/*! 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 11 A random sample of 46 adult coyo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A random sample of 46 adult coyotes in a region of northern Minnesota showed the average age to be \(\bar{x}=2.05\) years, with sample standard deviation \(s=0.82\) years (based on information from the book Coyotes: Biology, Bebavior and Management by M. Bekoff, Academic Press). However, it is thought that the overall population mean age of coyotes is \(\mu=1.75 .\) Do the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of \(1.75\) years? Use \(\alpha=0.01\).

Short Answer

Expert verified
The sample data indicates that coyotes in this region tend to live longer than 1.75 years.

Step by step solution

01

State the Hypotheses

We are conducting a hypothesis test to determine if the coyotes in the sampled region live longer than the population mean of 1.75 years. Set up the null and alternative hypotheses as follows:- Null hypothesis (H_0): \(\mu = 1.75\) (The mean age of coyotes is 1.75 years.)- Alternative hypothesis (H_a): \(\mu > 1.75\) (The mean age of coyotes is greater than 1.75 years.)This is a one-tailed test since we are only interested in whether they live longer.
02

Determine the Test Statistic

Since the population standard deviation is unknown, we will use the t-statistic for a one-sample t-test. The formula for the t-statistic is:\[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]Substitute in the given values:\(\bar{x} = 2.05\), \(\mu = 1.75\), \(s = 0.82\), and sample size \(n = 46\).
03

Calculate the Test Statistic

Calculate the test statistic using the formula:\[ t = \frac{2.05 - 1.75}{0.82/\sqrt{46}} \]First, calculate the standard error:\( \text{standard error} = \frac{0.82}{\sqrt{46}} \approx 0.1209 \).Then compute the t-value:\( t \approx \frac{0.30}{0.1209} \approx 2.48 \).
04

Determine the Critical Value and Decision Rule

For a significance level \(\alpha = 0.01\) and \(n - 1 = 45\) degrees of freedom, find the critical t-value from the t-distribution table. The critical value for a one-tailed test at \(\alpha = 0.01\) and 45 degrees of freedom is approximately 2.414. Our decision rule: Reject \(H_0\) if the calculated t-value is greater than 2.414.
05

Compare Test Statistic to Critical Value

Compare the calculated t-value to the critical value:- Calculated t-value: \(t = 2.48\)- Critical t-value: 2.414Since 2.48 > 2.414, we reject the null hypothesis.

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.

t-distribution
The t-distribution is an essential concept in statistics, particularly when dealing with small sample sizes or when the population standard deviation is unknown. It is a probability distribution that resembles the normal distribution but has heavier tails, allowing for the increased uncertainty that comes with smaller samples. This distribution is particularly useful when conducting hypothesis tests for small sample means.

The t-distribution is characterized by its number of degrees of freedom, which is typically calculated as the sample size minus one ( -1). The smaller the sample size, the heavier the tails of the distribution, which means that extreme values are more likely. As the sample size increases, the t-distribution approaches a normal distribution. When performing a hypothesis test, we use the t-distribution to find our critical value, helping us decide whether to reject the null hypothesis.
significance level
In hypothesis testing, the significance level, denoted by \(\alpha\), is the probability of rejecting the null hypothesis when it is true. It is a threshold set by the researcher before conducting the test. A common significance level used is 0.05, but in more stringent tests, such as the one in this coyotes example, a lower level like 0.01 is used.

The significance level helps us define the critical region of a test. Any test statistic that falls within this critical region will lead to the null hypothesis being rejected. A lower significance level reduces the chance of a Type I error (incorrectly rejecting a true null hypothesis), but it also makes it harder to detect a true effect if one exists.
  • Type I Error: Incorrectly rejecting the true null hypothesis.
  • Type II Error: Failing to reject a false null hypothesis.
Choosing the correct significance level is crucial as it balances the risks of these errors.
critical value
The critical value in a hypothesis test is the threshold that the calculated test statistic must exceed to reject the null hypothesis. It is derived from the t-distribution or normal distribution, depending on the sample size and known parameters. In a one-tailed test, like in our coyotes scenario, the critical value corresponds to the upper tail of the distribution.

