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Use a t-distribution and the given matched pair sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distribution of the differences is relatively normal. Assume that differences are computed using \(d=x_{1}-x_{2}\). Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}<\mu_{2}\) using the paired data in the following table: $$ \begin{array}{lllllllll} \hline \text { Treatment } 1 & 16 & 12 & 18 & 21 & 15 & 11 & 14 & 22 \\ \text { Treatment } 2 & 18 & 20 & 25 & 21 & 19 & 8 & 15 & 20 \\ \hline \end{array} $$

Short Answer

Expert verified
The null hypothesis \(H_{0}: \mu_{1}=\mu_{2}\) is rejected in favor of the alternative hypothesis \(H_{a}: \mu_{1}<\mu_{2}\), with t-statistic \(-2.0366\) and p-value less than \(0.05\). It's suggested that the mean of treatment 1 is less than treatment 2.

Step by step solution

01

Calculate differences

Calculate the differences between each pair of treatments 1 and 2 by using the formula \(d=x_{1}-x_{2}\). So, differences would be \(-2, -8, -7, 0, -4, 3, -1, 2\).
02

Compute mean of differences

Calculate the mean of these differences using the formula \(\bar{d} = \frac{1}{n} \sum_{i=1}^{n} d_{i}\). With the given differences and \(n=8\), the mean of differences, \(\bar{d}\), is \(-2.125\).
03

Calculate standard deviation of differences

Calculate the standard deviation of the differences using the formula \(s_{d} = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (d_{i}-\bar{d})^{2}}\). The standard deviation, \(s_{d}\), of these differences turns out to be approximately \(3.3066\).
04

Compute t-statistic

Calculate the t-statistic using the formula \(t = \frac{\bar{d}}{s_{d}/\sqrt{n}}\). The t-score is \(-2.0366\).
05

Determine p-value

For a one-tailed test with \(n-1=7\) degrees of freedom, the critical t-value for a significance level of \(0.05\) is approximately \(1.895\). The computed t-statistic is less than this, hence, the p-value is less than \(0.05\).
06

Make a decision

Because the p-value is less than \(0.05\), we reject the null hypothesis in favor of the alternative hypothesis, suggesting that there is a significant difference and the mean of treatment 1 is less than treatment 2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Paired Data Analysis
Paired data analysis is a statistical method used when we compare two related groups. Unlike independent samples, paired data consists of two related measures taken from the same participants, or matched participants. In this exercise, we're comparing two treatments on the same subjects, which makes it an ideal candidate for paired data analysis.

To analyze paired data, we calculate the difference between each pair of observations. This helps us determine whether there's a significant change or difference. For example, when comparing Treatment 1 and Treatment 2, differences are calculated as the values of Treatment 1 subtracted by those of Treatment 2.

This approach ensures that any differences we observe can be more confidently attributed to the treatment effect, rather than individual variability. Once the differences are computed, further statistical measures like mean and standard deviation of these differences can help us set up hypothesis testing.
  • Calculate differences: Use the formula \(d = x_1 - x_2\).
  • Ensures consistency: Reduces variability due to individual differences.
  • Facilitates hypothesis testing: Essential for subsequent statistical analysis.
Statistical Inference
Statistical inference refers to the process of drawing conclusions about a population's characteristics based on a sample. This involves using sample data to make educated guesses or estimates about the population.

In the context of hypothesis testing with t-distributions, it provides a framework to assess if there is sufficient evidence to support or reject a hypothesis. For this exercise, the hypothesis testing is formed as follows:

  • Null Hypothesis \(H_0\): \(\mu_1 = \mu_2\)
  • Alternative Hypothesis \(H_a\): \(\mu_1 < \mu_2\)
The null hypothesis suggests no difference in means, indicating that the treatments have similar effects. Meanwhile, the alternative hypothesis implies that Treatment 1 has a lesser mean compared to Treatment 2.
To decide if we can reject the null hypothesis, we calculate the t-statistic and compare it against a critical value, determined by a chosen significance level. A lower p-value than the significance threshold indicates we can favor the alternative hypothesis. This comparison forms the backbone of statistical inference.
  • Sample data informs decision: Provides expected characteristics of a whole population.
  • Framework for decision-making: Compares sample evidence against specified thresholds.
P-value Interpretation
The p-value is a critical concept in hypothesis testing. It helps determine the statistical significance of the observed result. A p-value tells us, under the assumption that the null hypothesis is true, how likely we are to observe a result as extreme as, or more extreme than, the one from our sample.

In this exercise, the p-value is used to judge whether the difference in the treatments is statistically significant. We calculated the t-statistic based on our paired data. The next step involves comparing it with critical values from a t-distribution with the appropriate degrees of freedom, which in this case is 7 (as there are 8 pairs).
If the p-value is less than the chosen significance level (usually 0.05), it infers that the observed difference is unlikely due to random chance, allowing us to reject the null hypothesis. This suggests that there is evidence in favor of the alternative hypothesis.
  • Indicator of significance: Shows the probability of observing such data under the null hypothesis.
  • Threshold-based decision: Helps to decide between null and alternative hypothesis.
  • Guide for inference: Integral to concluding about population based on sample data.

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Most popular questions from this chapter

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