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Hypothesis testing is a statistical method used to make decisions about the population based on sample data. In this particular problem, we test if there is a significant difference in heights between mothers and their first daughters. First, we set up our null hypothesis \(H_0\), which posits that there is no difference in the heights, i.e., \( \mu_d = 0 \). The alternative hypothesis \(H_1\) suggests that there is a difference in heights, i.e., \( \mu_d \eq 0 \).

We then choose a significance level, which in this exercise is 0.05. This indicates that we are willing to accept a 5% chance of rejecting the true null hypothesis. The steps involve calculating certain statistics from the data and comparing those statistics with critical values from statistical tables to make a decision.

Hypothesis testing allows us to quantify the evidence against \(H_0\), and decide if any observed differences are statistically significant.

The mean difference is a critical value in paired t-tests. It's the average of all individual differences between paired observations. By subtracting the height of each daughter from her mother’s height, we obtain a set of differences \(d_i \). These differences are: \(-0.5, 0.0, -2.5, -4.0, 1.0, -1.5, 0.0, -4.0, -1.0, 3.0 \).

We then calculate the mean of these differences using the formula: \(\bar{d} = \frac{\text{Sum of differences}}{n} \). For 10 mother-daughter pairs, this yields the average difference.

The mean difference \(\bar{d} \) helps in determining if the observed differences are substantial or merely due to random chance. In the context of the hypothesis test, if \(\bar{d} \) significantly deviates from 0, we may consider rejecting \(H_0\).

Standard deviation measures the dispersion or spread of differences from the mean. Higher standard deviations indicate greater variability among data points, while lower standard deviations reflect data that are more clustered around the mean difference.

To find the standard deviation \(s_d \) of the differences, use: \[\text{s}_d = \sqrt{\frac{ \sum (d_i - \bar{d})^2 }{n-1}} \].

This involves computing the differences between each individual difference and the mean difference, squaring them, summing them up, dividing by \(n-1 \) (with \ \ being the total number of pairs), and then taking the square root of the result.

The standard deviation is crucial because it normalizes the mean difference when calculating the t-statistic, making it comparable across different datasets.

The critical value is a threshold that helps us determine if the computed test statistic is statistically significant. It is based on the chosen significance level and the degrees of freedom (df). For our exercise, at the 0.05 significance level with 9 degrees of freedom (because there are 10 pairs), we look this value up in a t-distribution table.

If the absolute value of our calculated t-statistic exceeds the critical value, we reject \(H_0\). If it does not, we fail to reject \(H_0\). This comparison allows us to decide if the observed sample data are consistent with the null hypothesis.

Random samples are crucial in hypothesis testing because they ensure that each member of the population has an equal chance of being included. This randomness eliminates bias and allows for the generalization of the results to the broader population. In this exercise, the mother-daughter heights are assumed to be simple random samples, which means they are taken in such a way that each mother-daughter pair is independently and randomly chosen.

This randomization gives the paired t-test validity as it makes reasonable assumptions about the independence of observations and the normality of differences. Without random sampling, our conclusions could be skewed, limiting the reliability of the test outcomes.