To find the critical value, you reference the t-distribution table using the chosen significance level \(\alpha\) and the degrees of freedom (-1). These values define the critical region of our hypothesis test. If the test statistic exceeds this critical value, it indicates enough evidence to reject the null hypothesis in favor of the alternative hypothesis.
sample standard deviation
The sample standard deviation, represented by \(s\), measures the dispersion of data points in a sample. It provides an estimate of the variation or spread within a sample, serving as an approximation of the population standard deviation when the latter is unknown.

This measurement plays a crucial role in calculating the standard error, which is essential for hypothesis testing. The formula for standard error in the context of a sample mean is: \( \text{standard error} = \frac{s}{\sqrt{n}} \), where \(n\) is the sample size.

Understanding the sample standard deviation allows researchers to gauge how much variability is present within their sample, and it helps determine if their sample is representative of the population. This understanding underpins the reliability of conclusions drawn from hypothesis tests.

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

Consumer Reports stated that the mean time for a Chrysler Concorde to go from 0 to 60 miles per hour was \(8.7\) seconds. (a) If you want to set up a statistical test to challenge the claim of \(8.7\) seconds, what would you use for the null hypothesis? (b) The town of Leadville, Colorado, has an elevation over 10,000 feet. Suppose you wanted to test the claim that the average time to accelerate from 0 to 60 miles per hour is longer in Leadville (because of less oxygen). What would you use for the alternate hypothesis? (c) Suppose you made an engine modification and you think the average time to accelerate from 0 to 60 miles per hour is reduced. What would you use for the alternate hypothesis? (d) For each of the tests in parts (b) and (c), would the \(P\) -value area be on the left, on the right, or on both sides of the mean? Explain your answer in each case.

In the following data pairs, \(A\) represents birth rate and \(B\) represents death rate per 1000 resident population. The data are paired by counties in the Midwest. A random sample of 16 counties gave the following information. (Reference: County and City Data Book, U.S. Department of Commerce.) \(\begin{array}{l|cccccccc} \hline \text { A: } & 12.7 & 13.4 & 12.8 & 12.1 & 11.6 & 11.1 & 14.2 & 15.1 \\\ \hline B: & 9.8 & 14.5 & 10.7 & 14.2 & 13.0 & 12.9 & 10.9 & 10.0 \\ \hline \\ \hline A: & 12.5 & 12.3 & 13.1 & 15.8 & 10.3 & 12.7 & 11.1 & 15.7 \\ \hline B: & 14.1 & 13.6 & 9.1 & 10.2 & 17.9 & 11.8 & 7.0 & 9.2 \\ \hline \end{array}\) Do the data indicate a difference (either way) between population average birth rate and death rate in this region? Use \(\alpha=0.01\).

USA Today reported that the state with the longest mean life span is Hawaii, where the population mean life span is 77 years. A random sample of 20 obituary notices in the Honolulu Advertizer gave the following information about life span (in years) of Honolulu residents: \(\begin{array}{llllllllll} 72 & 68 & 81 & 93 & 56 & 19 & 78 & 94 & 83 & 84 \\ 77 & 69 & 85 & 97 & 75 & 71 & 86 & 47 & 66 & 27 \end{array}\) i. Use a calculator with mean and standard deviation keys to verify that \(\bar{x}=\) \(71.4\) years and \(s\) \& \(20.65\) years. ii. Assuming that life span in Honolulu is approximately normally distributed, does this information indicate that the population mean life span for Honolulu residents is less than 77 years? Use a \(5 \%\) level of significance.

Generally speaking, would you say that most people can be trusted? A random sample of \(n_{1}=250\) people in Chicago ages \(18-25\) showed that \(r_{1}=45\) said yes. Another random sample of \(n_{2}=280\) people in Chicago ages \(35-45\) showed that \(r_{2}=71\) said yes (based on information from the National Opinion Research Center, University of Chicago). Does this indicate that the population proportion of trusting people in Chicago is higher for the older group? Use \(\alpha=0.05\).

Consider a hypothesis test of difference of proportions for two independent populations. Suppose random samples produce \(r_{1}\) successes out of \(n_{1}\) trials for the first population and \(r_{2}\) successes out of \(n_{2}\) trials for the second population. What is the best pooled estimate \(\bar{p}\) for the population probability of success using \(H_{0}: p_{1}=p_{2} ?\)

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.