/*! 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 9 Suppose that \(n=100\) random sa... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that \(n=100\) random samples of water from a freshwater lake were taken and the calcium concentration (milligrams per liter) measured. A \(95 \%\) CI on the mean calcium concentration is \(0.49 \leq \mu \leq 0.82\) (a) Would a \(99 \%\) CI calculated from the same sample data be longer or shorter? (b) Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between 0.49 and \(0.82 .\) Is this statement correct? Explain your answer. (c) Consider the following statement: If \(n=100\) random samples of water from the lake were taken and the \(95 \%\) CI on \(\mu\) computed, and this process were repeated 1000 times, 950 of the CIs would contain the true value of \(\mu\). Is this statement correct? Explain your answer.

Short Answer

Expert verified
(a) Longer. (b) Incorrect. (c) Correct.

Step by step solution

01

Interpret Confidence Interval (CI) length for part (a)

A larger confidence level (e.g., 99%) generally means a wider confidence interval compared to a smaller confidence level (e.g., 95%). This is because a larger confidence level allows the data to encompass more variability, thereby widening the interval.
02

Evaluate the statement for part (b)

The statement 'There is a 95% chance that \( \mu \) is between 0.49 and 0.82.' is incorrect. In frequentist statistics, the calculated interval either contains the true mean \( \mu \) or it does not. The 95% confidence level means that if you were to repeat the sampling process many times, 95% of the calculated intervals would contain the parameter \( \mu \). The probability refers to the interval capturing \( \mu \) before the data are collected.
03

Evaluate the statement for part (c)

The statement is correct. A 95% confidence interval means that if you repeat the process of taking samples and calculating the confidence interval 1000 times, you would expect roughly 950 of those intervals to contain the true mean \( \mu \). This aligns with the intended frequentist interpretation of confidence intervals.

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 Inference
Statistical inference is the process of drawing conclusions about a population based on a sample taken from it. It helps us to use the information we gather from data and project it onto a larger context. The goal is to make estimates or predictions about unknown characteristics of a population. This involves several key methods, such as point estimation, interval estimation, and hypothesis testing.

For the given problem, we look at interval estimation, specifically confidence intervals, which give us a range within which we believe a population parameter (like the mean) lies. By using statistical inference, we can make informed guesses about a population without examining each individual member. This plays a crucial role in allowing us to make predictions and decisions based on available data.
Confidence Level
The confidence level is a critical piece of constructing confidence intervals. It reflects the percentage of intervals that would include the true population parameter if you repeated the sampling process many times. If we have a 95% confidence level, this indicates that we can be 95% confident that the interval calculated from our sample data contains the true mean of the population.

Think of it as a measure of reliability. A higher confidence level, such as 99%, makes us more assured that the interval covers the actual parameter, but it also means accepting a wider interval to accommodate for more variation. This is evident in the scenario where increasing the confidence level from 95% to 99% would result in a longer confidence interval, as it has to encompass more potential values.
Sampling Variability
Sampling variability refers to the natural tendency of sample statistics to differ from one sample to another. This is because every sample is drawn randomly, and the specific members in each sample can influence the results.

If you were to repeatedly sample from a population and calculate a statistic, such as the mean, those sample statistics would vary. However, over numerous trials, these statistics would form a distribution around the true population parameter. This concept is foundational because it explains why confidence intervals can be reliable estimators for population parameters. Despite variability, repeated sampling helps us to understand and estimate the true traits of a population.
Frequentist Statistics
Frequentist statistics is a framework for making inferences about populations using sample data. It is grounded in the assumption that probabilities are long-run frequencies of events. Under this view, for any given dataset, parameters are fixed but unknown constants, while the data are considered variable because they can change with each sampling.

This principle underlies the concept of confidence intervals in the problem we examined. In frequentist terms, a 95% confidence interval means that if you repeated your sampling process infinitely, 95% of the intervals you calculate would contain the true population mean. Importantly, according to frequentist statistics, once you've calculated a confidence interval, it either contains the true mean or it doesn't, with no probabilistic interpretation applied post data collection.

