91Ó°ÊÓ

The t-distribution is critical for confidence intervals, particularly when dealing with small sample sizes or unknown population variances. It resembles the more commonly known normal distribution, but it has heavier tails. This means it is more prone to producing values far from the mean. As sample size increases, the t-distribution becomes closer to a normal distribution. In our exercise, we are using a sample size of 10, which is relatively small.
This small size leads us to apply the t-distribution to estimate the confidence interval around our mean difference.When constructing a confidence interval, we use the formula:\[\bar{d} \pm t_{(n-1, \alpha / 2)} \times \frac{s}{\sqrt{n}}\]Here, \(\bar{d}\) is the sample mean difference, \(s\) is the standard deviation, and \(n\) is the sample size. The t-value is selected from the t-table based on our degrees of freedom \((n-1)\) and our desired confidence level. The confidence interval we calculated, (28.2, 70.8), shows the range in which we expect the true mean difference to fall, with 95% confidence.

The sample mean difference is a measure that helps us understand the average discrepancy between two sets of data. In this exercise, it's the average of how much the measured calorie content exceeded the stated content for the 10 meals. To calculate this, we first determine the difference for each individual meal and then average these differences.Mathematically speaking, the mean difference \(\bar{d}\) is calculated as:\[\bar{d} = \frac{\sum_{i=1}^{n} (d_i)}{n}\]where \(d_i\) represents each individual difference and \(n\) is the number of observations (in this case, 10). With a sample mean difference of 49.6, it tells us that, on average, the measured calorie values are 49.6 kcal higher than the stated values. This insight is essential to understand the overall trend between the two datasets.

Standard deviation is a statistical measure that tells us about the dispersion or spread of a dataset relative to its mean. It provides insight into how clustered or spread out the values are. In simpler terms, it's an indicator of how much the individual data points deviate from the average value.For our exercise, the standard deviation, denoted as \(s\), is derived from the differences between the measured and stated calorie contents. We calculate it using the square root of the variance:\[s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (d_i - \bar{d})^2}\]Here, \(\bar{d}\) is the mean of the differences, and \(d_i\) are the individual differences. In this case, the calculated standard deviation is approximately 29.97. A higher standard deviation would indicate more variability among the meals' differences, while a lower one would show that the differences are more consistently close to the mean.

The percentage increase is a useful metric that tells us how much one value has increased relative to another value, expressed in percentage terms. In this scenario, it's about understanding how much higher the measured calorie contents are compared to the stated values.To calculate the percentage increase:1. First, find the sample mean difference \(\bar{d}\), which is the average of all differences.2. Then determine the mean stated value of the energy contents.3. Use the formula:\[\text{Percentage increase} = \left(\frac{\bar{d}}{\text{mean stated value}}\right) \times 100\]In our exercise, with a mean stated value of 214 kcal, the percentage increase comes out to be approximately 23.18%. This means, on average, the measured energy contents are 23.18% higher than what was originally stated, pointing to a significant discrepancy between stated and measured values in these supermarket convenience meals.