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The paper "Alcohol Consumption, Sleep, and Academic Performance Among College Students" (Journal of Studies on Alcohol and Drugs [2009]: 355-363) describes a study of \(n=236\) students who were randomly selected from a list of students enrolled at a liberal arts college in the northeastern region of the United States. Each student in the sample responded to a number of questions about their sleep patterns. For these 236 students, the sample mean time spent sleeping per night was reported to be 7.71 hours and the sample standard deviation of the sleeping times was 1.03 hours. Suppose that you are interested in learning about the value of \(\mu,\) the population mean time spent sleeping per night for students at this college. The following table is similar to the table that appears in Example 12.4 . The "what you know" information has been provided. Complete the table by filling in the "how you know it" column.

Short Answer

Expert verified
The mean time spent sleeping per night (\(\mu\)) for students at this college lies between 7.58 hours and 7.84 hours with a 95% confidence interval.

Step by step solution

01

Identify the sample mean, standard deviation, and sample size

The sample size is \(n = 236\), sample mean is \(\bar{x} = 7.71\) hours, and sample standard deviation is \(s = 1.03\) hours.
02

Calculate Standard Error

We need to calculate the standard error, which is defined as \(\frac{s}{\sqrt{n}}\). In this case, \[ SE = \frac{1.03}{\sqrt{236}} \approx 0.067 \]
03

Determine Confidence Interval

Let's assume we want to estimate the population mean with a 95% confidence interval. To find the confidence interval, we will use the formula: \[ CI = \bar{x} \pm Z_{\frac{\alpha}{2}} \times SE \] Where \(Z_{\frac{\alpha}{2}}\) is the z-score for the desired confidence level (1.96 for a 95% confidence interval). So in this case, \[ CI = 7.71 \pm 1.96 \times 0.067 \]
04

Calculate Margin of Error

The margin of error is given by the term \(Z_{\frac{\alpha}{2}} \times SE\), in this case, \[ Margin\:of\:Error = 1.96 \times 0.067 \approx 0.131 \]
05

Calculate the Confidence Interval

Now let's calculate the confidence interval for the population mean time spent sleeping per night: \[ CI_{lower} = 7.71 - 0.131 \approx 7.58 \\ CI_{upper} = 7.71 + 0.131 \approx 7.84 \] So we can say that the mean time spent sleeping per night (\(\mu\)) for students at this college lies between 7.58 hours and 7.84 hours with a 95% confidence interval.

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Key Concepts

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

Confidence Interval
When we talk about the confidence interval (CI), we are referring to a range of values, derived from sample statistics, that is likely to contain the value of an unknown population parameter. In simpler terms, it provides an estimated range where you expect the actual population mean might lie. The CI has an associated confidence level that quantifies the level of certainty we have that the interval includes the parameter. This is often expressed in percentage, such as '95% confidence interval', which means you can be 95% certain the population mean falls within your calculated interval.

The CI is calculated as the sample mean \bar{x} plus or minus a margin of error, which is determined by the desired confidence level and the standard error of the mean. The formula commonly used for the CI when the population standard deviation is unknown and the sample size is above 30 (so the sample distribution is approximately normally distributed) is:
\[CI = \bar{x} \pm Z_{\frac{\alpha}{2}} \times SE\]
Here, \(Z_{\frac{\alpha}{2}}\) is the Z-score, which reflects the number of standard deviations from the mean your confidence level allows. For instance, a 95% confidence interval uses a Z-score of 1.96, which corresponds to the probability that the true population parameter lies within 1.96 standard deviations of the sample mean.
Standard Error
The standard error (SE) is essentially a measure of the variability or spread in a sample statistic from a sample to the true population parameter. If you're looking at the mean, for instance, the standard error of the mean indicates how much discrepancy you might expect between the sample mean and the actual population mean due to random sampling variability. A smaller SE means your sample statistic is likely closer to the population parameter. The standard error gives insight into the precision of your sample estimate: the smaller it is, the more precise your estimate is likely to be.

The standard error is calculated by dividing the sample standard deviation \(s\) by the square root of the sample size \(n\), mathematically represented as:\[SE = \frac{s}{\sqrt{n}}\]
This indicates that as your sample size increases, the standard error decreases. This is because larger samples tend to be more representative of the population, thus yielding more precise estimates of the population parameter.
Sample Mean
The sample mean, often denoted as \(\bar{x}\), is the average of all the data points in a sample. It is calculated by summing all the observations in the sample and then dividing by the number of observations. The sample mean serves as an unbiased estimator for the population mean \(\mu\), assuming that the sample is randomly drawn from the population.

