/*! 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 60 Moose Drool Makes Grass More App... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Moose Drool Makes Grass More Appetizing Different species can interact in interesting ways. One type of grass produces the toxin ergovaline at levels about 1.0 part per million in order to keep grazing animals away. However, a recent study \(^{27}\) has found that the saliva from a moose counteracts these toxins and makes the grass more appetizing (for the moose). Scientists estimate that, after treatment with moose drool, mean level of the toxin ergovaline (in ppm) on the grass is \(0.183 .\) The standard error for this estimate is 0.016 . (a) Give notation for the quantity being estimated, and define any parameters used. (b) Give notation for the quantity that gives the best estimate, and give its value. (c) Give a \(95 \%\) confidence interval for the quantity being estimated. Interpret the interval in context.

Short Answer

Expert verified
The parameter being estimated, indicated as \(\mu\), is the mean level of the toxin in the grass after being treated with moose drool. The best estimate of this parameter is given by the sample mean, denoted as \(\bar{x}\), which was found to be 0.183 ppm. The 95% confidence interval for the mean level of toxin was calculated to be between 0.152 and 0.214 ppm.

Step by step solution

01

Identify and Define Parameters

The parameter being estimated is the mean level of the toxin ergovaline in the grass after being treated with moose drool. In statistical notation, this parameter is usually denoted as \(\mu\).
02

Distinguish the Best Estimate

The quantity that provides the best estimate in this instance is the sample mean (denoted as \(\bar{x}\)). After the scientists' study, they were able to estimate the mean level of the toxin to be 0.183 ppm.
03

Calculate a 95% Confidence Interval

The confidence interval can be calculated using the following formula: \[\text{CI} = \(\bar{x}\) \pm Z*\frac{\(SE\)}{\sqrt{n}}\] Where CI designates the confidence interval, Z denotes the Z score (1.96 for a 95% confidence interval), SE represents standard error and n is the sample size. In this problem, we are only given \(\bar{x}\) (0.183) and SE (0.016), without the sample size; the formula simplifies to \[\text{CI} = 0.183 \pm 1.96*(0.016)\]
04

Compute & Interpret the Confidence Interval

After calculating the above, we get the Confidence Interval between 0.152 and 0.214. This means that we are 95% confident that the true mean level of toxin in the grass treated with moose drool is between 0.152 and 0.214 ppm. This interpretation of the interval reflects the uncertainty of our estimate due to using a sample instead of having data from all grazing situations.

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.

Statistical Estimation
Statistical estimation is a fundamental concept in statistics that entails predicting the population parameters based on a sample. In the moose drool example, we're estimating the mean level of toxin ergovaline in grass after treatment. The parameter we're interested in estimating is denoted by \( \mu \), which represents the true average level of toxin across all treated grass patches.
To estimate \( \mu \), scientists use sample data to calculate the sample mean, symbolized as \( \bar{x} \). Here, the sample mean serves as the best approximation of the population mean, \( \mu \). This allows researchers to make educated guesses about the entire population based on limited but well-chosen observations. This step is crucial because gathering data from the whole population can be impractical or impossible.
Understanding statistical estimation helps us appreciate the power of samples in making reliable generalizations about larger groups or populations. It is essential for drawing conclusions in numerous scientific studies and practical applications.
Standard Error
The standard error (SE) is a key statistical concept that measures the accuracy of the sample mean as an estimate of the population mean. It reflects how much the sample mean is expected to vary from the true mean if different samples were drawn from the population.
In the exercise involving moose drool, the standard error is given as 0.016. This value informs us how close the sample mean (0.183 ppm) might be to the true mean \( \mu \). A smaller standard error implies greater precision, indicating that the sample mean is likely to be near the population mean.
Calculating the standard error often involves dividing the standard deviation by the square root of the sample size. While we do not have the sample size in this example, the provided standard error still gives us a sense of the estimate's reliability. SE is crucial because it contributes directly to computing the confidence interval and helps gauge the precision of the estimated parameters.
Z Score
The Z score is a statistical measurement that describes the position of a data point in relation to the mean, measured in standard deviations. It enables determination of how far away a sample mean is from the expected mean under the null hypothesis. In the context of confidence intervals, the Z score represents the confidence level you wish to establish.
For instance, when a 95% confidence interval is calculated, the Z score used is 1.96. This value is derived from the standard normal distribution and indicates that about 95% of data points lie within 1.96 standard deviations from the mean. Therefore, in this exercise, the Z score of 1.96 helps in determining the range within which we expect the true population mean to fall.
Understanding the Z score is vital for constructing confidence intervals. These intervals offer insights into where the true population parameter likely resides, providing a way to communicate the uncertainty inherent in estimates derived from 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

What Proportion Believe in One True Love? In Data 2.1 on page 48 , we describe a study in which a random sample of 2625 US adults were asked whether they agree or disagree that there is "only one true love for each person." The study tells us that 735 of those polled said they agree with the statement. The standard error for this sample proportion is \(0.009 .\) Define the parameter being estimated, give the best estimate, the margin of error, and find and interpret a \(95 \%\) confidence interval.

Mean number of cell phone calls made or received per day by cell phone users. In a survey of 1917 cell phone users, the mean was 13.10 phone calls a day.

Predicting Election Results Throughout the US presidential election of \(2016,\) polls gave regular updates on the sample proportion supporting each candidate and the margin of error for the estimates. This attempt to predict the outcome of an election is a common use of polls. In each case below, the proportion of voters who intend to vote for each of two candidates is given as well as a margin of error for the estimates. Indicate whether we can be relatively confident that candidate A would win if the election were held at the time of the poll. (Assume the candidate who gets more than \(50 \%\) of the vote wins.) \(\begin{array}{lll}\text { (a) Candidate A: } 54 \% & \text { Candidate }\end{array}\) B: \(46 \%\) Margin of error: \(\pm 5 \%\) (b) Candidate A: \(52 \%\) Candidate B: \(48 \%\) Margin of error: \(\pm 1 \%\) \(\begin{array}{ll}\text { (c) Candidate A: } 53 \% & \text { Candidate }\end{array}\) B: \(47 \%\) Margin of error: \(\pm 2 \%\) \(\begin{array}{lll}\text { (d) Candidate A: } 58 \% & \text { Candidate }\end{array}\) B: \(42 \%\) Margin of error: \(\pm 10 \%\)

Number of Text Messages a Day A random sample of \(n=755\) US cell phone users age 18 and older in May 2011 found that the average number of text messages sent or received per day is 41.5 messages, 32 with standard error about \(6.1 .\) (a) State the population and parameter of interest. Use the information from the sample to give the best estimate of the population parameter. (b) Find and interpret a \(95 \%\) confidence interval for the mean number of text messages.

SKILL BUILDER 1 In Exercises 3.41 to \(3.44,\) data from a sample is being used to estimate something about a population. In each case: (a) Give notation for the quantity that is being estimated. (b) Give notation for the quantity that gives the best estimate. Random samples of organic eggs and eggs that are not organic are used to estimate the difference in mean protein level between the two types of eggs.

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.