91Ó°ÊÓ

Credit card companies earn a percent of the amount charged on their credit cards, paid by the stores that accept the card. A credit card company compares two proposals for increasing the amount that its customers charge on their credit cards. Proposal 1 offers to eliminate the annual fee for customers who charge \(\$ 1800\) or more during the year on their card. Proposal 2 offers a small percent of the total amount charged as a cash reward at the end of the year. The credit card company offers each proposal to an SRS of 100 of its existing customers. At the end of the year, the total amount charged by each customer is recorded. Here are the summary statistics. $$ \begin{array}{lccc} \hline \text { Group } & n & x^{-} \bar{x} & s \\ \hline \text { Proposal 1 } & 100 & \$ 1319 & \$ 261 \\ \hline \text { Proposal 2 } & 100 & \$ 1372 & \$ 274 \\ \hline \end{array} $$ (a) Do the data show a significant difference between the mean amounts charged by customers offered the two proposed plans? Give the null and alternative hypotheses, and calculate the two-sample \(t\) statistic. Obtain the \(P\)-value, using Option 2. State your practical conclusions. (b) The distributions of the amounts charged on credit cards are skewed to the right. However, outliers are prevented by the limits that the credit card companies impose on credit balances. Do you think that skewness threatens the validity of the text that you used in part (a)? Explain your answer.

Step by step solution

01

Define Hypotheses

Let's define the null and alternative hypotheses for part (a). We want to test whether there is a significant difference between the mean amounts charged by customers in Proposal 1 and Proposal 2.- **Null Hypothesis**: \( H_0 : \mu_1 = \mu_2 \) (No difference in the means)- **Alternative Hypothesis**: \( H_a : \mu_1 eq \mu_2 \) (There is a difference in the means)

02

Calculate Standard Error

To calculate the two-sample \(t\)-statistic, we first need to calculate the standard error of the difference in sample means. The formula is:\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]Where \(s_1 = 261\), \(s_2 = 274\), and both \(n_1 = n_2 = 100\).

03

Plug Values into SE Formula

Substitute the values into the standard error formula:\[ SE = \sqrt{\frac{261^2}{100} + \frac{274^2}{100}} = \sqrt{\frac{68121}{100} + \frac{75076}{100}} = \sqrt{681.21 + 750.76} = \sqrt{1431.97} \approx 37.83 \]

04

Calculate t-Statistic

Using the means and the standard error, the \(t\)-statistic is calculated as follows:\[ t = \frac{\bar{x_1} - \bar{x_2}}{SE} = \frac{1319 - 1372}{37.83} = \frac{-53}{37.83} \approx -1.40 \]

05

Determine P-Value

Since we are testing a two-tailed hypothesis, we check the t-distribution table or use software to find the \(P\)-value for \(t = -1.40\) with \(df = 198\) (approximately equal given large sample size). The \(P\)-value is larger than 0.05.

06

State Practical Conclusion

Since the \(P\)-value is greater than 0.05, we do not reject the null hypothesis. Thus, we conclude that there is not enough evidence to say there is a significant difference in the mean amounts charged by customers under the two proposals.

07

Address Skewness Validity

For part (b), skewness of the data can impact the validity of the \(t\)-test, especially in small samples. However, given the sample size of 100 in both proposals, the Central Limit Theorem suggests that the sampling distribution of the mean is approximately normal, which means skewness has a limited effect here. The presence of right-skewed distributions also suggests outliers are minimal given credit limits, reinforcing robustness of the \(t\)-test.

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Key Concepts

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

Null Hypothesis

In the context of statistical hypothesis testing, the null hypothesis (\( H_0 \)) serves as a starting point for any test or experiment. It's a formal way of saying "there is no effect" or "no difference." For our exercise about credit card proposals, the null hypothesis is:

• \( H_0: \mu_1 = \mu_2 \)

This means that, on average, customers offered Proposal 1 and Proposal 2 charge the same amount on their credit cards. The objective of hypothesis testing is to determine whether available data provides sufficient evidence to reject this assumption or not. In essence, we assume that both proposals produce the same result in terms of average amount charged, until proven otherwise. Remain unbiased and assume no change as our baseline hypothesis. Let's now explore its counterpart, the alternative hypothesis.

Alternative Hypothesis

An alternative hypothesis (\( H_a \)) challenges the null hypothesis by suggesting there is a statistical effect or difference. It represents what we aim to prove through our test. In this credit card proposal scenario, the alternative hypothesis is:

• \( H_a: \mu_1 \eq \mu_2 \)

This suggests that the mean amounts charged by customers under Proposal 1 is different from Proposal 2. The alternative hypothesis is crucial because it determines the type of statistical test used, in this case, a two-tailed t-test.If statistical analysis supports this hypothesis strongly enough by falling below a pre-determined significance level, we may reject the null hypothesis. Consider it the positive backstage actor in hypothesis testing, proposing a potentially meaningful discovery.

t-Statistic Calculation

The t-statistic helps evaluate how plausible the null hypothesis is. It indicates how far the sample data diverges from the null hypothesis. To calculate the t-statistic for our test on credit card proposals, we first find the standard error (\( SE \)) as follows:\[SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\]Given the summary data:

• \( s_1 = 261 \)
• \( s_2 = 274 \)
• \( n_1 = n_2 = 100 \)

We calculate as:\[SE \approx 37.83\]Next, using calculated standard error, determine the t-statistic:\[t = \frac{\bar{x_1} - \bar{x_2}}{SE} = \frac{1319 - 1372}{37.83} \approx -1.40\]The result, \(-1.40\), measures statistical significance. Closer to zero indicates weak evidence against the null hypothesis, while larger magnitudes suggest stronger evidence.

P-value Interpretation

The p-value shows how extreme the observed results are under the assumption that the null hypothesis is true. For the credit card proposal comparison, our calculated t-statistic was approximately \(-1.40\), and this corresponds to a p-value that we determine using statistical tables or software.In this exercise, the p-value was greater than 0.05. Thus:

• A p-value higher than 0.05 indicates there is insufficient evidence to reject the null hypothesis.
• Practically, this suggests that no significant difference exists between the average amounts charged under the two proposals.

P-values help frame our findings objectively. They show whether any observations could likely occur under the null hypothesis. Stay logical with evidenced-based conclusions, which guide informed decisions—critical in analyses like this comparison.