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Getting Enough Sleep? It is generally recommended that adults sleep at least 8 hours each night. One of the authors recently asked some of her students (undergraduate and graduate students at Harvard) how many hours each had slept the previous night, curious as to whether her students are getting enough sleep. The data are displayed in Figure 6.13 . The 12 students sampled averaged 6.2 hours of sleep with a standard deviation of 1.70 hours, Assuming this sample is representative of all her students, and assuming students need at least 8 hours of sleep a night, does this provide evidence that, on average, her students are not getting enough sleep?

Short Answer

Expert verified
To conclude whether the students are getting enough sleep, one needs to calculate the test statistic using the given sample mean, standard deviation, and sample size and compare this calculated t-value to the critical t-value for a specified confidence level. Decision will be made based on this comparison.

Step by step solution

01

Define the Hypotheses

The null hypothesis \(H_0\) is that the average student gets enough sleep, expressed as \(H_0: \mu = 8\). The alternative hypothesis \(H_a\), which we are testing, is that the average student does not get enough sleep, expressed as \(H_a: \mu < 8\) . Here, \( \mu \) represents the population mean.
02

Compute the Test Statistic

The test statistic for a one-sample mean test is a t-score, calculated using the formula \(t = \frac{ \overline{X} - \mu_{H_0}}{s / \sqrt{n}}\). Here, \(\overline{X} = 6.2\) is the sample mean, \( \mu_{H_0} = 8\) is the mean under the null hypothesis, \(s = 1.70\) is the sample standard deviation, and \(n = 12\) is the sample size. Plugging these values into the formula, we get \(t = \frac{6.2 - 8}{1.70 / \sqrt{12}}\).
03

Find the Critical Value

The critical value corresponds to a significance level (or alpha) of 0.05 with a sample size of 11 degrees of freedom (since df = n - 1). You can find this value on the t-distribution table or using a t-distribution calculator online. This will be the boundary which will determine whether we reject \(H_0\) or fail to reject \(H_0\).
04

Make the Decision

If the calculated test statistic is less than the critical value, then we reject the null hypothesis, confirming the alternative hypothesis. If not, we fail to reject the null hypothesis, meaning we do not have enough evidence to suggest the students are not getting enough sleep.

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Key Concepts

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

t-distribution
In statistical analyses, the t-distribution is often used when dealing with small sample sizes or when the population standard deviation is unknown. It's particularly handy in hypothesis testing for means.

The t-distribution is essential because it helps us account for variability in small samples.
  • It's similar to a normal distribution but with heavier tails. This means it accommodates more variability and works well when sample sizes are small.
  • As the sample size increases, the t-distribution approaches the normal distribution.
  • Degrees of freedom play a crucial role. In the example, since we have 12 students, our degrees of freedom are 11 (because df = n - 1).
This concept is crucial in determining the critical values for our t-test, allowing us to make informed decisions about our hypotheses.
null hypothesis
The null hypothesis is a fundamental part of hypothesis testing. It's a statement that indicates no effect or no difference, serving as a starting point for our analysis.

In our sleep study:
- The null hypothesis (\(H_0\)) states that students, on average, get 8 hours of sleep.- Symbolically, it is presented as \(H_0: \mu = 8\). Here, \( \mu \) represents the population mean.The null hypothesis is what we initially assume to be true. Our goal is to test whether this assumption can be rejected based on the data collected from the sample.

The null hypothesis serves as a baseline, and any departure from it observed in the data, if significant enough, leads us to consider the alternative hypothesis.
alternative hypothesis
The alternative hypothesis is the statement we are hoping to find evidence for through our statistical test. It represents the effect or difference we suspect is true.

In our example about sleep:- The alternative hypothesis (\(H_a\)) suggests that students, on average, get less than 8 hours of sleep.- It is expressed as \(H_a: \mu < 8\).This hypothesis stands in contrast to the null hypothesis and is only considered if the evidence against the null hypothesis is strong enough.

Our test will help us decide whether there is enough statistical evidence to support the claim that students are not getting the recommended amount of sleep.
one-sample mean test
A one-sample mean test is a statistical method used to determine if the sample mean is significantly different from a hypothesized population mean.

This test involves several steps:
  • We first define our null and alternative hypotheses, like in our sleep study.
  • Next, we calculate the test statistic, in this case, a t-score. The formula is given by \(t = \frac{ \overline{X} - \mu_{H_0}}{s / \sqrt{n}}\), where \(\overline{X}\) is the sample mean, \(\mu_{H_0}\) is the mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size.
  • We then determine the critical value using a t-distribution table or calculator, accounting for the degrees of freedom.
  • The decision is made by comparing the test statistic to the critical value. If the calculated statistic falls in the critical region, we reject the null hypothesis.
This test is perfect for assessing one group's mean against a known value, providing insight into whether there's enough evidence to suggest a significant difference.

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Most popular questions from this chapter

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Autism and Maternal Antidepressant Use A recent study \(^{41}\) compared 298 children with Autism Spectrum Disorder to 1507 randomly selected control children without the disorder. Of the children with autism, 20 of the mothers had used antidepressant drugs during the year before pregnancy or the first trimester of pregnancy. Of the control children, 50 of the mothers had used the drugs. (a) Is there a significant association between prenatal exposure to antidepressant medicine and the risk of autism? Test whether the results are significant at the \(5 \%\) level. (b) Can we conclude that prenatal exposure to antidepressant medicine increases the risk of autism in the child? Why or why not? (c) The article describing the study contains the sentence "No increase in risk was found for mothers with a history of mental health treatment in the absence of prenatal exposure to selective serotonin reuptake inhibitors [antidepressants]." Why did the researchers conduct this extra analysis?

Use the t-distribution to find a confidence interval for a difference in means \(\mu_{1}-\mu_{2}\) given the relevant sample results. Give the best estimate for \(\mu_{1}-\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed. A \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the sample results \(\bar{x}_{1}=5.2, s_{1}=2.7, n_{1}=10\) and \(\bar{x}_{2}=4.9, s_{2}=2.8, n_{2}=8 .\)

We examine the effect of different inputs on determining the sample size needed to obtain a specific margin of error when finding a confidence interval for a proportion. Find the sample size needed to give a margin of error to estimate a proportion within \(\pm 3 \%\) with \(99 \%\) confidence. With \(95 \%\) confidence. With \(90 \%\) confidence. (Assume no prior knowledge about the population proportion \(p\).) Comment on the relationship between the sample size and the confidence level desired.

Use the t-distribution and the given 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 distributions are relatively normal. Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2}\) using the sample results \(\bar{x}_{1}=15.3, s_{1}=11.6\) with \(n_{1}=100\) and \(\bar{x}_{2}=18.4, s_{2}=14.3\) with \(n_{2}=80\).

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