/*! 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 13 A random sample of 46 adult coyo... [FREE SOLUTION] | 91Ó°ÊÓ

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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, Behavior 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
Yes, coyotes in northern Minnesota live longer than 1.75 years, at the 0.01 significance level.

Step by step solution

01

State the Hypotheses

To determine if the coyotes in northern Minnesota live longer than the average of 1.75 years, we set up our null and alternative hypotheses. The null hypothesis is that the mean age of coyotes is 1.75 years, \[ H_0: \mu = 1.75 \]. The alternative hypothesis is that the mean age is greater than 1.75 years, \[ H_a: \mu > 1.75 \]. This is a one-tailed test.
02

Determine the Test Statistic

We use a t-test for the sample mean because the population standard deviation is unknown and the sample size is below 30. The test statistic is calculated as: \[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \] Substitute \( \bar{x} = 2.05 \), \( \mu = 1.75 \), \( s = 0.82 \), and \( n = 46 \) into the formula to get \[ t = \frac{2.05 - 1.75}{\frac{0.82}{\sqrt{46}}} \].
03

Calculate the Test Statistic

Continuing from Step 2, the calculation becomes: \[ t = \frac{0.30}{\frac{0.82}{\sqrt{46}}} = \frac{0.30}{0.1208} \approx 2.48 \]. So the computed t value is approximately 2.48.
04

Determine the Critical Value

Using a t-distribution table with \( \alpha = 0.01 \) and \( df = n - 1 = 45 \), the critical value for a one-tailed test is approximately 2.414. This is the value that separates the rejection region from the non-rejection region.
05

Compare Test Statistic to Critical Value

Compare the computed t value (2.48) to the critical value (2.414). Since 2.48 is greater than 2.414, we reject the null hypothesis.
06

Conclusion

Since the computed t value exceeds the critical value, we reject the null hypothesis. There is sufficient evidence at the \( \alpha = 0.01 \) significance level to support the claim that the mean age of coyotes in northern Minnesota is greater than 1.75 years.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the t-test
The **t-test** is a statistical method used to determine if there is a significant difference between the means of two groups. This test is particularly useful when dealing with small sample sizes or when the population standard deviation is unknown. The test compares the sample mean to the known population mean to determine if the difference observed is likely to occur by random chance or if it suggests a real difference exists.

In the context of our coyote study, a t-test helps us determine whether the average age of the coyotes in northern Minnesota significantly differs from the well-known population mean age of 1.75 years. We perform a one-tailed test here because we have a specific direction of interest - testing if the mean age is greater.
  • This test uses the formula: \[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]
  • Where: \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.
The computed t-value tells us how many standard deviations our sample mean is away from the population mean, thus helping to assess its significance.
Exploring the Sample Mean
At the heart of the t-test, we have the **sample mean** or \(\bar{x}\). The sample mean is an average of all observations in the sample. For our coyotes, this is calculated as \(\bar{x} = 2.05\) years. This number serves as a crucial comparison point against the population mean (\(\mu = 1.75\) years) to examine if the observed coyotes tend to live longer.

The sample mean gives us important insight into the specific set of coyotes being studied and estimates the central value of their ages. However, due to natural variability in data, the sample mean can vary from sample to sample. The t-test helps us decide whether the observed difference is meaningful, given this variability.

As a statistical measure, the sample mean:
  • Combines all data points to provide a single figure representing the survey group’s average.
  • Is pivotal for conducting hypothesis tests to make inferences about the larger population.
Demystifying Confidence Level
The **confidence level** in hypothesis testing helps us understand the probability that our result is not due to random chance. It is an expression of how certain we are about our conclusions. For our exercise with coyotes, we use a confidence level of 99%, which corresponds with an \(\alpha = 0.01\). This means we have a high degree of belief that the observed data reflects a true difference, not just a random one.

