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When comparing two populations, we often look at the **mean difference** between paired samples. In this scenario, we have values from Population 1 and Population 2 for five different elements. To find the mean difference, we first compute the difference for each pair.
For example, for Element 1 we subtract the value from Population 2 from Population 1: 21 - 20 = 1. We repeat this for all elements, giving us the list of differences: 1, 2, 0, 0, 2.
Next, we calculate the average of these differences. The mean difference formula is:\[ \bar{d} = \frac{\sum{d}}{n} \]Here, \(\sum{d} = 5\), and \(n\) (the number of elements) is 5.
This means the mean difference \(\bar{d}\) is 1.0, which indicates that Population 1 tends to have slightly higher values than Population 2.

The **standard deviation** of differences helps us understand the variability among the differences. It's a measure of how much the individual differences \(d_i\) deviate from the mean difference \(\bar{d}\).
We start by figuring out how each difference compares to the mean difference. For example, with the differences (1, 2, 0, 0, 2) and \(\bar{d} = 1\), each element deviation \((d_i - \bar{d})\) is calculated: -1 for 0, 0 for 1, and 1 for 2.
We then square these deviations, sum them up, and divide by \(n-1\) (degrees of freedom) to get the variance.\[ s_d = \sqrt{\frac{\sum (d - \bar{d})^2}{n-1}} \]
By plugging the numbers, we find that \(s_d = 1.0\), indicating a relatively low variability among difference values.

The **critical t-value** is a threshold determining the statistical significance in hypothesis testing. In our case, we're performing a one-tailed test to see if the mean difference \(\mu_d\) is greater than zero.
For this, we calculate the *t-statistic* using:\[ t = \frac{\bar{d} - \mu_0}{s_d / \sqrt{n}} \]where \(\mu_0\) (hypothesized mean difference) is 0. Our calculated t-statistic is 2.236, and to decide whether this result is statistically significant, we need the critical t-value from a t-distribution table for 4 degrees of freedom (\(n-1\)) at a significance level of 0.05.
This value is 2.132.
Since our calculated t-value exceeds this threshold, we have reason to reject the null hypothesis, suggesting a significant mean difference.

In hypothesis testing, a **one-tailed test** checks if a parameter is greater or less than a value. Here, we're testing if the mean difference \(\mu_d\) is greater than zero, leading us to a one-tailed test.
The hypotheses set up are:\[ H_{0}: \mu_{d} \leq 0 \text{ and } H_{a}: \mu_{d} > 0 \]
By focusing only on one direction (greater than), we require a strong enough test result (t-value) to reject the null hypothesis unequivocally.
This method is powerful when the hypothesis is in a specific direction, making our test precise and streamlining decision-making when the sample evidence supports the alternative hypothesis.