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Paired Differences Test For a random sample of 36 data pairs, the sample mean of the differences was 0.8. The sample standard deviation of the differences was \(2 .\) At the \(5 \%\) level of significance, test the claim that the population mean of the differences is different from 0. (a) Check Requirements Is it appropriate to use a Student's \(t\) distribution for the sample test statistic? Explain. What degrees of freedom are used? (b) State the hypotheses. (c) Compute the sample test statistic and corresponding \(t\) value. (d) Estimate the \(P\) -value of the sample test statistic. (e) Do we reject or fail to reject the null hypothesis? Explain. (f) Interpretation What do your results tell you?

Short Answer

Expert verified
We reject the null hypothesis, indicating a significant difference from zero.

Step by step solution

01

Checking Requirements and Degrees of Freedom

Since the population standard deviation is unknown and the sample size is less than 30, it's appropriate to use a Student's t-distribution provided the sample is approximately normally distributed. Since the sample size is 36 (greater than 30), we can relax the normality requirement. The degrees of freedom are calculated as the sample size minus one, so the degrees of freedom are \(36 - 1 = 35\).
02

Stating the Hypotheses

We need to set up the null and alternative hypotheses. The null hypothesis \(H_0\) states that the population mean difference \(\mu_d = 0\), and the alternative hypothesis \(H_a\) states that \(\mu_d eq 0\). Mathematically:- \(H_0: \mu_d = 0\)- \(H_a: \mu_d eq 0\)
03

Computing the Test Statistic

The test statistic \(t\) is calculated using the formula:\[ t = \frac{\bar{x}_d - \mu_0}{s_d / \sqrt{n}} \]where \(\bar{x}_d = 0.8\) is the sample mean difference, \(\mu_0 = 0\) is the hypothesized population mean difference, \(s_d = 2\) is the sample standard deviation, and \(n = 36\). Substitute the values into the formula:\[ t = \frac{0.8 - 0}{2 / \sqrt{36}} = \frac{0.8}{2/6} = \frac{0.8}{0.3333} \approx 2.4 \]
04

Estimating the P-Value

With \(t = 2.4\) and 35 degrees of freedom, we refer to the t-distribution table or use a calculator to find the p-value. The p-value corresponds to the probability of observing a test statistic as extreme as 2.4 or more extreme in both tails of the distribution. The approximate p-value is 0.02.
05

Decision to Reject or Fail to Reject the Null Hypothesis

The p-value (0.02) is less than the significance level (0.05), so we reject the null hypothesis. This indicates there is sufficient evidence to support the claim that the population mean of the differences is not zero.
06

Interpretation of Results

The rejection of the null hypothesis suggests that there is a statistically significant difference between the sample mean of the differences and zero. Therefore, the evidence supports the claim that the true population mean of the differences is different from zero.

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Key Concepts

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

Student's t-distribution
Student's t-distribution is a probability distribution used in statistical inferences when the sample size is small, and the population standard deviation is unknown. It is especially useful in hypothesis testing, as it helps estimate the probability of observing a calculated sample statistic. The t-distribution varies depending on the sample size, becoming more similar to the normal distribution as the sample size increases. It has heavier tails than the normal distribution, which accounts for the extra variability expected to come with smaller samples. When evaluating the significance of a sample mean, like in paired differences tests, the t-distribution provides a way to assess how likely a given mean difference would occur if the null hypothesis were true.
Degrees of freedom
Degrees of freedom ( extit{d.f.}) is a concept used to describe the number of values in a calculation that are free to vary. In the context of t-distribution, the degrees of freedom are calculated as the sample size minus one ( extit{n-1}). For the given exercise, a random sample of 36 data pairs means that the degrees of freedom is 36 - 1 = 35. It is crucial to use the correct degrees of freedom when referring to statistical tables or software to find critical values and p-values. Degrees of freedom play a vital role in hypothesis testing, as they influence the shape of the t-distribution. The larger the degrees of freedom, the closer the t-distribution resembles the normal distribution.
Null hypothesis
The null hypothesis ( extit{H}_0) is the statement we test in statistical hypothesis testing. It represents the assumption that there is no effect or difference, and it is the statement we seek to reject or fail to reject based on the sample evidence. In the exercise provided, the null hypothesis is that the population mean difference is zero, which means that any observed difference in sample means is due to random sampling variability. The null hypothesis is considered the default position, similar to the assumption of innocence before evidence is given. The aim in hypothesis testing is to collect data that either bolster this assumption or provide enough evidence to reject it.
Alternative hypothesis
The alternative hypothesis ( extit{H}_a) is the statement that contradicts the null hypothesis. It reflects the possibility that there is a real effect or difference. In hypothesis tests, the alternative hypothesis is what we aim to demonstrate. For the exercise at hand, the alternative hypothesis posits that the population mean difference is not zero. This would indicate that there is a significant difference in the means beyond mere chance. Choosing the correct alternative hypothesis is crucial as it drives the direction of the test (two-tailed, one-tailed), which in turn determines how we interpret the results on the basis of the p-value or critical value obtained from the test.

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Most popular questions from this chapter

When conducting a test for the difference of means for two independent populations \(x_{1}\) and \(x_{2},\) what alternate hypothesis would indicate that the mean of the \(x_{2}\) population is larger than that of the \(x_{1}\) population? Express the alternate hypothesis in two ways.

For one binomial experiment, 200 binomial trials produced 60 successes. For a second independent binomial experiment, 400 binomial trials produced 156 successes. At the \(5 \%\) level of significance, test the claim that the probability of success for the second binomial experiment is greater than that for the first. (a) Compute the pooled probability of success for the two experiments. (b) Check Requirements What distribution does the sample test statistic follow? Explain. (c) State the hypotheses. (d) Compute \(\hat{p}_{1}-\hat{p}_{2}\) and the corresponding sample distribution value. (e) Find the \(P\) -value of the sample test statistic. (f) Conclude the test. (g) Interpret the results.

When using the Student's \(t\) distribution to test \(\mu,\) what value do you use for the degrees of freedom?

To test \(\mu\) for an \(x\) distribution that is mound-shaped using sample size \(n \geq 30,\) how do you decide whether to use the normal or the Student's \(t\) distribution?

Basic Computation: Testing \(p\) A random sample of 30 binomials trials resulted in 12 successes. Test the claim that the population proportion of successes does not equal \(0.50 .\) Use a level of significance of 0.05 (a) Check Requirements Can a normal distribution be used for the \(\hat{p}\) distribution? Explain. (b) State the hypotheses. (c) Compute \(\hat{p}\) and the corresponding standardized sample test statistic. (d) Find the \(P\) -value of the test statistic. (e) Do you reject or fail to reject \(H_{0}\) ? Explain. (f) Interpretation What do the results tell you?

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