/*! 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] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 45

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 solution will involve applying statistical concepts related to calculating the sample mean, the sample standard deviation, the standard error, finding corresponding critical value and finally calculating the 95% confidence interval. The final result will be an interval that estimates the average number of months since the last dentist visit for the population of students participating in the free checkup program.

Step by step solution

01

Calculating the sample mean

First, calculate the mean (average) of the provided data. This can be done by adding up the months recorded for all five students and then dividing by the number of students. In this case, the calculation is as follows: (6+17+11+22+29)/5.
02

Calculating the standard error

Next, calculate the standard error of the mean. This is the standard deviation of the sample divided by the square root of the sample size. Since this is a small sample and we don't know the population standard deviation, we must first calculate the sample standard deviation using the formula for the sample standard deviation. Then proceed with the calculation of the standard error.
03

Finding the critical value

Then, find the critical value that corresponds to a 95% confidence level. Since the sample size is small and the population standard deviation is unknown, use the t-distribution. For a 95% confidence level and 4 degrees of freedom (since n-1=5-1), the critical value (also called t-star) is commonly 2.776.
04

Constructing the confidence interval

The formula for a confidence interval is: Confidence Interval = sample mean 卤 (critical value * standard error). Substitute the values previously calculated into this formula to obtain the 95% confidence interval.

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 91影视!

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

In a survey of 1000 randomly selected adults in the United States, participants were asked what their most favorite and what their least favorite subject was when they were in school (Associated Press, August 17,2005\() .\) In what might seem like a contradiction, math was chosen more often than any other subject in both categories! Math was chosen by 230 of the 1000 as the favorite subject, and it was also chosen by 370 of the 1000 as the least favorite subject. a. Construct a \(95 \%\) confidence interval for the proportion of U.S. adults for whom math was the favorite subject in school. b. Construct a \(95 \%\) confidence interval for the proportion of U.S. adults for whom math was the least favorite subject.

For each of the following choices, explain which would result in a wider large-sample confidence interval for \(\pi\) : a. \(90 \%\) confidence level or \(95 \%\) confidence level b. \(n=100\) or \(n=400\)

Fat contents (in percentage) for 10 randomly selected hot dogs were given in the article "Sensory and Mechanical Assessment of the Quality of Frankfurters" (Journal of Texture Studies \([1990]: 395-409\) ). Use the following data to construct a \(90 \%\) confidence interval for the true mean fat percentage of hot dogs: \(\begin{array}{lllllllllllll}25.2 & 21.3 & 22.8 & 17.0 & 29.8 & 21.0 & 25.5 & 16.0 & 20.9 & 19.5\end{array}\)

One thousand randomly selected adult Americans participated in a survey conducted by the Associated Press (June, 2006). When asked "Do you think it is sometimes justified to lie or do you think lying is never justified?" \(52 \%\) responded that lying was never justified. When asked about lying to avoid hurting someone's feelings, 650 responded that this was often or sometimes OK. a. Construct a \(90 \%\) confidence interval for the proportion of adult Americans who think lying is never justified. b. Construct a \(90 \%\) confidence interval for the proportion of adult American who think that it is often or sometimes OK to lie to avoid hurting someone's feelings. c. Based on the confidence intervals from Parts (a) and (b), comment on the apparent inconsistency in the responses given by the individuals in this sample.

Suppose that a random sample of 50 bottles of a par. ticular brand of cough medicine is selected and the alcohol content of each bottle is determined. Let \(\mu\) denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the sample of 50 results in a \(95 \%\) confidence interval for \(\mu\) of \((7.8,9.4)\). a. Would a \(90 \%\) confidence interval have been narrower or wider than the given interval? Explain your answer. b. Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between \(7.8\) and \(9.4\). Is this statement correct? Why or why not? c. Consider the following statement: If the process of selecting a sample of size 50 and then computing the corresponding \(95 \%\) confidence interval is repeated 100 times, 95 of the resulting intervals will include \(\mu\). Is this statement correct? Why or why not?
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation
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.