10. Hypothesis Testing for Two Samples

Two Means - Matched Pairs (Dependent Samples)

10. Hypothesis Testing for Two Samples

Two Means - Matched Pairs (Dependent Samples): Videos & Practice Problems

Topic summary

Matched pairs are samples that are related in a unique way, often through a before-and-after comparison of the same individual or related individuals. To conduct hypothesis tests with matched pairs, replace sample means and standard deviations with their difference equivalents. The null hypothesis typically states no difference, while the alternative hypothesis indicates a significant change. Confidence intervals for mean differences use the sample mean difference as a point estimator, with the margin of error calculated using the critical t-value and the standard deviation of the differences.

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Introduction to Matched Pairs

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Introduction to Matched Pairs Video Summary

In statistical analysis, understanding the concept of matched pairs is crucial when conducting hypothesis tests involving two samples. Matched pairs occur when two samples are related in a specific way, allowing for a one-to-one pairing of values. This relationship can manifest in various forms, such as before-and-after comparisons of the same individual, comparisons between related individuals (like siblings or coworkers), or contrasting self-reported data with measured data.

To determine if two samples are matched pairs, several criteria must be met. First, the sample sizes must be equal, ensuring that each value from one sample can be paired with a corresponding value from the other sample. Second, the samples must exhibit a relationship as outlined above. For instance, if we analyze heart rates of individuals before and after a specific event, we can establish a clear before-and-after relationship, confirming that the samples are indeed matched pairs. Lastly, the values must be paired in a one-to-one relationship, meaning that each value from one sample directly corresponds to a specific value in the other sample.

For example, consider a study measuring heart rates of nine adults before and after sleeping. Each individual's heart rate before sleeping can be directly compared to their heart rate after sleeping, fulfilling all criteria for matched pairs. Conversely, if we compare heart rates between two different groups, such as males and females, even if the sample sizes are equal, the lack of a direct relationship means these samples are independent rather than matched pairs.

When working with matched pairs, one common calculation involves determining the differences between paired values, denoted as $d$. This is done by subtracting one value from its paired counterpart, maintaining a consistent order throughout the calculations. For instance, if the heart rate before sleeping is 84 bpm and after is 80 bpm, the difference would be $d = 84 - 80 = 4$. After calculating the differences for all pairs, the mean difference, represented as $\bar{d}$, can be computed by summing all differences and dividing by the number of pairs.

Additionally, the standard deviation of these differences, denoted as $SD_d$, provides insight into the variability of the differences. This statistical approach allows researchers to analyze the effects of interventions or changes over time effectively, making matched pairs a powerful tool in hypothesis testing.

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Introduction to Matched Pairs Example 1

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Introduction to Matched Pairs Example 1 Video Summary

Determining whether two samples are independent or matched pairs is crucial in hypothesis testing, as it influences the type of statistical analysis to be conducted. To differentiate between these two types of samples, several criteria can be evaluated.

In the first example, a nutritionist measures the blood pressure of 30 patients before and after they start a new diet. Here, the sample sizes are equal, with 30 measurements taken before and 30 after the diet. This equality is a strong indicator of a matched pairs scenario. Additionally, the samples are related because they involve the same individuals measured at two different times, creating a clear before-and-after relationship. Each measurement before the diet can be paired with a corresponding measurement after the diet, confirming a one-to-one relationship. Thus, this scenario is classified as matched pairs.

In the second example, a teacher administers a practice exam to 40 students and then gives a different group of 40 students the same exam a week later. Although the sample sizes are equal, the key distinction lies in the lack of a direct relationship between the two groups. Each student's score cannot be paired with a specific score from the other group, as they are different individuals. Therefore, this scenario represents independent samples.

The third example involves the same teacher giving the same 40 students a similar graded exam a week after the initial practice exam. Again, the sample sizes are equal, and there is a relationship since the same students are being tested. Each student's score from the first exam can be directly paired with their score from the second exam, establishing a one-to-one relationship. Consequently, this scenario is also classified as matched pairs.

In summary, when assessing whether samples are independent or matched pairs, consider the equality of sample sizes, the relationship between the samples, and whether the values can be paired one-to-one. This systematic approach will guide you in determining the appropriate statistical tests to apply.

Problem

A personal trainer is studying whether a new stretching routine improves flexibility. She records the forward reach (in cm) of 6 clients before and after a 4-week program. Calculate the difference (after - before) for each client, the mean difference, and standard deviation.

Difference for each client: $A=4,B=4,C=3,D=-1,E=0,F=2$ ; The Mean Difference = $2$ ; and Standard Deviation = $1.91$

Difference for each client: $A=4,B=4,C=3,D=-1,E=0,F=2$ ; The Mean Difference = $2.5$ ; and Standard Deviation = $1.91$

Difference for each client: $A=4,B=4,C=3,D=-1,E=0,F=2$ ; The Mean Difference = $2.5$ ; and Standard Deviation = $2.10$

Difference for each client: $A=4,B=4,C=3,D=-1,E=0,F=2$ ; The Mean Difference = $2$ ; and Standard Deviation = $2.10$

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Matched Pairs: Hypothesis Tests

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Matched Pairs: Hypothesis Tests Video Summary

Matched pairs hypothesis testing is a statistical method used to compare two related samples, often in a before-and-after scenario. This approach simplifies the process of hypothesis testing by focusing on the differences between paired observations rather than the individual means. In this context, we replace standard variables with their difference equivalents: the sample mean difference is denoted as $ \bar{d} $, the population mean difference as $ \mu_d $, and the standard deviation of the differences as $ s_d $.

