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Paired data analysis is a statistical method used to compare two sets of observations. In our exercise, each item from set M has a corresponding partner in set N, thus forming pairs, which is why it's called "paired data." This kind of analysis is useful when we want to see how these paired observations differ from each other under an intervention or over time.

When working with paired data, we first calculate the differences between each pair, as shown in Step 1 of the solution. From M and N, we compute each difference using the formula:

• d_i = M_i - N_i,

and we get a set of differences. These differences are what we use in further statistical analyses, like the t-test, to determine if there's a significant mean difference between the paired observations.

Paired data analysis is particularly helpful when the assumptions of normal distribution are met, as indicated here. Each pair is dependent on its fellow, not on other pairs, keeping the data statistically manageable and often more robust against assumption violations like non-normality.

The t-test is a great way to test hypotheses about the difference between means in paired data. Specifically, the paired t-test is used here to determine if the mean difference of the pairs is significantly different from zero. The hypothesized outcome here is \(\mu_{d} < 0\), meaning the differences are expected to be less than zero.

The calculation involves crucial steps, carried out meticulously, usually with the help of a calculator or statistical software. First, after obtaining the mean (

• \(\bar{d} = -2.33\),

) and the standard deviation of differences (

• \(s_d = 3.5\)

), we compute the observed t-statistic using:

• \(t_{obs} = \frac{\bar{d} - \mu_0}{\frac{s_d}{\sqrt{n}}}\).

Here, \(\mu_0\) is typically zero as we test if the mean difference is zero. The t-statistic indicates how much the observed data differs from what we'd expect under the null hypothesis. The goal is to see if this data is significantly far off, given the standard deviation and sample size.

The significance level, represented by the Greek letter alpha (\(\alpha\)), is a threshold set by the researcher. It helps determine if an observed effect is statistically significant. In this exercise, we have \(\alpha = 0.02\). This means a 2% chance of incorrectly rejecting the null hypothesis when it is actually true – a Type I error.

Choosing \(\alpha\) is up to the researcher's discretion, depending on how much risk of Type I error they can tolerate. Commonly used levels are 0.05, 0.01, and 0.10, but in this exercise, the chosen level is tighter, being 0.02.

With a specified \(\alpha\), we find a critical t-value using t-distribution tables or calculators. This value, here identically -2.571 due to degrees of freedom calculated from the sample size, acts as our comparison benchmark. If the calculated t-statistic surpasses this negative critical threshold, the null hypothesis is rejected.

However, in this exercise, our observed statistic does not exceed the critical value threshold, indicating insufficient evidence to conclude a true difference exists at the \(\alpha = 0.02\) level. So, the hypothesis about the differences being less than zero isn't supported.