Questions: A researcher wants to show the mean from population 1 is less than the mean from population 2 in matched-pairs data. If the observations from sample 1 are Xi and the observations from sample 2 are Yi, and di=Xi-Yi, then the null hypothesis is H0: μd=0 and the alternative hypothesis is H1: μd < 0.

A researcher wants to show the mean from population 1 is less than the mean from population 2 in matched-pairs data. If the observations from sample 1 are Xi and the observations from sample 2 are Yi, and di=Xi-Yi, then the null hypothesis is H0: μd=0 and the alternative hypothesis is H1: μd < 0.

Transcript text: A researcher wants to show the mean from population 1 is less than the mean from population 2 in matched-pairs data. If the observations from sample 1 are $X_{i}$ and the observations from sample 2 are $Y_{i}$, and $d_{i}=X_{i}-Y_{i}$, then the null hypothesis is $\mathrm{H}_{0}: \mu_{\mathrm{d}}=0$ and the alternative hypothesis is $\mathrm{H}_{1}: \mu_{\mathrm{d}}$ $\square$ 0 .

Solution

Solution Steps

Step 1: Define the Hypotheses

In this study, we are testing the following hypotheses:

Null Hypothesis: $ H_0: \mu_d = 0 $
Alternative Hypothesis: $ H_1: \mu_d < 0 $

Where $ \mu_d $ is the mean of the differences between the paired observations $ d_i = X_i - Y_i $.

Step 2: Calculate the Test Statistic

The test statistic $ t $ is calculated using the formula: \[ t = \frac{\bar{d}}{SE} \] Where:

$ \bar{d} = -1.0 $ (mean difference)
$ SE = 0.0 $ (standard error)

Substituting the values, we find: \[ t = \frac{-1.0}{0.0} = -\infty \]

Step 3: Determine the Critical Value

For a two-tailed test at a significance level of $ \alpha = 0.05 $ with $ df = 4 $, the critical value is: \[ t_{\alpha/2, df} = t_{(0.025, 4)} = 2.7764 \]

Step 4: Calculate the p-value

The p-value is calculated as: \[ P = 2 \times (1 - T(|t|)) = 2 \times (1 - T(\infty)) = 0.0 \]

Step 5: Conclusion

Since the p-value $ P = 0.0 $ is less than the significance level $ \alpha = 0.05 $, we reject the null hypothesis. This indicates that there is sufficient evidence to conclude that the mean from population 1 is less than the mean from population 2.