91Ó°ÊÓ

According to the credit rating agency Equifax, credit limits on newly issued credit cards increased between January 2011 and May 2011 (money.cnn.com/2011/08/19/pf/credit_card_issuance/index.htm). Suppose that random samples of 400 credit cards issued in January 2011 and 500 credit cards issued in May 2011 had average credit limits of \(\$ 2635\) and \(\$ 2887\), respectively. Suppose that the sample standard deviations for these two samples were \(\$ 365\) and \(\$ 412\), respectively, and the assumption that the population standard deviations are equal for the two populations is reasonable. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the average credit limits on all credit cards issued in January 2011 and in May 2011 , 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 limit for all new credit cards issued in January 2011 was lower than the corresponding average for all credit cards issued in May 2011 ? Use both the \(p\) -value and the critical-value approaches to make this test.

Step by step solution

01

Calculate Point Estimate for \(\mu_{1} - \mu_{2}\)

The point estimate of \(\mu_{1} - \mu_{2}\) is the difference between the sample means of the two samples, \(\bar{x}_{1} - \bar{x}_{2}\) which equals $2635 - $2887 = -$252.

02

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

To construct the confidence interval, we need the standard error, which is calculated as: \(SE = \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}\), where \(s_{1}\) and \(s_{2}\) are the sample standard deviations and \(n_{1}\) and \(n_{2}\) are the sample sizes of the two groups. Substituting the provided values: \(SE = \sqrt{\frac{(365)^{2}}{400} + \frac{(412)^{2}}{500}}\). The z-value for a 98% confidence interval is 2.33. Hence, the confidence interval is given by: \(CI = (\bar{x}_{1} - \bar{x}_{2}) \pm z_{\alpha/2} * SE\)

03

Perform Hypothesis Testing

Firstly, formulate the null and alternative hypothesis. \(H_{0}: \mu_{1} - \mu_{2} = 0\), the average credit limits are equal. \(H_{1}: \mu_{1} - \mu_{2} < 0\), the average credit limit of January 2011 is lower. We need to conduct a two-tailed test with 1% significance level. The testing statistic, z, is calculated as: \(z = \frac{(\bar{x}_{1} - \bar{x}_{2}) - D}{SE}\), where D is the difference in population means under the null hypothesis (0). Compute the p-value and the critical value and compare.

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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 method used to make decisions about population parameters based on sample data. Here, we want to see if the average credit limit for credit cards in January 2011 is lower than in May 2011.

We start by defining our hypotheses:

• Null Hypothesis (\(H_0\)): There is no difference in the average credit limits, or \( \mu_1 - \mu_2 = 0 \).
• Alternative Hypothesis (\(H_1\)): The average credit limit in January is lower, or \( \mu_1 - \mu_2 < 0\).

Next, we calculate the test statistic using the formula:
\( \text{z} = \frac{(\bar{x}_1 - \bar{x}_2) - D}{SE}\)\, where \(D\) is 0 under the null hypothesis. The calculated z-value will help us find the p-value, which tells us the probability of observing our data if the null hypothesis is true.

If the p-value is less than 0.01 (our significance level), or if our z-value is less than the critical z-value at 1%, we reject the null hypothesis and accept that there is a difference.

Confidence Interval

A confidence interval provides a range of values for a population parameter based on sample data. Here, we're looking for a 98% confidence interval for the difference in credit limits from January to May.The confidence interval is calculated with:
\( \text{CI} = (\bar{x}_1 - \bar{x}_2) \pm z_{\alpha/2} \times SE\)\,
where \(z_{\alpha/2}\) for a 98% confidence level is 2.33. The standard error (SE) is crucial here, informing us of the variability in our samples.

This interval gives us a range where we expect the true difference in population means to lie, 98% of the time. If the interval doesn't include 0, it suggests a significant difference.

Point Estimate

The point estimate provides a single best guess for a population parameter based on sample data. In this exercise, it's the difference in average credit limits between two periods. The point estimate is determined by the difference in sample means: \( \bar{x}_1 - \bar{x}_2 = 2635 - 2887 = -252\)\.

This means, on average, credit limits were \( \$252\) lower in January compared to May. This single value is critical as it represents the observed effect in the data.

Sample Standard Deviation

The sample standard deviation is a measure of the variability or spread in a data set. Here, it tells us how much the credit limits vary in each sample month.For January 2011, the standard deviation is \( \\(365\), and for May 2011, it's \( \\)412\). These values are used to compute the standard error:

• January: \(s_1 = 365\)
• May: \(s_2 = 412\)

A larger standard deviation indicates more variability in credit limits. By converting these into a standard error, we get an understanding of how reliable our sample means are in estimating the true population means.