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According to a recent survey, Americans get a mean of 7 hours of sleep per night. A random sample of 50 students at West Virginia University revealed the mean number of hours slept last night was 6 hours and 48 minutes \((6.8\) hours \() .\) The standard deviation of the sample was 0.9 hours. Is it reasonable to conclude that students at West Virginia sleep less than the typical American? Compute the \(p\) -value.

Short Answer

Expert verified
No, the p-value (0.0582) indicates it's not reasonable to conclude WVU students sleep less.

Step by step solution

01

Define the Hypotheses

We formulate our null and alternative hypotheses as follows:- Null Hypothesis (\(H_0\)): The mean number of hours students at WVU sleep is equal to 7 hours. - Alternative Hypothesis (\(H_a\)): The mean number of hours students at WVU sleep is less than 7 hours.
02

Define the Test Statistic

Use the formula for the test statistic for a sample mean:\[z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\]Here, \(\bar{x} = 6.8\) hours, \(\mu = 7\) hours, \(\sigma = 0.9\) hours, and \(n = 50\).
03

Calculate the Test Statistic

Substitute the values into the formula:\[z = \frac{6.8 - 7}{\frac{0.9}{\sqrt{50}}} = \frac{-0.2}{\frac{0.9}{7.07}} = \frac{-0.2}{0.127} \approx -1.57\]
04

Determine the p-value

Using z-tables or a calculator, find the p-value for \(z = -1.57\). Since it's a left-tailed test, the p-value is the area to the left of \(z = -1.57\) in the standard normal distribution. This area (p-value) is approximately 0.0582.
05

Make a Conclusion

Compare the p-value to the significance level, typically \(\alpha = 0.05\). Since the p-value (0.0582) is greater than \(\alpha = 0.05\), we do not reject the null hypothesis. It's not reasonable to conclude that WVU students sleep less than the typical American based on this sample.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis is like the starting assumption you aim to test. It is a statement that suggests there is no effect or difference. For this exercise, the null hypothesis (\(H_0\)) states that the mean number of hours students at West Virginia University (WVU) sleep is equal to 7 hours, just like the average American. The null hypothesis is an essential component as it serves as the basis for statistical testing. It proposes that any observed variation is due to random chance rather than a significant effect or difference. In practical terms, the null hypothesis is assumed true until we have enough evidence to suggest otherwise. Often symbolized as "equals to," it acts as the benchmark against which the test results are compared.
Alternative Hypothesis
The alternative hypothesis offers a contrasting perspective to the null hypothesis. Instead of assuming no change or effect, it suggests that a specific condition or difference exists. In this scenario, the alternative hypothesis (\(H_a\)) posits that students at WVU sleep less than the typical American, or in other words, the mean sleep duration is less than 7 hours.The role of the alternative hypothesis is to show that there is an effect or difference worth examining. It's the claim that the researcher truly believes in and wants to verify statistically. The ultimate goal of hypothesis testing is often to gather enough evidence to accept the alternative hypothesis by rejecting the null hypothesis.
p-value
The p-value is a critical measure in hypothesis testing that helps you determine the strength of the evidence against the null hypothesis. It tells you how likely it is to observe the test results, or something more extreme, assuming that the null hypothesis is true.In this exercise, after calculating the z-score, we found the p-value to be 0.0582. This means there is a 5.82% chance of observing the sample mean of 6.8 hours or less if students at WVU truly sleep 7 hours like the typical American.A smaller p-value signals stronger evidence against the null hypothesis. Traditionally, a p-value less than the significance level (\(\alpha = 0.05\)) indicates sufficient evidence to reject the null hypothesis. Since our p-value of 0.0582 is greater than 0.05, it suggests that the evidence isn't strong enough to state confidently that students at WVU sleep less than the average American.
z-test
The z-test is a statistical method used to determine if there is a significant difference between the sample mean and a known population mean. It's particularly useful when the sample size is large, typically over 30, as it assumes a normal distribution of the sample means.For this exercise, we used the z-test to evaluate whether the mean sleep duration for WVU students is less than 7 hours. We calculated the z-score, which compares the sample mean (\(\bar{x} = 6.8\) hours) to the population mean (\(\mu = 7\) hours), taking into account the standard deviation (\(\sigma = 0.9\) hours) and sample size (\(n = 50\)).The resulting z-score of -1.57 tells us how far the sample mean is from the population mean in terms of standard errors. With this z-score, we determine the probability of observing such a value under the null hypothesis, which leads us to the p-value.
Sample Mean
The sample mean is the average value of a set of data collected from a sample, serving as an estimate of the population mean. In this exercise, the sample mean is calculated from the sleep data of 50 WVU students, resulting in a mean sleep time of 6.8 hours. Understanding the sample mean is crucial for hypothesis testing as it plays a central role in the z-test. The sample mean is compared to the population mean to see if there is a statistically significant difference. It's important to note that while the sample mean gives a snapshot of the sample data, the reliability of the sample mean as a true reflection of the entire population improves with larger sample sizes and less variability in the sample data. This makes the statistical findings more robust and reliable.

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Most popular questions from this chapter

The liquid chlorine added to swimming pools to combat algae has a relatively short shelf life before it loses its effectiveness. Records indicate that the mean shelf life of a 5 -gallon jug of chlorine is 2,160 hours (90 days). As an experiment, Holdlonger was added to the chlorine to find whether it would increase the shelf life. A sample of nine jugs of chlorine had these shelf lives (in hours): $$\begin{array}{lllllllll}2,159 & 2,170 & 2,180 & 2,179 & 2,160 & 2,167 & 2,171 & 2,181 & 2,185 \\\\\hline\end{array}$$ At the .025 level, has Holdionger increased the shelf life of the chlorine? Estimate the \(p\) -value.

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