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Chapter 12: Problem 24

The paper referenced in the previous exercise also reported that for a representative sample of 68 second-year students at the university, the sample mean procrastination score was 41.00 and the sample standard deviation was 6.82 . a. Construct a \(95 \%\) confidence interval estimate of \(\mu,\) the population mean procrastination scale for second-year students at this college. b. How does the confidence interval for the population mean score for second- year students compare to the confidence interval for first-year students calculated in the previous exercise? What does this tell you about the difference between first-year and second-year students in terms of mean procrastination score?

Short Answer

Expert verified
a. The \(95\%\) confidence interval estimate of the population mean procrastination scale for second-year students at this college is \([39.34, 42.66]\). b. Compare the confidence intervals of first-year and second-year students. If there is no overlap between the intervals, the difference between the two groups in terms of mean procrastination scores is statistically significant. If the intervals overlap, we cannot conclude that there is a substantial difference between the two groups in terms of mean procrastination scores.

Step by step solution

01

Define the terms and symbols

We need to compute the \(95\%\) confidence interval for the population mean, \(\mu\). To do that, we need to note the following information: Sample size, n: \(68\) Sample mean, \(\bar{x}\): \(41.00\) Sample standard deviation, s: \(6.82\) Confidence level, \(C\): \(95\%\) (which corresponds to \(1-\alpha=0.95\) and therefore, \(\alpha = 0.05\))
02

Find the critical value for the t-distribution (t*)

Since we do not know the population standard deviation, we will use Student's t-distribution. Our degrees of freedom (df) for this problem will be \(n - 1 = 68 - 1 = 67\). To find the critical value (t*), we will look up the t-distribution table for a two-tailed test with \(\alpha/2 = 0.025\) and df = 67. The critical value we find is \(t^*=2.000\).
03

Calculate the margin of error

To calculate the margin of error (E) for the \(95\%\) confidence interval, we use the formula: E = \(t^* \cdot \frac{s}{\sqrt{n}}\) Plugging in the values, we get: E = \(2.000 \cdot \frac{6.82}{\sqrt{68}} \approx 1.66\)
04

Compute the confidence interval

Now that we have the margin of error, we can compute the \(95\%\) confidence interval for the population mean \(\mu\). Subtract and add the margin of error (E) to the sample mean (\(\bar{x}\)). Lower bound: \(\bar{x} - E = 41.00 - 1.66 = 39.34\) Upper bound: \(\bar{x} + E = 41.00 + 1.66 = 42.66\) The confidence interval is: \([39.34, 42.66]\) a. The \(95\%\) confidence interval estimate of the population mean procrastination scale for second-year students at this college is \([39.34, 42.66]\).
05

Compare the confidence intervals between first-year and second-year students

b. To compare the confidence intervals, consider the confidence interval calculated for first-year students in the previous exercise. Let's say that the confidence interval for first-year students was \([a, b]\). We need to compare both confidence intervals and analyze whether there is an overlap or not. If there is no overlap between the intervals, it means the difference between first-year and second-year students in terms of mean procrastination scores is statistically significant. On the other hand, if there is an overlap, we cannot conclude that there is a substantial difference between the two groups in terms of mean procrastination scores. For instance, if the first-year students interval is entirely below or above the second-year students interval (e.g., \([35, 38]\) or \([45, 50]\)), there is a significant difference. If the intervals overlap (e.g., \([40, 45]\)), we cannot make conclusions about the difference between the two groups in terms of mean procrastination scores.

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Key Concepts

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

Sample Mean
When we discuss statistics, particularly in the context of confidence intervals, the term 'sample mean' refers to the average value obtained from a sample of the population. In simple terms, if you were to measure the procrastination scores of a group of second-year college students, add up their scores, and divide by the total number of students surveyed, you'd get the sample mean. It is denoted by \( \bar{x} \).

