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People in a random sample of 236 students enrolled at a liberal arts college were asked questions about how much sleep they get each night ("Alcohol Consumption, Sleep, and Academic Performance Among College Students," Journal of Studies on Alcohol and Drugs [2009]: 355-363). The sample mean sleep duration (average hours of daily sleep) was 7.71 hours and the sample standard deviation was 1.03 hours. The recommended number of hours of sleep for college-age students is 8.4 hours per day. Is there convincing evidence that the population mean daily sleep duration for students at this college is less than the recommended number of 8.4 hours? Test the relevant hypotheses using \(\alpha\) \(=0.01\)

Short Answer

Expert verified
At a significance level of 伪 = 0.01, there is convincing evidence that the population mean daily sleep duration for students at this college is less than the recommended number of 8.4 hours, as the t-value (-9.338) is less than the critical value (-2.344).

Step by step solution

01

State the null and alternative hypotheses

Null hypothesis (H鈧): 渭 = 8.4 hours Alternative hypothesis (H鈧): 渭 < 8.4 hours
02

Determine the t-value

To calculate the t-value, we will use the formula: \[t = \frac{\bar{x} - 渭鈧}{s/鈭歯}\] where - \(\bar{x}\) is the sample mean, - \(渭鈧\) is the hypothesized population mean, - s is the sample standard deviation, and - n is the sample size. Using the given information: - \(\bar{x} = 7.71\), - \(渭鈧 = 8.4\), - s = 1.03, and - n = 236. Calculate the t-value: \[t = \frac{7.71 - 8.4}{1.03/鈭236} 鈮 -9.338\]
03

Determine the critical value

We are given a significance level of 伪 = 0.01, and this is a one-tailed test because our alternative hypothesis is 渭 < 8.4 hours. To find the critical value, we need to determine the degrees of freedom, which is: df = n - 1 = 236 - 1 = 235 Using the t-distribution table, we find the critical value for df = 235 and 伪 = 0.01: t_critical 鈮 -2.344
04

Compare the t-value to the critical value

Now, we compare the calculated t-value to the critical value: - t-value = -9.338 - t_critical = -2.344 Since t-value < t_critical (-9.338 < -2.344), we reject the null hypothesis.
05

Conclusion

Considering the significance level of 伪 = 0.01, there is convincing evidence that the population mean daily sleep duration for students at this college is less than the recommended number of 8.4 hours.

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Key Concepts

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

Student's t-test
The Student's t-test is a powerful statistical tool used to determine if there is a significant difference between the means of two groups. It is particularly useful when the sample size is small and the population standard deviation is unknown. In our example, we use the Student's t-test to compare the average sleep duration of a sample of college students to the recommended 8.4 hours.

In the t-test, we calculate a t-value using our sample data. This involves comparing the sample mean to a hypothesized population mean and accounting for the variability in the data, measured by the sample standard deviation. The formula for the t-value is as follows:
\[t = \frac{\bar{x} - 渭鈧}{s/\sqrt{n}}\]
  • \(\bar{x}\) is the sample mean
  • \(渭鈧\) is the hypothesized population mean
  • \(s\) is the sample standard deviation
  • \(n\) is the sample size
By computing this t-value, we can subsequently compare it to critical values from the t-distribution to determine if the difference is statistically significant.
Null Hypothesis
The null hypothesis, often denoted as \(H鈧\), is a statement that there is no effect or no difference in the context of hypothesis testing. It serves as a starting point for statistical analysis and is presumed true until evidence suggests otherwise.

In our sleep duration example, the null hypothesis is that the average daily sleep duration for the college students is equal to the recommended 8.4 hours (\(H鈧: \mu = 8.4\)).
We test this hypothesis by looking for evidence to the contrary. The aim is to determine whether there is sufficient evidence to suggest that the true average sleep duration is different from the claimed 8.4 hours, specifically less than this value as per our alternative hypothesis. If the results of the t-test demonstrate that the null hypothesis is unlikely, we reject it in favor of the idea that the mean sleep duration is indeed less than 8.4 hours.
Significance Level
The significance level, represented by \(\alpha\), is a threshold set by the researcher that determines the probability of rejecting the null hypothesis when it is actually true. It effectively defines how rigorous the test is. A common choice for \(\alpha\) is 0.05, but it can vary depending on how strict we want to be. In our test, we use a significance level of 0.01.

Choosing a lower significance level, like 0.01, means that we require even stronger evidence to reject the null hypothesis. This reduces the likelihood of making a Type I error, which is when we falsely declare a difference or effect that doesn鈥檛 actually exist.
In practical terms, a significance level of 0.01 indicates that there is only a 1% risk of concluding that there is an effect when there is none. By setting such a high standard, our conclusions can be made with greater confidence.
One-tailed Test
A one-tailed test is used when we are interested in detecting an effect in a specific direction. It is ideal when the research hypothesis states that one group is expected to be greater or lesser than the other. In our sleep study, the alternative hypothesis (\(H鈧:\mu < 8.4\)) suggests we are looking to see if students sleep less than the recommended amount.

The one-tailed test only considers the area in one tail of the distribution curve, providing a more powerful test for detecting specific differences. It has more statistical power to detect effects in one direction compared to a two-tailed test, which checks for differences in both directions.
By focusing on whether the sample mean is less than the specified population mean, we are only interested in evidence that supports this particular result. This focus on directionality allows for a clearer decision on whether the null hypothesis should be rejected based on the calculated t-value and its relation to a critical threshold, which in our case, indicated evidence for a lower average sleep duration.

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Most popular questions from this chapter

The Economist collects data each year on the price of a Big Mac in various countries around the world. A sample of McDonald's restaurants in Europe in July 2016 resulted in the following Big Mac prices (after conversion to U.S. dollars): \(\begin{array}{llll}4.44 & 3.15 & 2.42 & 3.96\end{array}\) \(\begin{array}{llll}4.51 & 4.17 & 3.69 & 4.62\end{array}\) \(\begin{array}{lll}3.80 & 3.36 & 3.85\end{array}\) The mean price of a Big Mac in the U.S. in July 2016 was \$5.04. For purposes of this exercise, you can assume it is reasonable to regard the sample as representative of European McDonald's restaurants. Does the sample provide convincing evidence that the mean July 2016 price of a Big Mac in Europe is less than the reported U.S. price? Test the relevant hypotheses using \(\alpha=0.05 .\) (Hint: See Example 12.12.)

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