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

Of 1000 randomly selected cases of lung cancer, 823 resulted in death within 10 years. (a) Calculate a \(95 \%\) two-sided confidence interval on the death rate from lung cancer. (b) Using the point estimate of \(p\) obtained from the preliminary sample, what sample size is needed to be \(95 \%\) confident that the error in estimating the true value of \(p\) is less than \(0.03 ?\) (c) How large must the sample be if we wish to be at least \(95 \%\) confident that the error in estimating \(p\) is less than \(0.03,\) regardless of the true value of \(p ?\)

The yield of a chemical process is being studied. From previous experience, yield is known to be normally distributed and \(\sigma=3\). The past five days of plant operation have resulted in the following percent yields: \(91.6,88.75,90.8,89.95,\) and \(91.3 .\) Find a \(95 \%\) two-sided confidence interval on the true mean yield.

An article in the Journal of Agricultural Science [ "The Use of Residual Maximum Likelihood to Model Grain Quality Characteristics of Wheat with Variety, Climatic and Nitrogen Fertilizer Effects" (1997, Vol. 128, pp. \(135-142\) ) ] investigated means of wheat grain crude protein content (CP) and Hagberg falling number (HFN) surveyed in the UK. The analysis used a variety of nitrogen fertilizer applications (kg N/ha), temperature \(\left({ }^{\circ} \mathrm{C}\right),\) and total monthly rainfall \((\mathrm{mm})\). The data shown below describe temperatures for wheat grown at Harper Adams Agricultural College between 1982 and \(1993 .\) The temperatures measured in June were obtained as follows: $$ \begin{array}{llllll} 15.2 & 14.2 & 14.0 & 12.2 & 14.4 & 12.5 \\ 14.3 & 14.2 & 13.5 & 11.8 & 15.2 & \end{array} $$ Assume that the standard deviation is known to be \(\sigma=0.5\). (a) Construct a \(99 \%\) two-sided confidence interval on the mean temperature. (b) Construct a \(95 \%\) lower-confidence bound on the mean temperature. (c) Suppose that we wanted to be \(95 \%\) confident that the error in estimating the mean temperature is less than 2 degrees Celsius. What sample size should be used? (d) Suppose that we wanted the total width of the two-sided confidence interval on mean temperature to be 1.5 degrees Celsius at \(95 \%\) confidence. What sample size should be used?

A civil engineer is analyzing the compressive strength of concrete. Compressive strength is normally distributed with \(\sigma^{2}=1000(\mathrm{psi})^{2}\). A random sample of 12 specimens has a mean compressive strength of \(\bar{x}=3250\) psi. (a) Construct a \(95 \%\) two-sided confidence interval on mean compressive strength. (b) Construct a \(99 \%\) two-sided confidence interval on mean compressive strength. Compare the width of this confidence interval with the width of the one found in part (a).

During the 1999 and 2000 baseball seasons, there was much speculation that the unusually large number of home runs that were hit was due at least in part to a livelier ball. One way to test the "liveliness" of a baseball is to launch the ball at a vertical surface with a known velocity \(V_{L}\) and measure the ratio of the outgoing velocity \(V_{O}\) of the ball to \(V_{L}\). The ratio \(R=V_{O} / V_{L}\) is called the coefficient of restitution. Following are measurements of the coefficient of restitution for 40 randomly selected baseballs. The balls were thrown from a pitching machine at an oak surface. $$ \begin{array}{llllll} 0.6248 & 0.6237 & 0.6118 & 0.6159 & 0.6298 & 0.6192 \\ 0.6520 & 0.6368 & 0.6220 & 0.6151 & 0.6121 & 0.6548 \\ 0.6226 & 0.6280 & 0.6096 & 0.6300 & 0.6107 & 0.6392 \\ 0.6230 & 0.6131 & 0.6223 & 0.6297 & 0.6435 & 0.5978 \\ 0.6351 & 0.6275 & 0.6261 & 0.6262 & 0.6262 & 0.6314 \\ 0.6128 & 0.6403 & 0.6521 & 0.6049 & 0.6170 & \\ 0.6134 & 0.6310 & 0.6065 & 0.6214 & 0.6141 & \end{array} $$ (a) Is there evidence to support the assumption that the coefficient of restitution is normally distributed? (b) Find a \(99 \% \mathrm{CI}\) on the mean coefficient of restitution. (c) Find a \(99 \%\) prediction interval on the coefficient of restitution for the next baseball that will be tested. (d) Find an interval that will contain \(99 \%\) of the values of the coefficient of restitution with \(95 \%\) confidence. (e) Explain the difference in the three intervals computed in parts (b), (c), and (d).

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.