s), \mu + (t_{\alpha/2}*s)] \), where \( t_{\alpha/2}\) is the t-value for a 95% confidence interval with degrees of freedom equal to \( n-1 \) for our two-sample case where n (sample size) is 200. Here, \(\alpha = 1 - 0.95 = 0.05\). Use t-table to find the t-value corresponding to 199 degrees of freedom and \(\alpha/2 = 0.025\)

04

Hypothesis Test

Perform a hypothesis test at the 2.5% significance level to decide whether the mean credit limit in 2008 was higher than in 2009. The null hypothesis is that the means are equal. The alternative hypothesis is that the 2008 mean is larger. Use the t-statistic formula \( t = \mu / (s/ \sqrt{n}) \). Compare calculated t-value with the critical t-value from the t-table for \(\alpha = 0.025\). Reject null hypothesis if calculated t-value is greater than critical t-value.

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Key Concepts

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

Hypothesis Testing

Hypothesis testing is a statistical method used to decide if there is enough evidence to reject a null hypothesis in favor of an alternative hypothesis. It's like a process for making decisions based on data. In our exercise, the null hypothesis (often denoted as \(H_0\)) claims that the mean credit limits for cards in 2008 and the first part of 2009 are equal.

The alternative hypothesis (\(H_a\)) suggests that the mean credit limit for cards issued in 2008 is greater than those in 2009. Using a significance level of 2.5%, we measure the probability of observing data as extreme as what we actually obtained, assuming the null hypothesis is true. This probability is compared against our significance level, to decide whether to accept or reject the null hypothesis.

• If the computed t-value is greater than the critical t-value from the t-table, the null hypothesis can be rejected, indicating sufficient evidence to support the alternative hypothesis.

Sample Standard Deviation

The sample standard deviation is a measure that quantifies the amount of variation or dispersion in a set of sample data. In simpler terms, it tells us how much the values in our sample fluctuate around the mean. In our scenario, the credit limits of the sampled cards from each year (2008 and 2009) have associated sample standard deviations of \(\\(485\) and \(\\)447\) respectively.

Calculating sample standard deviation is essential because it provides insight into how spread out our sample data is. This, in turn, affects the confidence interval and the hypothesis test results. A larger sample standard deviation indicates a wider dispersion of data points and potentially a less precise estimate of the true population mean.

• Sample standard deviation is calculated using the formula \( s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \), where \(x_i\) is each individual sample point, \(\bar{x}\) is the sample mean, and \(n\) is the sample size.

Pooled Standard Deviation

The pooled standard deviation is used when you assume that two populations have the same variance but differ in means. It provides a single measure of spread for two datasets that can be used in calculating t-statistics or constructing confidence intervals.

In our case, the pooled standard deviation considers the variability of credit limits for both 2008 and 2009 samples. It's calculated by combining the standard deviations from both samples into one, weighted by the sizes of each sample.

The formula for pooled standard deviation is given by \[ s_{pooled} = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2 }{n_1+n_2-2}} \] This ensures we account for the overall variability of both samples rather than looking at them separately. Using a pooled standard deviation is particularly useful in hypothesis testing and confidence interval estimates when sample sizes are small.

T-Distribution

The t-distribution is a probability distribution that is symmetric and bell-shaped, much like the normal distribution but with heavier tails. It is especially useful when dealing with small sample sizes, which is why it is often used in conjunction with hypothesis tests and confidence intervals.

In this exercise, the t-distribution helps account for variability in our estimate of the population standard deviation, particularly because we lack sufficient data on the population itself.

• The shape of the t-distribution is determined by degrees of freedom, which in two-sample tests is typically \( n_1 + n_2 - 2 \).
• A t-table can be used to find critical values needed for determining significance in hypothesis testing based on our particular level of confidence and sample size.

By referencing the t-distribution, we can calculate critical t-values that guide us in rejecting or failing to reject a null hypothesis, as well as in constructing meaningful confidence intervals over our sample data.