/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 144 It is generally recommended that... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(6.18 .\) The 12 students sampled averaged $$ \begin{aligned} &\text { 40 Belguermi, A., "Pigeons discriminate between human feeders," }\\\ &\text { Animal Cognition, } 2011 ; 14: 909-914 . \end{aligned} $$ 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
The precise conclusion depends on the calculation of the t-value and comparison with the critical value. Unfortunately, without a computational tool at hand, the exact answer can't be provided right now. However, the direction for the calculation and interpretation has been thoroughly explained in the steps above.

Step by step solution

01

State the Null and Alternative Hypotheses

The null hypothesis (H0) is: µ = 8 (the students are getting an average of 8 hours sleep per night). The alternative hypothesis (HA) is: µ < 8 (the students are getting less than 8 hours sleep per night) 'µ' refers to the population mean.
02

Calculate the Test Statistic

To conduct a t-test, compute the t-value using the formula: t = (x̄ - µ) / (s /√n) where x̄ is the sample mean (6.2 hours), µ is the hypothesized population mean (8 hours), s is the standard deviation (1.70 hours), and n is the sample size (12).
03

Determine the Degrees of Freedom

Degrees of freedom (df) are calculated as n - 1, where n is the sample size. Hence, in this case, df = 12 - 1 = 11.
04

Find the Critical Value

From a t-distribution table, using a significance level of 0.05 (commonly used), look for the t-value associated with the calculated df (11 in this case) and one-tailed test (since the alternative hypothesis is µ < 8). This will be the critical value.
05

Compare the Test Statistic to the Critical Value

If the absolute value of the calculated t-value is greater than the critical value, reject the null hypothesis. Otherwise, there isn't sufficient evidence to reject it.
06

Conclude

Based on the comparison made in step 5, make a conclusion about whether there is enough evidence to support the claim that students are not getting sufficient sleep.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

t-test
The t-test is a statistical method used to determine if there is a significant difference between the means of two groups. It's particularly useful when dealing with small sample sizes, as it accounts for the variability and distribution of the sample data.

In the context of our problem, we want to see if the average sleep duration of students (6.2 hours) is significantly different from the recommended 8 hours. Essentially, we are checking if this difference is due to random chance or if it truly reflects a lower sleep amount for the population (all students).

The formula to calculate the t-statistic in a one-sample t-test is:
\[ t = \frac{(\bar{x} - \mu)}{(s / \sqrt{n})} \]
Here, \(\bar{x}\) is the sample mean, \(\mu\) is the population mean (hypothesized value), \(s\) is the sample standard deviation, and \(n\) is the sample size.
  • \(\bar{x} = 6.2\)
  • \(\mu = 8\)
  • \(s = 1.70\)
  • \(n = 12\)
This statistical tool helps us determine whether the observed sample mean significantly deviates from the expected population mean.
Null and Alternative Hypotheses
In hypothesis testing, we start by stating two hypotheses; the null hypothesis and the alternative hypothesis.

The **null hypothesis** (H0) generally reflects the status quo or a statement of 'no effect'. It's what we assume to be true until we have evidence to suggest otherwise. In this sleep study, the null hypothesis states that students, on average, are getting 8 hours of sleep per night, expressed as:
\[ H_0: \mu = 8 \]

The **alternative hypothesis** (HA), on the other hand, represents what we want to prove. It suggests that a significant effect exists. Here, it claims that students, on average, are getting less than the stipulated 8 hours:
\[ H_A: \mu < 8 \]

The process is a bit like a courtroom trial, where the null hypothesis is assumed true unless the evidence collected from the data suggests with a high degree of confidence that we should reject it.
Degrees of Freedom
Degrees of freedom are an essential concept in statistics, particularly when working with sample data. They are associated with the number of independent values or quantities that can vary in an analysis, without violating any given constraints.

In the context of a t-test, degrees of freedom are important for determining the critical value from a t-distribution table. The degrees of freedom in a one-sample t-test are calculated using the formula:
\[ df = n - 1 \]
where \(n\) is the sample size. In our example of sleep data, \(n = 12\), thus:
  • \(df = 12 - 1 = 11\)

The degrees of freedom determine the shape of the t-distribution curve. This, in turn, helps to identify the critical t-value, which is used to decide whether to reject the null hypothesis. A higher number of degrees of freedom results in a t-distribution that closely approximates a normal distribution. This convergence is why larger sample sizes provide more reliable results.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

We see in the AllCountries dataset that the percent of the population that is over 65 is 13.4 in Australia and 12.5 in New Zealand. Suppose we take random samples of size 500 from Australia and size 300 from New Zealand, and compute the difference in sample proportions \(\hat{p}_{A}-\hat{p}_{N Z},\) where \(\hat{p}_{A}\) represents the sample proportion of elderly in Australia and \(\hat{p}_{N Z}\) represents the sample proportion of elderly in New Zealand. Find the mean and standard deviation of the differences in sample proportions.

Standard Error from a Formula and a Bootstrap Distribution In Exercises 6.19 to \(6.22,\) use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of survey respondents who say exercise is important, with \(n=1000\) and \(\hat{p}=0.753\)

Exercise B.5 on page 305 introduces a study examining the effect of diet cola consumption on calcium levels in women. A sample of 16 healthy women aged 18 to 40 were randomly assigned to drink 24 ounces of either diet cola or water. Their urine was collected for three hours after ingestion of the beverage and calcium excretion (in mg) was measured. The summary statistics for diet cola are \(\bar{x}_{C}=56.0\) with \(s_{C}=4.93\) and \(n_{C}=8\) and the summary statistics for water are \(\bar{x}_{W}=49.1\) with \(s_{W}=3.64\) and \(n_{W}=8\). Figure 6.26 shows dotplots of the data values. Test whether there is evidence that diet cola leaches calcium out of the system, which would increase the amount of calcium in the urine for diet cola drinkers. In Exercise B.5, we used a randomization distribution to conduct this test. Use a t-distribution here, after first checking that the conditions are met and explaining your reasoning. The data are stored in ColaCalcium.

A \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired data in the following table: $$ \begin{array}{lcc} \hline \text { Case } & \text { Situation 1 } & \text { Situation 2 } \\ \hline 1 & 77 & 85 \\ 2 & 81 & 84 \\ 3 & 94 & 91 \\ 4 & 62 & 78 \\ 5 & 70 & 77 \\ 6 & 71 & 61 \\ 7 & 85 & 88 \\ 8 & 90 & 91 \\ \hline \end{array} $$

Random samples of the given sizes are drawn from populations with the given means and standard deviations. For each scenario: (a) Find the mean and standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 25 from Population 1 with mean 6.2 and standard deviation 3.7 and samples of size 40 from Population 2 with mean 8.1 and standard deviation 7.6

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.