To illustrate this, consider a study evaluating a new medication's effectiveness in reducing blood pressure. Researchers collect data from 37 patients, measuring their blood pressure before and 15 days after taking the medication. The sample mean difference in blood pressure is found to be $ \bar{d} = 3.2 $ mmHg, with a standard deviation of $ s_d = 6.7 $ mmHg. The goal is to determine if the medication significantly reduces blood pressure, using a significance level of $ \alpha = 0.05 $.

The first step in hypothesis testing involves formulating the null and alternative hypotheses. The null hypothesis ($ H_0 $) posits that there is no difference in blood pressure, or $ \mu_d = 0 $. Conversely, the alternative hypothesis ($ H_a $) suggests that the medication does reduce blood pressure, represented as $ \mu_d > 0 $. This indicates that the mean difference in blood pressure after medication is greater than before.

Next, we calculate the test statistic using the formula for the t-score, which is given by:

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

Substituting the known values, we have:

$ t = \frac{3.2 - 0}{6.7 / \sqrt{37}} $

Calculating this yields a t-score of approximately $ t = 2.91 $. The degrees of freedom for this test is calculated as $ n - 1 = 36 $.

To determine the p-value, we assess the probability of observing a t-score greater than 2.91 in a right-tailed test. Using statistical tables or software, we find that the p-value is approximately $ 0.0031 $.

Finally, we compare the p-value to the significance level. Since $ 0.0031 < 0.05 $, we reject the null hypothesis. This result indicates sufficient evidence to conclude that the medication is effective in reducing blood pressure.

In summary, matched pairs hypothesis testing allows for a straightforward analysis of differences between related samples, following a systematic approach to hypothesis formulation, test statistic calculation, and p-value evaluation. This method is particularly useful in clinical studies where the same subjects are measured before and after an intervention.

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Matched Pairs: Hypothesis Tests Example 2

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Matched Pairs: Hypothesis Tests Example 2 Video Summary

In hypothesis testing, particularly with matched pairs, the goal is to determine if there is a significant difference between two related groups. In this scenario, a professor randomly selects 12 students to compare their scores on a pop quiz and a scheduled quiz, calculating the differences between the two scores for each student. The mean difference, denoted as $ \bar{d} $, is found to be -1.1, with a standard deviation of 1.8.

To set up the hypothesis test, we define the null hypothesis $ H_0 $ and the alternative hypothesis $ H_a $. The null hypothesis typically states that there is no difference in performance, which translates to $ \mu_d = 0 $, where $ \mu_d $ represents the population mean of the differences. In this case, the alternative hypothesis is that students perform worse on the pop quizzes than on the scheduled quizzes, leading to $ H_a: \mu_d < 0 $. This indicates a left-tailed test, as we are interested in whether the mean difference is less than zero.

Next, we calculate the test statistic using the formula for matched pairs, which is similar to that used for one-sample tests. The formula is given by:

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

In this formula, $ \bar{d} $ is the sample mean difference, $ \mu_d $ is the hypothesized population mean difference (which is 0), $ s_d $ is the standard deviation of the differences, and $ n $ is the number of pairs. Plugging in the values:

$ t = \frac{-1.1 - 0}{1.8 / \sqrt{12}} $

Calculating this yields a test statistic of approximately -2.12. This value indicates how many standard deviations the sample mean difference is from the hypothesized population mean difference under the null hypothesis. The next steps would typically involve determining the p-value and making a conclusion based on the significance level, but those steps are not covered here.

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Matched Pairs: Confidence Intervals

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Matched Pairs: Confidence Intervals Video Summary

In statistical analysis, matched pairs are used to compare two related samples, and creating a confidence interval for the mean difference is a common task. When constructing a confidence interval for matched pairs, the focus is on the difference between two values, specifically the population difference denoted as $ \mu_d $. The sample mean difference serves as the point estimator, which is the best estimate of this population parameter.

To illustrate this, consider a study evaluating the effectiveness of a new blood pressure medication. In this study, blood pressure readings were taken from 37 patients before and 15 days after medication. The sample mean difference ($ \bar{d} $) was found to be 3.2, with a standard deviation of the differences ($ s_d $) equal to 6.7. To create a 90% confidence interval, we first verify that the conditions for matched pairs are met: the samples must be random and paired, and the sample size should ideally be greater than 30 to assume normality.

The next step involves determining the critical value from the t-distribution, as we use the t-score for means rather than the z-score. For a 90% confidence level, the critical value $ t_{\alpha/2} $ is calculated using the degrees of freedom, which is $ n - 1 $ (in this case, 36). The critical t-value is found to be approximately 1.688.