The confidence level tells us how often, in the long run, our conclusions drawn from the test would be correct. A higher confidence level corresponds to stricter criteria, making the test more robust against random errors, thus:
  • A 99% confidence level implies that in 99 out of 100 cases, our findings will be accurate.
  • The choice of confidence level affects the calculation of the critical value, influencing whether to reject the null hypothesis.
This level helps balance between identifying genuine differences and avoiding incorrect rejections of the null hypothesis due to chance.
Interpreting the Critical Value
The **critical value** in our statistical test functions as the decision threshold. It helps determine whether the observed test statistic is considered extreme enough to reject the null hypothesis. In practice, it comes from the t-distribution table matched with the chosen confidence level and degrees of freedom (df).

For our example with the coyotes, the critical value for a one-tailed t-test at \(\alpha = 0.01\) with \(df = 45\) was calculated to be 2.414. This means if our t-statistic (calculated from the sample data) exceeds this critical value, we conclude that there is a significant difference.

The importance of the critical value lies in its role to:
  • Establish the boundary between the rejection and non-rejection regions of the null hypothesis.
  • Function as a benchmark to determine if the test statistic reflects significant results when compared against this threshold.
By comparing our calculated t-value to this critical value, we can confidently decide if our results support the alternative hypothesis.

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Most popular questions from this chapter

A random sample of 49 measurements from a population with population standard deviation 3 had a sample mean of 10\. An independent random sample of 64 measurements from a second population with population standard deviation 4 had a sample mean of \(12 .\) Test the claim that the population means are different. Use level of significance \(0.01\). (a) What distribution does the sample test statistic follow? Explain. (b) State the hypotheses. (c) Compute \(\bar{x}_{1}-\bar{x}_{2}\) and the corresponding sample test statistic. (d) Find the \(P\) -value of the sample test statistic. (e) Conclude the test. (f) The results.

Socially conscious investors screen out stocks of alcohol and tobacco makers, firms with poor environmental records, and companies with poor labor practices. Some examples of "good," socially conscious companies are Johnson and Johnson, Dell Computers, Bank of America, and Home Depot. The question is, are such stocks overpriced? One measure of value is the \(\mathrm{P} / \mathrm{E}\), or. price-to-earnings, ratio. High \(\mathrm{P} / \mathrm{E}\) ratios may indicate a stock is overpriced. For the S\&P stock index of all major stocks, the mean \(\mathrm{P} / \mathrm{E}\) ratio is \(\mu=19.4\). A random sample of 36 "socially conscious" stocks gave a \(\mathrm{P} / \mathrm{E}\) ratio sample mean of \(\bar{x}=17.9\), with sample standard deviation \(s=5.2\) (Reference: Morningstar, a financial analysis company in Chicago). Does this indicate that the mean \(\mathrm{P} / \mathrm{E}\) ratio of all socially conscious stocks is different (either way) from the mean \(\mathrm{P} / \mathrm{E}\) ratio of the \(S \& P\) stock index? Use \(\alpha=0.05\).

Please read the Focus Problem at the beginning of this chapter. Recall that Benford's Law claims that numbers chosen from very large data files tend to have " \(1 "\) as the first nonzero digit disproportionately often. In fact, research has shown that if you randomly draw a number from a very large data file, the probability of getting a number with " 1 " as the leading digit is about \(0.301\) (see the reference in this chapter's Focus Problem). Now suppose you are an auditor for a very large corporation. The revenue report involves millions of numbers in a large computer file. Let us say you took a random sample of \(n=215\) numerical entries from the file and \(r=46\) of the entries had a first nonzero digit of \(1 .\) Let \(p\) represent the population proportion of all numbers in the corporate file that have a first nonzero digit of \(1 .\) j. Test the claim that \(p\) is less than \(0.301\). Use \(\alpha=0.01\). ii. If \(p\) is in fact less than \(0.301\), would it make you suspect that there are not enough numbers in the data file with leading 1 's? Could this indicate that the books have been "cooked" by "pumping up" or inflating the numbers? Comment from the viewpoint of a stockholder. Comment from the perspective of the Federal Bureau of Investigation as it looks for money laundering in the form of false profits. iii. Comment on the following statement: "If we reject the null hypothesis at level of significance \(\alpha\), we have not proved \(H_{0}\) to be false. We can say that the probability is \(\alpha\) that we made a mistake in rejecting \(H_{0} . "\) Based on the outcome of the test, would you recommend further investigation before accusing the company of fraud?

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