Question

A paired-difference experiment consists of \(n=18\) pairs, \(\bar{d}=5.7,\) and \(s_{d}^{2}=256 .\) Suppose you wish to detect \(\mu_{d} > 0\). a. Give the null and alternative hypotheses for the test. b. Conduct the test and state your conclusions.

Short answer

Answer: Yes, there is enough evidence to support the claim that the mean difference (µd) is greater than 0 at the 5% significance level.

Step-by-step solution

H0 and Ha: Hypotheses

To test whether the mean difference (\(\mu_{d}\)) is greater than 0, we will define our null and alternative hypotheses as follows: H0 (null hypothesis): \(\mu_{d} = 0\) Ha (alternative hypothesis): \(\mu_{d} > 0\) Now that we have our hypotheses defined, we can proceed with the hypothesis test. #b. Conducting the hypothesis test and stating conclusions#

Test statistic

The first step in conducting a t-test is to calculate the test statistic. This is done using the following formula: \(t = \frac{\bar{d} - \mu_{d}}{s_{d}/\sqrt{n}}\) Given the problem's information, we have \(\bar{d}=5.7\), \(s_{d}^{2}=256\) (which gives \(s_d=16\)), and \(n=18\). We're testing against the null hypothesis which states that \(\mu_{d} = 0\). Therefore, we can calculate the test statistic as: \(t = \frac{5.7 - 0}{16/\sqrt{18}} \approx 2.01\)

Degrees of freedom and critical value

Next, we will find the critical value for our test. Since this is a one-tailed test (Ha: \(\mu_{d} > 0\)), we will need to find the critical value \(t_{\alpha}\) corresponding to our chosen level of significance, denoted by \(\alpha\). Typically, \(\alpha = 0.05\) is used for a 95% confidence level. First, we need to find the degrees of freedom for our t-distribution, which is given by: \(df = n - 1 = 18 - 1 = 17\) Next, we'll use a t-table or calculator to find the critical value \(t_{\alpha}\) with a one-tailed test at \(0.05\) significance level and \(17\) degrees of freedom: \(t_{\alpha} \approx 1.740\)

Comparing the test statistic to the critical value

Now we can compare our test statistic (\(t \approx 2.01\)) to the critical value (\(t_{\alpha} \approx 1.740\)). To make our decision, we look at the following: - If \(t > t_{\alpha}\), reject the null hypothesis (H0) in favor of the alternative hypothesis (Ha). - If \(t \leq t_{\alpha}\), fail to reject the null hypothesis (H0). Since \(2.01 > 1.740\), we reject the null hypothesis, stating: There is enough evidence, at the 5% significance level, to support the claim that the mean difference (\(\mu_{d}\)) is greater than 0.

Key concepts

Null and Alternative Hypotheses

When starting a hypothesis test in a paired-difference experiment, it's crucial to clearly define the null and alternative hypotheses. These hypotheses help us understand what we are testing for and guide us through the analysis.

In general, the **null hypothesis** \((H_0)\) represents a statement of no effect or no difference. It often includes a specific value for the parameter being tested. In the context of a paired-difference experiment, the null hypothesis might propose that the mean difference \(\mu_d\) between paired observations is zero (\(H_0: \mu_d = 0\)).

Meanwhile, the **alternative hypothesis** \((H_a)\) challenges the null hypothesis. It suggests there is an effect or a difference. In this case, we want to determine if the mean difference is greater than zero (\(H_a: \mu_d > 0\)).

Writing clear hypotheses is essential because they form the basis of the testing procedure. If the evidence supports the alternative hypothesis, we may reject the null hypothesis, indicating a potential effect or difference in the context of our study.

T-test

A t-test is a statistical test used to compare the means of two groups or to compare a sample mean with a known value. When you have paired data, as in a paired-difference experiment, a t-test can help determine if there is a significant difference between the paired values.

In a paired-difference test, we perform a **t-test** to check if the mean of the differences \(\bar{d}\) between paired observations is significantly different from zero. The t-test uses the sample data to determine whether the differences are statistically significant.

This process usually involves formulating a hypothesis, calculating a t statistic, and determining whether this statistic supports the null or alternative hypothesis based on a comparison with a critical value.

Test Statistic

The **test statistic** is a key component in hypothesis testing that allows us to compare our sample data against the null hypothesis. It is a numerical summary of the data that is used to determine whether or not to reject the null hypothesis.

In a paired t-test, the test statistic \(t\) is calculated using the formula:

\[ t = \frac{\bar{d} - \mu_d}{s_d/\sqrt{n}} \]

Where:

• \(\bar{d}\) is the mean of the differences between paired observations.
• \(\mu_d\) is the mean difference under the null hypothesis.
• \(s_d\) is the standard deviation of the differences.
• \(n\) is the number of paired differences.

The test statistic is used to determine where our sample mean falls in relation to the distribution expected under the null hypothesis. We use this value to find if our results are statistically significant by comparing it to a critical value from the t-distribution, based on our desired level of confidence and degrees of freedom.

Degrees of Freedom

**Degrees of freedom** refer to the number of values in a calculation that are free to vary while estimating statistical parameters. They are crucial in determining the precise shape of the t-distribution used in a hypothesis test.

For a paired-difference t-test, the degrees of freedom (df) are calculated as one less than the number of pairs. If there are \(n = 18\) pairs, then the degrees of freedom are:

\[ df = n - 1 = 18 - 1 = 17 \]

Degrees of freedom affect the critical value used to decide whether to reject the null hypothesis. The smaller the sample size, the larger the critical value must be to reach the same confidence level. This is because smaller samples provide less certainty about the population parameters.