91Ó°ÊÓ

Please provide the following information for Problems. (a) What is the level of significance? State the null and alternate hypotheses. (b) Check Requirements What sampling distribution will you use? What assumptions are you making? Compute the sample test statistic and corresponding \(z\) or \(t\) value as appropriate. (c) Find (or estimate) the \(P\) -value. Sketch the sampling distribution and show the area corresponding to the \(P\) -value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \(\alpha ?\) (e) Interpret your conclusion in the context of the application. Note: For degrees of freedom \(d . f .\) not in the Student's \(t\) table, use the closest \(d . f .\) that is smaller. In some situations, this choice of \(d . f .\) may increase the \(P\) -value a small amount and therefore produce a slightly more "conservative" answer. Medical: REM Sleep REM (rapid eye movement) sleep is sleep during which most dreams occur. Each night a person has both REM and non-REM sleep. However, it is thought that children have more REM sleep than adults (Reference: Secrets of Sleep by Dr. A. Borbely). Assume that REM sleep time is normally distributed for both children and adults. A random sample of \(n_{1}=10\) children (9 years old) showed that they had an average REM sleep time of \(\bar{x}_{1}=2.8\) hours per night. From previous studies, it is known that \(\sigma_{1}=0.5\) hour. Another random sample of \(n_{2}=10\) adults showed that they had an average REM sleep time of \(\bar{x}_{2}=2.1\) hours per night. Previous studies show that \(\sigma_{2}=0.7\) hour. Do these data indicate that, on average, children tend to have more REM sleep than adults? Use a \(1 \%\) level of significance.

Step by step solution

01

Define Level of Significance and Hypotheses

The level of significance given is \( \alpha = 0.01 \). We need to state the null and alternative hypotheses. Let \( \mu_1 \) denote the mean REM sleep time for children and \( \mu_2 \) for adults. The null hypothesis \( H_0 \) is \( \mu_1 = \mu_2 \) (no difference in mean REM sleep time). The alternative hypothesis \( H_a \) is \( \mu_1 > \mu_2 \) (children have more REM sleep than adults).

02

Determine the Sampling Distribution and Assumptions

We will use the normal distribution (\( z \)-distribution) since the sample sizes are small (\( n_1 = n_2 = 10 \)), but the population standard deviations are known, making it suitable to use a Z-test. We assume that REM sleep time is normally distributed as per the problem statement.

03

Calculate the Test Statistic

The formula for the test statistic in a two-sample Z-test with known standard deviations is: \[ z = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]Substitute the given values:\[ z = \frac{(2.8 - 2.1)}{\sqrt{\frac{0.5^2}{10} + \frac{0.7^2}{10}}} = \frac{0.7}{\sqrt{0.025 + 0.049}} = \frac{0.7}{\sqrt{0.074}} = \frac{0.7}{0.272} \approx 2.574 \]

04

Find the P-value

Use the Z-table to find the P-value corresponding to \( z = 2.574 \). Checking the Z-table, a Z-value of 2.574 corresponds to a P-value of about 0.005. This is a one-tailed test since \( H_a \) suggests \( \mu_1 > \mu_2 \).

05

Make a Decision

Since the P-value (0.005) is less than the level of significance (0.01), we reject the null hypothesis \( H_0 \). This indicates that the data is statistically significant at the 1% level.

06

Interpret the Conclusion

In context, this means there is strong evidence to support the claim that, on average, children tend to have more REM sleep than adults, with the evidence being statistically significant at the 1% level.

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Key Concepts

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

Z-test

A Z-test is a statistical method used to determine if there is a significant difference between the means of two groups. In our REM sleep example, a Z-test is appropriate because we know the population standard deviations (given as \( \sigma_1 = 0.5 \) and \( \sigma_2 = 0.7 \)), and the sample sizes are not very large but sufficient (each sample having \( n = 10 \)).
The Z-test works by converting the difference between the sample means to a standard normal variable, called the Z-score. The formula for calculating this Z-score is:

• \( z = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \)

Once we have the Z-score, we compare it against a standard normal distribution to determine the probability of observing such a result, which helps in making a decision regarding the null hypothesis.
In this exercise, a calculated Z-score of approximately 2.574 was used to assess the REM sleep data.

Level of Significance

The Level of Significance is an important concept in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true. In simpler terms, it's risks you're willing to take to be wrong when concluding that a difference exists. In this exercise, we have set the level of significance at \( \alpha = 0.01 \), or 1%.
This implies that we only accept a 1% chance of committing a Type I error, which is rejecting a true null hypothesis. A 1% level is quite strict, indicating that we need very strong evidence from the data to reject the null hypothesis.
The choice of significance level often depends on the context. For high-stakes decisions, like medical or safety-related hypotheses, a low alpha level such as 0.01 is preferred, ensuring a higher level of certainty in support of the results.

P-value

A P-value is a number that helps you determine the strength of your results in a hypothesis test. It represents the probability of obtaining a test statistic as extreme as, or more extreme than, the observed value, assuming that the null hypothesis is true.
In this scenario, the P-value is derived from the Z-score. A lookup in the Z-table shows us that a Z-value of 2.574 corresponds approximately to a P-value of 0.005.
This tells us that there is a 0.5% chance of observing a difference as large as 0.7 hours (or more) in REM sleep just by random chance if there truly were no difference. Since the P-value (0.005) is less than our significance level of 0.01, it provides strong evidence against the null hypothesis, suggesting that the difference observed is unlikely due to random variation.
A small P-value like 0.005 suggests that the observed data is significant at the 1% level.

Null and Alternative Hypotheses

The null and alternative hypotheses are the cornerstones of hypothesis testing, providing a structured way to make inferences based on sample data. In our problem, we were tasked with comparing REM sleep between children and adults.
The null hypothesis \( H_0 \) states that there is no difference in the mean REM sleep time between children and adults:

• \( H_0: \mu_1 = \mu_2 \)

Conversely, the alternative hypothesis \( H_a \) expresses what we are trying to prove: children have more REM sleep than adults:

• \( H_a: \mu_1 > \mu_2 \)

The objective of hypothesis testing is to assess which of these statements is supported by the sample data. Based on the decision rule that compares the P-value to the significance level, we either reject or fail to reject the null hypothesis.
In this exercise, the evidence (P-value = 0.005) led us to reject the null hypothesis, indicating that there is a statistically significant difference in REM sleep, with children experiencing more than adults.