/*! 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 30 Travel Time to School A random s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Travel Time to School A random sample of \(50 \mathrm{1} 2\) th-grade students was asked how long it took to get to school. The sample mean was \(16.2\) minutes, and the sample standard deviation was \(12.3\) minutes. (Source: AMSTAT Census at School) a. Find a \(95 \%\) confidence interval for the population mean time it takes 12 th-grade students to get to school. b. Would a \(90 \%\) confidence interval based on this sample data be wider or narrower than the \(95 \%\) confidence interval? Explain. Check your answer by constructing a \(90 \%\) confidence interval and comparing this width of the interval with the width of the \(95 \%\) confidence interval you found in part a.

Short Answer

Expert verified
The 95% confidence interval for the population mean time it takes 12th-grade students to get to school is approximately 12.79 to 19.61 minutes. A 90% confidence interval would be narrower and is found to be around 13.34 to 19.06 minutes.

Step by step solution

01

Find the z-score for the 95% confidence interval

Z-scores for a 95% confidence interval can be found in a standard z-table, or calculated using online resources. A 95% confidence interval is \(\pm 1.96\) standard deviations away from the mean in a normal distribution.
02

Compute the standard error

The standard error can be calculated by dividing the standard deviation (\(12.3\) minutes) by the square root of the sample size (\(50\)). Thus, \( SE = \frac{12.3}{\sqrt{50}} = 1.74\) minutes.
03

Find the 95% confidence interval

The formula to calculate a confidence interval is \( CI = \mu \pm (Z * SE) \). But for here, substitute \(\mu = 16.2\) minutes, \(Z = 1.96\), and \(SE = 1.74\) minutes, which gives \( CI = 16.2 \pm (1.96 * 1.74) \). Therefore, the 95% confidence interval is \( 16.2 \pm 3.41 \) minutes, or \(12.79 to 19.61\) minutes.
04

Find the z-score for the 90% confidence interval

As with the 95% interval, a z-table or online resources can be used to locate the z-score for a 90% confidence interval. A 90% confidence interval corresponds to a z-score of \(\pm 1.645\).
05

Find the 90% confidence interval

Use the same confidence interval formula as Step 3, but this time use \(Z = 1.645\). So \( CI = 16.2 \pm (1.645 * 1.74) \) gives the 90% confidence interval, which is \( 16.2 \pm 2.86 \) minutes, or \(13.34 to 19.06\) minutes.
06

Compare the two intervals

The 90% confidence interval is narrower than the 95% confidence interval, as anticipated. Confidence intervals are narrower at lower levels of confidence because there is more uncertainty about where the true population parameter lies.

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.

Population Mean
In statistical terms, the population mean is the average of a set of characteristics or numbers for the entire group being studied. When trying to discover how long 12th-grade students take to get to school, the population mean gives us the central tendency or typical travel time across all students. However, instead of surveying every single student, we often rely on a sample.
For the given exercise, our sample mean is 16.2 minutes. This sample mean serves as an estimate of the actual population mean, which would be the true average travel time if we were able to collect data from every student in the population.
Using a sample mean provides a reasonable estimate, but with a level of uncertainty. This is why constructing a confidence interval is helpful, as it gives us a range where the true population mean is likely to lie.
Sample Standard Deviation
The sample standard deviation is a measure that indicates how much variability or dispersion exists in a set of sample data. In this context, it helps us understand how travel times differ among the 50 sampled 12th-grade students.
In our exercise, the sample standard deviation is 12.3 minutes. This tells us that there is quite a bit of variation in the time it takes different students to reach school. Some might take much less or much more time than the average.
Understanding the standard deviation is crucial because it plays a directly influential role in calculating the standard error. A higher standard deviation would lead to greater uncertainty about the location of the population mean, leading to a wider confidence interval.
Z-score
A z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. When calculating confidence intervals, the z-score represents how many standard deviations away from the sample mean our true population mean is likely to fall.
For different confidence levels, different z-scores are used. For a 95% confidence interval, the z-score is typically 1.96, which reflects that our range will cover 95% of the potential data distribution.
Similarly, at a 90% confidence level, the z-score drops to 1.645. This change impacts the width of the confidence interval, making it narrower at 90% than at 95%. This occurs because we are accepting more risk of the mean lying outside our interval boundaries at a lower confidence level.
Standard Error
The standard error estimates the variability of the sample mean by considering the sample size and the sample standard deviation. It essentially measures how much the sample mean is expected to fluctuate from the true population mean.
In the given exercise, the standard error is calculated using the formula:
  • Standard Error = \( \frac{\text{Sample Standard Deviation}}{\sqrt{\text{Sample Size}}} \)
Substituting our values, it becomes \( \frac{12.3}{\sqrt{50}} \approx 1.74 \) minutes.
A smaller standard error implies that the sample mean is a more accurate estimate of the population mean. Hence, it's pivotal in accurately determining the span of the confidence interval for estimating the population mean.

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

Random Numbers If you take samples of 40 lines from a random number table and find that the confidence interval for the proportion of odd-numbered digits captures \(50 \%\). 37 times out of the 40 lines, is it the confidence interval or confidence level you are estimating with the 37 out of 40 ?

Presidents' Ages at Inauguration A \(95 \%\) confidence interval for the ages of the first six presidents at their inaugurations is \((56.2,59.5) .\) Either interpret the interval or explain why it should not be interpreted.

RBls (Example 11) A random sample of 25 baseball players from the 2017 Major League Baseball season was taken and the sample data was used to construct two confidence intervals for the population mean. One interval was \((22.0,42.8)\). The other interval was \((19.9,44.0)\). (Source: mlb.com) a. One interval is a \(95 \%\) interval, and one is a \(90 \%\) interval. Which is which, and how do you know? b. If a larger sample size was used, for example, 40 instead of 25 , how would this affect the width of the intervals? Explain.

Ages A study of all the students at a small college showed a mean age of \(20.7\) and a standard deviation of \(2.5\) years. Are these numbers statistics or parameters? Explain. b. Label both numbers with their appropriate symbol (such as \(\bar{x}, \mu, s\), or \(\sigma)\).

Student Ages The mean age of all 2550 students at a small college is \(22.8\) years with a standard deviation is \(3.2\) years, and the distribution is right- skewed. A random sample of 4 students' ages is obtained, and the mean is \(23.2\) with a standard deviation of \(2.4\) years. a. \(\mu=? \quad \sigma=? \quad \bar{x}=? \quad s=?\) b. Is \(\mu\) a parameter or a statistic? \(c\). Are the conditions for using the CLT fulfilled? What would be the shape of the approximate sampling distribution of many means, cach from a sample of 4 students? Would the shape be right-skewed, Normal, or left-skewed?

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.