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A confidence interval gives you a range of values that is likely to contain the true population parameter. In other words, it tells us how much uncertainty there is around the estimated difference between two population means, like the annual credit card charges for groceries and dining out in this exercise.
To construct a confidence interval for the mean difference, we use the sample mean difference \( \bar{d} = 850 \), the sample standard deviation \( s_d = 1123 \), and the sample size \( n = 42 \). With this information, we can calculate the 95% confidence interval.
The formula for the confidence interval is:\[ \bar{d} \pm t^* \frac{s_d}{\sqrt{n}} \]Here, \( t^* \) is the critical value from the t-distribution corresponding to the desired level of confidence (95% in this case). For 41 degrees of freedom, \( t^* \approx 2.02 \).
Plugging the values into the formula, the confidence interval becomes approximately \((501.51, 1198.49)\). This means we are 95% confident that the true difference in mean credit card charges between groceries and dining out lies within this interval.
This confidence interval helps us get a sense of how reliably the sample mean difference of \( 850 \) reflects the true difference in the population.

The T-Test helps in determining if there are significant differences between the means of two groups — in this case, the credit card charges for groceries versus dining out.
To perform this test, we start by formulating the null hypothesis \( H_0: \mu_d = 0 \) which assumes no difference between the population means, and the alternative hypothesis \( H_a: \mu_d eq 0 \), stating there is a difference.
The test statistic is calculated using the formula:\[ t = \frac{\bar{d} - \mu_d}{s_d / \sqrt{n}} \]In our exercise, \( \mu_d = 0 \), \( \bar{d} = 850 \), \( s_d = 1123 \), and \( n = 42 \), which results in \( t \approx 4.88 \). This t-value tells us how far our sample mean is from the hypothesized population mean (under \( H_0 \)).
The higher the absolute t-value, the stronger the evidence against the null hypothesis. The calculated t-value of 4.88 indicates a significant difference given the conventional cutoff values for significance.

The p-value is a key value in hypothesis testing, helping us determine the significance of our results. It quantifies the probability of observing results at least as extreme as the actual results, assuming the null hypothesis is true.
In this exercise, after calculating the t-statistic, we obtain a p-value \( \approx 0.00002 \). This is a very small value, much less than the significance level \( \alpha = 0.05 \) set in the test.
A small p-value implies that the observed data is unlikely under the assumption that the null hypothesis is true.
Consequently, because the p-value is so small, we reject the null hypothesis \( H_0 \) and conclude that there is a significant difference between the mean annual credit card charges for groceries as compared to dining out.
Understanding the p-value is crucial since it helps in making a decision about the hypotheses and determining whether the observed results have statistical significance.