With the point estimator $ \bar{d} = 3.2 $ established, we calculate the margin of error using the formula:

Margin of Error = $ t_{\alpha/2} \times \frac{s_d}{\sqrt{n}} $

Substituting the values, we find:

Margin of Error = $ 1.688 \times \frac{6.7}{\sqrt{37}} \approx 1.85 $

Now, we can construct the confidence interval by adding and subtracting the margin of error from the point estimator:

Lower Bound = $ 3.2 - 1.85 = 1.35 $

Upper Bound = $ 3.2 + 1.85 = 5.05 $

Thus, the 90% confidence interval for the mean difference in blood pressure is $ (1.35, 5.05) $.

This interval suggests that we are 90% confident that the true mean difference in blood pressure before and after taking the medication lies between 1.35 and 5.05. Since this interval does not include zero, we can reject the null hypothesis, which posits that there is no difference in blood pressure after medication. Therefore, there is sufficient evidence to conclude that the medication has a significant effect on blood pressure.

Problem

Construct a 95% confidence interval for the mean difference of the population given the following information. Would you reject or fail to reject the claim that there is no difference in the mean?
$d̄=-0.728$
$s_{d}=1.34$
$n=10$

The 95% confidence interval = $\left(-1.152,-0.304\right)$ ; Fail to reject the claim that there is no difference in the mean.

The 95% confidence interval = $\left(-1.686,0.230\right)$ ; Fail to reject the claim that there is no difference in the mean.

The 95% confidence interval = $\left(-1.686,0.230\right)$ ; Reject the claim that there is no difference in the mean.

The 95% confidence interval = $\left(-1.152,-0.304\right)$ ; Reject the claim that there is no difference in the mean.

Here’s what students ask on this topic:

What are matched pairs in statistics, and how do you identify them?

Matched pairs in statistics refer to two samples that are related in a unique way, often through a one-to-one pairing. Common examples include before-and-after measurements on the same individual or related individuals, such as siblings or coworkers. To identify matched pairs, check three criteria: (1) The sample sizes must be the same, ensuring a one-to-one relationship. (2) The samples must be related, such as through a before-and-after scenario or a connection like siblings. (3) The values must be paired logically, meaning each value in one sample corresponds directly to a value in the other. For example, comparing the heart rate of the same person before and after sleeping would qualify as matched pairs, while comparing unrelated groups, like males and females, would not.

How do you perform a hypothesis test for matched pairs?

To perform a hypothesis test for matched pairs, follow these steps: (1) State the null hypothesis (H₀) as no difference in the population mean difference (μ_d = 0) and the alternative hypothesis (H_a) as a significant difference (e.g., μ_d ≠ 0, μ_d > 0, or μ_d < 0). (2) Calculate the test statistic using the formula:

$t = \frac{{\bar{d}}_{d} - μ_{d}}{\frac{s_{d}}{\sqrt{n}}}$

where d̄ is the sample mean difference, μ_d is the hypothesized mean difference (usually 0), s_d is the standard deviation of the differences, and n is the number of pairs. (3) Determine the p-value using the t-distribution with degrees of freedom (n - 1). (4) Compare the p-value to the significance level (α). If p < α, reject H₀, indicating a significant difference. Otherwise, fail to reject H₀.

How do you calculate a confidence interval for the mean difference in matched pairs?

To calculate a confidence interval for the mean difference in matched pairs, follow these steps: (1) Use the sample mean difference (d̄) as the point estimator. (2) Calculate the margin of error (ME) using the formula:

$ME = t_{α / 2} \times \frac{s_{d}}{\sqrt{n}}$

where t_α/2 is the critical t-value for the confidence level, s_d is the standard deviation of the differences, and n is the number of pairs. (3) Construct the confidence interval as:

$[{\bar{d}}_{d} - ME, {\bar{d}}_{d} + ME]$

For example, if d̄ = 3.2, s_d = 6.7, n = 37, and the critical t-value is 1.688 for a 90% confidence level, the margin of error is 1.85. The confidence interval would be [1.35, 5.05].

What is the difference between matched pairs and independent samples?

The key difference between matched pairs and independent samples lies in the relationship between the data points. Matched pairs involve two samples that are related in a unique way, such as before-and-after measurements on the same individual or paired data from related individuals (e.g., siblings). Each value in one sample corresponds directly to a value in the other, forming a one-to-one relationship. In contrast, independent samples consist of two groups with no inherent relationship between their data points. For example, comparing the test scores of two different classes would involve independent samples. Matched pairs require paired t-tests, while independent samples use two-sample t-tests. The choice depends on whether the data points are related or independent.

How do you interpret the results of a matched pairs confidence interval?

To interpret a matched pairs confidence interval, focus on whether the interval includes the value 0. If the interval does not include 0, it suggests a significant difference in the population mean difference (μ_d). For example, if a 90% confidence interval for blood pressure differences is [1.35, 5.05], it indicates that the true mean difference is likely between 1.35 and 5.05, and there is evidence of a significant change. If the interval includes 0, it suggests no significant difference. Always state the confidence level and the context of the data when interpreting the results.