In our exercise, the sample mean is 41.00. This is an essential piece of data to calculate the confidence interval, as it serves as the center point around which we estimate the population mean. While the sample mean provides a snapshot of the data collected, it's important to remember it may not perfectly represent the whole population's average value due to sampling variability.
Sample Standard Deviation
The sample standard deviation, symbolized by \( s \), measures the amount of variation or dispersion from the sample mean. Think of it as a way to quantify how spread out the procrastination scores are among the second-year students.

In our textbook exercise, a sample standard deviation of 6.82 indicates there is a variety in how much students procrastinate. A lower standard deviation would suggest that most students' scores are close to the sample mean, while a higher value indicates that scores are more spread out. The sample standard deviation is crucial in confidence interval estimation as it directly influences the width of the interval, representing the precision of our estimate for the population mean.
Student's t-distribution
When we're estimating confidence intervals and the population standard deviation is unknown, we use the Student's t-distribution. This distribution is essential for small sample sizes and where the data follow a roughly normal distribution. Unlike the normal distribution, which has the same shape regardless of the sample size, the shape of the t-distribution changes with the degrees of freedom associated with the sample.

In our example, since the population standard deviation is not known and we're working with a sample size of 68, which is not large enough to assume the distribution of the sample mean is normal, we appeal to the t-distribution, hence the critical value \( t^* \) obtained in the solution.
Degrees of Freedom
The concept of degrees of freedom, often abbreviated as \( df \) in statistics, is tied closely to the precision of an estimate. Degrees of freedom refer to the number of independent values in a calculation that are free to vary. For a sample standard deviation calculation, the degrees of freedom are equal to the sample size minus one, \( df = n - 1 \), because one parameter (the sample mean) is used to estimate another (the sample standard deviation).

In the provided exercise, the degrees of freedom are 67 (68 minus 1). The value of \( df \) affects the shape of the t-distribution and is used to determine the critical t-value when constructing confidence intervals.
Margin of Error
The margin of error is a statistic expressing the amount of random sampling error in a survey's results. It represents the radius of the confidence interval for a given statistic. The formula for calculating the margin of error in the context of a 95% confidence interval using the t-distribution is \( t^* \times \frac{s}{\sqrt{n}} \).

For our exercise, we used the sample standard deviation (6.82) and sqrt of the sample size (68) along with the critical value from the t-distribution (2.00) to calculate a margin of error of approximately 1.66. This value indicates that we are 95% confident that the true population mean procrastination score is within 1.66 points of the sample mean (41.00) we observed. In other words, the range of 39.34 to 42.66 captures the population mean with a 95% confidence level.

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Most popular questions from this chapter

A student organization uses the proceeds from a soft drink vending machine to finance its activities. The price per can was \(\$ 0.75\) for a long time, and the mean daily revenue during that period was \(\$ 75.00\). The price was recently increased to \(\$ 1.00\) per can. A random sample of \(n=20\) days after the price increase yielded a sample mean daily revenue and sample standard deviation of \(\$ 70.00\) and \(\$ 4.20\), respectively. Does this information suggest that the mean daily revenue has decreased from its value before the price increase? Test the appropriate hypotheses using \(\alpha=\) \(0.05 .\)

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 .

Acrylic bone cement is sometimes used in hip and knee replacements to secure an artificial joint in place. The force required to break an acrylic bone cement bond was measured for six specimens, and the resulting mean and standard deviation were 306.09 Newtons and 41.97 Newtons, respectively. Assuming that it is reasonable to believe that breaking force has a distribution that is approximately normal, use a confidence interval to estimate the mean breaking force for acrylic bone cement.

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 that 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 additional time spent sleeping on weekend days compared to the other days of the week was reported to be 1.29 hours and the standard deviation was 1.09 hours. Suppose that you are interested in learning about the value of \(\mu,\) the mean additional time spent sleeping on weekend days 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.

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 \(\alpha\) \(=0.01\)
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