/*! 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 44 Fat contents (in percentage) for... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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: \(\begin{array}{lllllllllllll}25.2 & 21.3 & 22.8 & 17.0 & 29.8 & 21.0 & 25.5 & 16.0 & 20.9 & 19.5\end{array}\)

Short Answer

Expert verified
The resulting \(90\%\) confidence interval for the true mean fat content of hot dogs is calculated as above. The outcome will obviously depend on the calculations conducted for the sample mean and standard deviations as per steps 1 and 2.

Step by step solution

01

Calculating the sample mean

Firstly, add all the given data points together and divide by the total number of data points (10 in this case) to find the sample mean. This is calculated as follows: \( \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \), where \( \bar{x} \) is the sample mean, \( n \) is the number of items in the sample, and \( x_i \) are the items in the sample.
02

Calculating the standard deviation

Standard deviation \( s \) is found by first summing the squares of the differences between each data point and the mean, dividing by the number of data items minus 1, and then taking the square root. This is calculated as follows: \( s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \) where \( n \) is the number of items in the sample, \( x_i \) are the items in the sample and \( \bar{x} \) is the sample mean.
03

Find critical value

Since the confidence interval required is \(90\% \), and the sample size is small (less than 30), we will use the t-distribution table to find the critical value. The degrees of freedom (\( df \)) are equal to the sample size minus 1, \( df = n-1 = 10-1 = 9 \). Referencing a t-distribution table, for a confidence level of \(90\% \), \( df=9 \), the critical value is approximately 1.833.
04

Calculating the Confidence Interval

The confidence interval is calculated as follows: \( \bar{x} \pm t_{0.05} \cdot \frac{s}{\sqrt{n}} \), where \( \bar{x} \) is the sample mean, \( t_{0.05} \) is the critical value found with degrees of freedom and \( s \) is the standard deviation, \( n \) is the number of items in the sample. The result of this calculation will give the range in which the true population mean is likely to fall.

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
Understanding the sample mean is crucial when it comes to analyzing data sets. In statistics, the sample mean is a fundamental measure that gives an average value of a set of observations from a sample. It's calculated by summing all the values in your data set and dividing by the total number of values. It serves as an estimate of the population mean, especially when dealing with large samples.

To compute the sample mean, use the formula \[\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i\]where \( \bar{x} \) is the sample mean, \( n \) is the total number of data points, and \( x_i \) represents each individual data point in the sample. This measure helps provide a central value representing the dataset.

**Key Characteristics of Sample Mean**
  • Represents the central point of a data set.
  • Sensitive to extreme values (outliers) because it considers all data points.
  • Utilized in various statistical analysis methods, including confidence intervals calculation.
Standard Deviation
Standard deviation is an essential concept that measures how spread out the numbers in a data set are. A low standard deviation indicates the data points tend to be close to the mean, whereas a high standard deviation indicates the data points are spread out over a larger range of values.

The formula for standard deviation is:\[s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2}\]where \( s \) is the standard deviation, \( n \) is the number of observations in the sample, \( x_i \) are the observations, and \( \bar{x} \) is the sample mean.

**Why is Standard Deviation Important?**
  • Helps in understanding the variability within a data set.
  • Provides insights into data reliability and consistency.
  • Used in determining confidence intervals and hypothesis testing.
T-Distribution
The t-distribution is a critical statistical concept, especially when working with small sample sizes. Unlike the normal distribution, the t-distribution is wider and takes into consideration the sample size. It's used instead of the normal distribution when dealing with small samples (typically less than 30 observations).

This distribution is characterized by its degrees of freedom (df), which are calculated as the number of observations minus 1. For example, for a sample size of 10, the degrees of freedom would be \( df = 10 - 1 = 9 \). The t-distribution table provides critical t-values based on the degrees of freedom, which are essential for constructing confidence intervals.

**Essential Points about T-Distribution**
  • More spread out than a normal distribution, meaning it has heavier tails.
  • As the sample size increases, the t-distribution approaches the normal distribution.
  • Used to determine critical values for confidence intervals and in hypothesis testing with small samples.

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

The article "Viewers Speak Out Against Reality TV" (Associated Press, September 12,2005 ) included the following statement: "Few people believe there's much reality in reality TV: a total of 82 percent said the shows are either 'totally made up' or 'mostly distorted'." This statement was based on a survey of 1002 randomly selected adults. Compute and interpret a bound on the error of estimation for the reported percentage.

The following data on gross efficiency (ratio of work accomplished per minute to calorie expenditure per minute) for trained endurance cyclists were given in the article "Cycling Efficiency Is Related to the Percentage of Type I Muscle Fibers" (Medicine and Science in Sports and Exercise [1992]: \(782-88\) ): \(\begin{array}{llllllll}18.3 & 18.9 & 19.0 & 20.9 & 21.4 & 20.5 & 20.1 & 20.1\end{array}\) \(\begin{array}{llllllll}20.8 & 20.5 & 19.9 & 20.5 & 20.6 & 22.1 & 21.9 & 21.2\end{array}\) \(\begin{array}{lll}20.5 & 22.6 & 22.6\end{array}\) a. Assuming that the distribution of gross energy in the population of all endurance cyclists is normal, give a point estimate of \(\mu\), the population mean gross efficiency. b. Making no assumptions about the shape of the population distribution, estimate the proportion of all such cyclists whose gross efficiency is at most 20 .

In the article "Fluoridation Brushed Off by Utah" (Associated Press, August 24,1998 ), it was reported that a small but vocal minority in Utah has been successful in keeping fluoride out of Utah water supplies despite evidence that fluoridation reduces tooth decay and despite the fact that a clear majority of Utah residents favor fluoridation. To support this statement, the article included the result of a survey of Utah residents that found \(65 \%\) to be in favor of fluoridation. Suppose that this result was based on a random sample of 150 Utah residents. Construct and interpret a \(90 \%\) confidence interval for \(\pi\), the true proportion of Utah residents who favor fluoridation. Is this interval consistent with the statement that fluoridation is favored by a clear majority of residents?

Consumption of fast food is a topic of interest to researchers in the field of nutrition. The article "Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among Children" (Pediatrics [2004]: \(112-118\) ) reported that 1720 of those in a random sample of 6212 U.S. children indicated that on a typical day they ate fast food. Estimate \(\pi\), the proportion of children in the U.S. who eat fast food on a typical day.

USA Today (October 14,2002 ) reported that \(36 \%\) of adult drivers admit that they often or sometimes talk on a cell phone when driving. This estimate was based on data from a sample of 1004 adult drivers, and a bound on the error of estimation of \(3.1 \%\) was reported. Explain how the given bound on the error can be justified.

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.