The sample mean is a very important statistic because it is used as a central value around which other calculations are made, particularly in estimating the population mean. It is also used to calculate the standard error and confidence intervals. The fact that it integrates information from every single data point in the sample makes it highly responsive to the presence of outliers, which can sometimes skew its representation of the central location of the data.
Sample Standard Deviation
The sample standard deviation (often denoted as \(s\)) measures the amount of variation or dispersion of a set of values in a sample. Unlike the sample mean, which provides a measure of central tendency, the standard deviation indicates how spread out the numbers are in your sample. It's the square root of the variance, which is the average of the squared differences from the Mean.

To calculate the sample standard deviation, you take each observation in your sample, subtract the sample mean, and square the result. Then sum up all those squared values, divide by the number of observations minus one (to correct for bias in a sample), and finally, take the square root of that. Mathematically, it looks like this:\[s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2}\]
The sample standard deviation is used in various calculations in statistics, including the standard error, and plays a key role in hypothesis testing, confidence intervals, and other statistical analyses.

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Most popular questions from this chapter

Students in a representative sample of 65 first-year students selected from a large university in England participated in a study of academic procrastination ("Study Goals and Procrastination Tendencies at Different Stages of the Undergraduate Degree," Studies in Higher Education [2016]: 2028-2043). Each student in the sample completed the Tuckman Procrastination Scale, which measures procrastination tendencies. Scores on this scale can range from 16 to \(64,\) with scores over 40 indicating higher levels of procrastination. For the 65 first-year students in this study, the mean score on the procrastination scale was 37.02 and the standard deviation was 6.44 . a. Construct a \(95 \%\) confidence interval estimate of \(\mu,\) the mean procrastination scale for first-year students at this college. (Hint: See Example 12.7.) b. Based on your interval, is 40 a plausible value for the population mean score? What does this imply about the population of first-year students?

The formula used to calculate a confidence interval for the mean of a normal population is $$ \bar{x} \pm(t \text { critical value }) \frac{s}{\sqrt{n}} $$ What is the appropriate \(t\) critical value for each of the following confidence levels and sample sizes? a. \(90 \%\) confidence, \(n=12\) b. \(90 \%\) confidence, \(n=25\) c. \(95 \%\) confidence, \(n=10\)

The report "Majoring in Money: How American College Students Manage Their Finances" (SallieMae, \(2016,\) www.news/salliemae.com, retrieved December 24,2106\()\) includes data from a survey of college students. Each person in a representative sample of 793 college students was asked if they had one or more credit cards and if so, whether they paid their balance in full each month. There were 500 who paid in full each month. For this sample of 500 students, the sample mean credit card balance was reported to be \(\$ 825 .\) The sample standard deviation of the credit card balances for these 500 students was not reported, but for purposes of this exercises, suppose that it was \(\$ 200\). Is there convincing evidence that college students who pay their credit card balance in full each month have mean balance that is lower than \(\$ 906\), the value reported for all college students with credit cards? Carry out a hypothesis test using a significance level of 0.01 .

A manufacturing process is designed to produce bolts with a diameter of 0.5 inches. Once each day, a random sample of 36 bolts is selected and the bolt diameters are recorded. If the resulting sample mean is less than 0.49 inches or greater than 0.51 inches, the process is shut down for adjustment. The standard deviation of bolt diameters is 0.02 inches. What is the probability that the manufacturing line will be shut down unnecessarily? (Hint: Find the probability of observing an \(\bar{x}\) in the shutdown range when the actual process mean is 0.5 inches.)

The time that people have to wait for an elevator in an office building has a uniform distribution over the interval from 0 to 1 minute. For this distribution, \(\mu=0.5\) and \(\sigma=0.289\) a. If \(\bar{x}\) is the average waiting time for a random sample of \(n=16\) waiting times, what are the values of the mean and standard deviation of the sampling distribution of \(\bar{x} ?\) b. Answer Part (a) for a random sample of 50 waiting times. Draw a picture of the approximate sampling distribution of \(\bar{x}\) when \(n=50\).

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