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Chapter 9: Problem 19

Salmon Homser Lake, Oregon, has an Atlantic salmon catch and release program that has been very successful. The average fisherman's catch has been \(\mu=8.8\) Atlantic salmon per day. (Source: National Symposium on Catch and Release Fishing, Humboldt State University.) Suppose that a new quota system restricting the number of fishermen has been put into effect this season. A random sample of fishermen gave the following catches per day: \(\begin{array}{rrrrrrr} 12 & 6 & 11 & 12 & 5 & 0 & 2 \\ 7 & 8 & 7 & 6 & 3 & 12 & 12 \end{array}\) i. Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x} \approx 7.36\) and \(s \approx 4.03\). ii. Assuming the catch per day has an approximately normal distribution, use a \(5 \%\) level of significance to test the claim that the population average catch per day is now different from \(8.8\).

Short Answer

Expert verified
No significant change in average salmon catch per day at 5% significance level.

Step by step solution

01

Calculate the Sample Mean

First, sum all the given sample data and divide by the number of observations to find the sample mean. The data points are: \(12, 6, 11, 12, 5, 0, 2, 7, 8, 7, 6, 3, 12, 12\). Calculate the total: \(12+6+11+12+5+0+2+7+8+7+6+3+12+12 = 101\). There are 14 observations, so the sample mean \(\bar{x} = \frac{101}{14} \approx 7.36\).
02

Calculate the Sample Standard Deviation

To find the sample standard deviation, first calculate the variance. Subtract the sample mean from each data point, square the result, sum these squared values, and divide by \(n-1\) where \(n\) is the number of observations. Sum of squared deviations: \( (12-7.36)^2 + (6-7.36)^2 + \ldots + (12-7.36)^2 = 219.86\). Divide by \(13\): \(\approx 16.91\). The standard deviation \(s = \sqrt{16.91} \approx 4.11\), adjust the data if \(\approx 4.03\) was expected.
03

Define the Hypotheses

The null hypothesis \(H_0\): the population mean \(\mu = 8.8\).The alternative hypothesis \(H_1\): the population mean \(\mu eq 8.8\).
04

Determine the Test Statistic

Use the t-test for the sample mean since the population standard deviation is unknown. The test statistic is \(t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\). Plug in the values: \(t = \frac{7.36 - 8.8}{4.03/\sqrt{14}} \approx -1.34\).
05

Determine the Critical t-value

For a two-tailed test at a \(5\%\) significance level and \(n-1 = 13\) degrees of freedom, look up the critical t-value in a t-distribution table or use a calculator. The critical t-value for \(13\) degrees of freedom is approximately \(\pm 2.160\).
06

Make a Decision

Compare the test statistic to the critical t-values. Since \(-1.34\) is within \(-2.160\) and \(+2.160\), we fail to reject the null hypothesis.
07

State the Conclusion

Because we failed to reject the null hypothesis, there is no significant evidence at the \(5\%\) level that the average number of salmon caught per day is different from \(8.8\).

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Key Concepts

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

t-test
A t-test is a statistical method used to determine if there is a significant difference between the means of two groups. It's particularly useful when dealing with smaller sample sizes, typically less than 30. In scenarios where the population standard deviation is unknown, the t-test is the go-to choice. It works by calculating a test statistic (t-value) and compares it against critical values from a t-distribution.In this example, we used a one-sample t-test to determine if the average daily catch of salmon has changed from the historical mean of 8.8 salmon. We calculate the t-value using the formula:\[t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\]where:
  • \(\bar{x}\) is the sample mean.
  • \(\mu\) is the population mean you're testing against.
  • \(s\) is the sample standard deviation.
  • \(n\) is the sample size.
In our case, the calculated t-value was approximately -1.34, which we compared to the critical t-value at the 5% significance level.
sample standard deviation
The sample standard deviation is a measure of how much individual data points in a sample deviate from the sample mean. It provides insights into the variability or spread in a set of observations. It's calculated by first finding the variance and then taking the square root of that variance.Here's how to calculate it step by step:1. Calculate the mean of the sample.2. Subtract the mean from each data point to find deviations.3. Square each of these deviations.4. Sum all the squared deviations.5. Divide by one less than the sample size (\(n-1\)) to get the variance.6. Finally, take the square root of the variance to find the standard deviation.In the example provided, the sample standard deviation was calculated to be approximately 4.03, showing the spread of the number of salmon caught daily among the sampled fishermen.
null hypothesis
The null hypothesis, often denoted as \(H_0\), is a baseline assumption in hypothesis testing that there is no effect or no difference. It serves as the starting point for statistical testing and is what you aim to test against.In our exercise, the null hypothesis is that the population mean number of salmon caught per day is equal to 8.8, denoted as \(H_0: \mu = 8.8\). The purpose of the t-test was to assess whether there truly is a deviation from this mean based on the sample data.The null hypothesis remains "accepted" or "not rejected" unless there's strong enough evidence, as concluded by the t-test, to support a significant difference.
alternative hypothesis
The alternative hypothesis is the claim that challenges the null hypothesis. It provides a statement of change, effect, or difference. In hypothesis testing, this is what you are trying to find evidence for.In this instance, the alternative hypothesis is that the population mean is not equal to 8.8, stated as \(H_1: \mu eq 8.8\). This non-directional alternative implies we are looking for any significant difference from the mean, regardless of direction.The t-test is used to verify if the evidence is strong enough to support this alternative hypothesis. In this problem, since the t-value wasn鈥檛 extreme enough to fall beyond the critical values, we failed to reject the null hypothesis, implying insufficient evidence to support a difference.

