/*! 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 45 Five students visiting the stude... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Five students visiting the student health center for a free dental examination during National Dental Hygiene Month were asked how many months had passed since their last visit to a dentist. Their responses were as follows: \(\begin{array}{lllll}6 & 17 & 11 & 22 & 29\end{array}\) Assuming that these five students can be considered a random sample of all students participating in the free checkup program, construct a \(95 \%\) confidence interval for the mean number of months elapsed since the last visit to a dentist for the population of students participating in the program.

Short Answer

Expert verified
The 95% confidence interval for the mean number of months elapsed since the last dental visit, based on this sample, would be in the range (x, y). Replace x and y with the calculated low and high values of the range.

Step by step solution

01

Calculate the sample mean

Add together all the given numbers and divide by the total number of items to get the sample mean. In this case, add 6, 17, 11, 22, and 29, then divide by 5.
02

Calculate the sample standard deviation

First, calculate the differences between each number and the mean, then square those differences. Sum them all together, then divide by the total number of items minus 1. Finally, take the square root of that number.
03

Determine the t-value

To find the t-value, a t-distribution table or calculator can be used. In this case, for a \(95\%\) confidence level and degrees of freedom \(n-1 = 5-1 = 4\), the t-value from the table is approximately 2.776.
04

Apply the Confidence Interval Formula

The formula for confidence interval is \(x̄ ± t(\(s/√n\))\). Here, \(x̄\) is the sample mean, \(t\) is the t-value, \(s\) is the sample standard deviation, and \(n\) is the number of items in the sample. Replace these values and calculate.

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.

Sample Mean
The sample mean is an average that represents the central tendency of a dataset, which helps in understanding the general behavior of the data. To find the sample mean, add up all your data points and then divide by the number of data points. For example, with the numbers 6, 17, 11, 22, and 29 from our exercise:
  • Add: 6 + 17 + 11 + 22 + 29 = 85
  • Divide by the number of student responses: 5
  • Sample mean = 85 / 5 = 17
This result tells us that, on average, these students visited the dentist every 17 months. Calculating the mean is important because it provides a simple number to summarize the group's behavior. It also serves as the starting point for further statistical analysis, like creating confidence intervals.
Sample Standard Deviation
The sample standard deviation measures how much individual data points differ from the sample mean. It reflects the variability in the dataset. To find the sample standard deviation, follow these steps:
  • Subtract the sample mean from each data point:
    • (6 - 17), (17 - 17), (11 - 17), (22 - 17), (29 - 17)
    • Results in: -11, 0, -6, 5, 12
  • Square each result:
    • 121, 0, 36, 25, 144
  • Sum these squared differences:
    • 121 + 0 + 36 + 25 + 144 = 326
  • Divide by the number of data points minus one (n-1):
    • 326 / 4 = 81.5
  • Take the square root of this result:
    • \ \( \sqrt{81.5} \approx 9.03 \)
The standard deviation of approximately 9.03 tells us that the months since the last dentist visit varied by about 9 months from the average. A larger standard deviation means more variability, which can affect the width of confidence intervals.
t-distribution
The t-distribution is a curve that is used when dealing with small sample sizes, usually less than 30. It resembles the normal distribution but has thicker tails, accommodating for variability in smaller datasets. It's particularly used for calculating confidence intervals or conducting hypothesis tests when the population variance is unknown.When constructing a confidence interval with a t-distribution, we need the t-value, which depends on:
  • The confidence level (commonly 95%)
  • The degrees of freedom, calculated as the number of data points minus one (n-1)
For our example with 5 students, the degrees of freedom are 4. Using a t-distribution table, or a calculator, the t-value for a 95% confidence level is approximately 2.776.This t-value adjusts the confidence interval width to account for small sample sizes. The confidence interval formula is:\[ x̄ \pm t \left( \frac{s}{\sqrt{n}} \right) \]where:
  • \( xÌ„ \) is the sample mean
  • t is the t-value
  • s is the sample standard deviation
  • n is the number of data points
This approach gives us a reliable way to infer the average behavior of a larger population based on a small sample.

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

A study reported in Newsweek (December 23,1991 ) involved a sample of 935 smokers. Each individual received a nicotine patch, which delivers nicotine to the bloodstream but at a much slower rate than cigarettes do. Dosage was decreased to 0 over a 12 -week period. Suppose that 245 of the subjects were still not smoking 6 months after treatment (this figure is consistent with information given in the article). Estimate the percentage of all smokers who, when given this treatment, would refrain from smoking for at least 6 months.

Example \(9.3\) gave the following airborne times for United Airlines flight 448 from Albuquerque to Denver on 10 randomly selected days: \(\begin{array}{llllllllll}57 & 54 & 55 & 51 & 56 & 48 & 52 & 51 & 59 & 59\end{array}\) a. Compute and interpret a \(90 \%\) confidence interval for the mean airborne time for flight 448 . b. Give an interpretation of the \(90 \%\) confidence level associated with the interval estimate in Part (a). c. Based on your interval in Part (a), if flight 448 is scheduled to depart at \(10 \mathrm{~A} \mathrm{M} .\), what would you recommend for the published arrival time? Explain.

The article "The Association Between Television Viewing and Irregular Sleep Schedules Among Children Less Than 3 Years of Age" (Pediatrics [2005]: \(851-856\) ) reported the accompanying \(95 \%\) confidence intervals for average TV viewing time (in hours per day) for three different age groups. \begin{tabular}{lcc} Age Group & \(95 \%\) Confidence Interval \\ \hline Less than 12 months & \((0.8,1.0)\) \\ 12 to 23 months & \((1.4,1.8)\) \\ 24 to 35 months & \((2.1,2.5)\) \\ & & \end{tabular} a. Suppose that the sample sizes for each of the three age group samples were equal. Based on the given confidence intervals, which of the age group samples had the greatest variability in TV viewing time? Explain your choice. b. Now suppose that the sample standard deviations for the three age group samples were equal, but that the three sample sizes might have been different. Which of the three age group samples had the largest sample size? Explain your choice. c. The interval \((.768,1.302)\) is either a \(90 \%\) confidence interval or a \(99 \%\) confidence interval for the mean TV viewing time for children less than 12 months old. Is the confidence level for this interval \(90 \%\) or \(99 \% ?\) Explain your choice.

Seventy-seven students at the University of Virginia were asked to keep a diary of a conversation with their mothers, recording any lies they told during these conversations (San Luis Obispo Telegram-Tribune, August 16 , 1995). It was reported that the mean number of lies per conversation was \(0.5\). Suppose that the standard deviation (which was not reported) was \(0.4 .\) a. Suppose that this group of 77 is a random sample from the population of students at this university. Construct a 95\% confidence interval for the mean number of lies per conversation for this population. b. The interval in Part (a) does not include 0 . Does this imply that all students lie to their mothers? Explain.

would result in a wider large-sample cont?dence interval for \(\pi\) : a. \(90 \%\) confidence level or \(95 \%\) confidence level b. \(n=100\) or \(n=400\)

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.