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Confidence intervals help us estimate an unknown population parameter by providing a range of plausible values based on sample data. For instance, in our problem, we constructed a 95% confidence interval to estimate the mean energy expenditure. This interval is calculated using the formula: \[\bar{x} \pm t_{\alpha/2, df} \cdot \frac{s}{\sqrt{n}} \] where \(\bar{x}\) is the sample mean, \(s\) is the sample standard deviation, \(n\) is the sample size, and \(t_{\alpha/2, df}\) is the critical t-value for the chosen confidence level with appropriate degrees of freedom (\(df\)).
For our example, the 95% confidence interval is (1507.86, 1728.14). This means we are 95% confident that the true mean expenditure lies within this interval. The interpretation of the interval impacts our decision-making by showing that 1493 is not within this range, suggesting a significant change in expenditures.

A t-test is a statistical test used to compare the mean of a sample to a known value or another sample's mean. This test helps determine if there is a significant difference between the two means. In our problem, we used a t-test to see if the mean energy expenditure in a recent year is significantly different from that in 2001.
The test statistic for a one-sample t-test is calculated as: \[\frac{\bar{x} - \mu}{s / \sqrt{n}} \] where \(\bar{x}\) is the sample mean, \(\text{mu}\) is the population mean, \(\text{s}\) is the sample standard deviation, and \(n\) is the sample size. In our case, we found the t-value to be approximately 2.112, which we compared to the critical t-value (approximately \(\pm 2.032\) for \(df = 34\) at \(\text{alpha} = 0.05\)).
Since 2.112 is greater than 2.032, we rejected the null hypothesis and concluded that there is a significant change in mean expenditure.

The level of significance, denoted \(\alpha\), is the threshold for deciding whether to reject the null hypothesis. Commonly set at 0.05, it represents a 5% risk of concluding that a difference exists when there is none (Type I error).
In hypothesis testing, we compare the p-value or test statistic to the level of significance to determine the result. For our problem, we set \(\alpha = 0.05\) and compared the calculated t-value (2.112) to the critical value (2.032).
If the test statistic exceeds the critical value, we reject the null hypothesis. In our case, since 2.112 > 2.032, we concluded that the mean expenditure has changed significantly. This use of \(\alpha\) ensures our results are statistically significant and not due to random chance.