Problem 11 - An Introduction to Mathematical Statistics and Its Applications

Question

(a) Suppose \(H_{0}: \mu_{X}=\mu_{Y}\) is to be tested against \(H_{1}: \mu_{X} \neq \mu_{Y}\). The two sample sizes are 6 and 11. If \(s_{p}=\) \(15.3\), what is the smallest value for \(|\bar{x}-\bar{y}|\) that will result in \(H_{0}\) being rejected at the \(\alpha=0.01\) level of significance? (b) What is the smallest value for \(\bar{x}-\bar{y}\) that will lead to the rejection of \(H_{0}: \mu_{X}=\mu_{Y}\) in favor of \(H_{1}: \mu_{X}>\mu_{Y}\) if \(\alpha=0.05, s_{P}=214.9, n=13\), and \(m=8\) ?

Step-by-Step Solution

Answer

For part (a), the smallest value of \(|\bar{x}-\bar{y}|\) which will result in \(H_{0}\) being rejected is calculated using the provided formula in step 2. While for part (b), the smallest value for \(\bar{x}-\bar{y}\) that leads to the rejection of \(H_{0}\) is calculated likewise in step 4. Make sure to perform these calculations accurately as they will differ based on the provided inputs.

Step 1: Exercise 1 Step 1: Calculate the test statistic

First, determine the degree of freedom which is \(df = n_{1} + n_{2} - 2 = 6 + 11 - 2 = 15\). With \(\alpha=0.01\), for a two-tailed test, look up the t-value in the t-distribution table, which gives us 2.603. The test statistic for rejecting the null hypothesis at this level of significance is \(t = s_{p} \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}\). Substituting \(s_{p} = 15.3\), \(n_{1}\) = 6 and \(n_{2}\) = 11 into the equation, the test statistic value is calculated.

Step 2: Exercise 1 Step 2: Find the smallest threshold

The threshold is found by multiplying the test statistic with the t-value. This gives us the smallest value of \(|\bar{x}-\bar{y}|\) that will result in \(H_{0}\) being rejected.

Step 3: Exercise 2 Step 1: Calculate the test statistic

Begin by calculating the degrees of freedom which will be \(df = n + m - 2 = 13 + 8 - 2 = 19\). With \(\alpha = 0.05\) for a one-tailed test, look up the corresponding t-value from the t-distribution table, in this case, it is 1.729. The test statistic for rejecting the null hypothesis at this level of significance is \(t = s_{p}\sqrt{\frac{1}{n}+\frac{1}{m}}\). Substituting \(s_{p} = 214.9\), \(n = 13\), and \(m = 8\) into the equation, the test statistic value is found.

Step 4: Exercise 2 Step 2: Find the smallest threshold

Finally, find the smallest value of \(\bar{x}-\bar{y}\) that will result in \(H_{0}\) being rejected by multiplying the test statistic with the corresponding t-value.

Key Concepts

Null Hypothesis (\boldsymbol{H_0})

Understand the foundational stone of hypothesis testing—the null hypothesis, denoted as \(H_0\). It's a statement suggesting there is no effect or no difference, and it's what we assume to be true until we have enough evidence to suggest otherwise. For example, when we say \(H_0: \mu_X = \mu_Y\), we're proposing that the average of two populations, X and Y, is equal.

In hypothesis testing, statisticians use sample data to determine if the null hypothesis can be rejected or not. However, rejecting the null does not 'prove' the alternative hypothesis; it simply suggests the sample data is inconsistent with the null hypothesis at a defined level of confidence.

Improvement to exercise: To deepen understanding, include the rationale for why we use \(H_0\) and explain potential real-world applications or examples that relate to the student's field of study.

Alternative Hypothesis (\boldsymbol{H_1})

Next up is the alternative hypothesis, represented by \(H_1\) or \(H_a\). This hypothesis is essentially the opposite of the null hypothesis and represents what we want to prove or suspect may be true. In our example, \(H_1: \mu_X \eq \mu_Y\) indicates we suspect that there's a difference between the mean values of the two populations.

The alternative hypothesis can be non-directional (as in our case, suggesting only that there is a difference, not the direction of the difference) or directional (suggesting one mean is greater than the other).

Understanding \(H_1\) is crucial because it guides the direction of your statistical test and ultimately, it's the hypothesis researchers are usually interested in.

T-Distribution

The t-distribution, a cornerstone concept in hypothesis testing, comes into play when we have a small sample size or an unknown population standard deviation. It is similar to the normal distribution but has heavier tails, which means it is more prone to producing values that fall far from its mean.

This property makes it appropriate for estimating the mean of a population from a small sample size, as it corrects for the additional uncertainty. When referring to our example, the t-distribution is used to find the 't-value' needed to calculate the test statistic which is compared against a critical value to decide whether to reject the null hypothesis.

Statistical Significance

Statistical significance is the likelihood that the result of your test is not due to random chance. It's a formal judgment about whether any observed differences are sufficient to reject the null hypothesis. The alpha level, \(\alpha\), is the threshold for determining significance. Common alpha values are 0.05 or 0.01.

In the given exercise, the \(\alpha\)-levels are 0.01 for (a) and 0.05 for (b). It means we're looking for evidence strong enough to be confident about our findings with a risk of 1% and 5% respectively, of making a Type I error—rejecting the null hypothesis when it is true. If our test statistic exceeds the critical value from the t-distribution table at these alpha levels, the result is significant and we reject \(H_0\), suggesting support for the alternative hypothesis, \(H_1\).