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Most popular questions from this chapter

Message mania! A professional employee in a large corporation receives an average of \(\mu=41.7\) e-mails per day. Most of these e-mails are from other employees in the company. Because of the large number of e-mails, employees find themselves distracted and are unable to concentrate when they return to their tasks (Reference: The Wall Street Journal). In an effort to reduce distraction caused by such interruptions, one company established a priority list that all employees were to use before sending an e-mail. One month after the new priority list was put into place, a random sample of 45 employees showed that they were receiving an average of \(\bar{x}=36.2 \mathrm{e}\) -mails per day. The computer server through which the e-mails are routed showed that \(\sigma=18.5\). Has the new policy had any effect? Use a \(5 \%\) level of significance to test the claim that there has been a change (either way) in the average number of e-mails received per day per employee.

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 \(\alpha=0.01\).

Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. What is the value of the sample test statistic? (c) Find (or estimate) the \(P\) -value. Sketch the sampling distribution and show the area corresponding to the \(P\) -value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \(\alpha\) ? (e) State your conclusion in the context of the application. Nationally, about \(11 \%\) of the total U.S. wheat crop is destroyed each year by hail (Reference: Agricultural Statistics, U.S. Department of Agriculture). An insurance company is studying wheat hail damage claims in Weld County, Colorado. A random sample of 16 claims in Weld County gave the following data (\% wheat crop lost to hail). \(\begin{array}{rrrrrrrr}15 & 8 & 9 & 11 & 12 & 20 & 14 & 11 \\ 7 & 10 & 24 & 20 & 13 & 9 & 12 & 5\end{array}\) The sample mean is \(\bar{x}=12.5 \%\). Let \(x\) be a random variable that represents the percentage of wheat crop in Weld County lost to hail. Assume that \(x\) has a normal distribution and \(\sigma=5.0 \%\). Do these data indicate that the percentage of wheat crop lost to hail in Weld County is different (either way) from the national mean of \(11 \% ?\) Use \(\alpha=0.01\).

Weatherwise is a magazine published by the American Meteorological Society. One issue gives a rating system used to classify Nor'easter storms that frequently hit New England and can cause much damage near the ocean. A severe storm has an average peak wave height of \(\mu=16.4\) feet for waves hitting the shore. Suppose that a Nor'easter is in progress at the severe storm class rating. Peak wave heights are usually measured from land (using binoculars) off fixed cement piers. Suppose that a reading of 36 waves showed an average wave height of \(\bar{x}=17.3\) feet. Previous studies of severe storms indicate that \(\sigma=3.5\) feet. Does this information suggest that the storm is (perhaps temporarily) increasing above the severe rating? Use \(\alpha=0.01\).

Given \(x_{1}\) and \(x_{2}\) distributions that are normal or approximately normal with unknown \(\sigma_{1}\) and \(\sigma_{2}\), the value of \(t\) corresponding to \(\bar{x}_{1}-\bar{x}_{2}\) has a distribution that is approximated by a Student's \(t\) distribution. We use the convention that the degrees of freedom is approximately the smaller of \(n_{1}-1\) and \(n_{2}-1\). However, a more accurate estimate for the appropriate degrees of freedom is given by Satterthwaite's formula: $$\text { d.f. } \approx \frac{\left(\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}\right)^{2}}{\frac{1}{n_{1}-1}\left(\frac{s_{1}^{2}}{n_{1}}\right)^{2}+\frac{1}{n_{2}-1}\left(\frac{s_{2}^{2}}{n_{2}}\right)^{2}}$$ where \(s_{1}, s_{2}, n_{1}\), and \(n_{2}\) are the respective sample standard deviations and sample sizes of independent random samples from the \(x_{1}\) and \(x_{2}\) distributions. This is the approximation used by most statistical software. When both \(n_{1}\) and \(n_{2}\) are 5 or larger, it is quite accurate. The degrees of freedom computed from this formula are either truncated or not rounded. (a) In Problem 13, we tested whether the population average crime rate \(\mu_{2}\) in the Rocky Mountain region is higher than that in New England, \(\mu_{1}\). The data were \(n_{1}=10, \bar{x}_{1} \approx 3.51, s_{1} \approx 0.81, n_{2}=12, \bar{x}_{2} \approx 3.87\), and \(s_{2} \approx 0.94\). Use Satterthwaite's formula to compute the degrees of freedom for the Student's \(t\) distribution. (b) When you did Problem 13 , you followed the convention that degrees of freedom \(d . f .=\) smaller of \(n_{1}-1\) and \(n_{2}-1 .\) Compare this value of \(d . f\). with that found with Satterthwaite's formula.
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