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Pyramid Lake is on the Paiute Indian Reservation in Nevada. The lake is famous for cutthroat trout. Suppose a friend tells you that the average length of trout caught in Pyramid Lake is \(\mu=19\) inches. However, the Creel Survey (published by the Pyramid Lake Paiute Tribe Fisheries Association) reported that of a random sample of 51 fish caught, the mean length was \(\bar{x}=18.5\) inches, with estimated standard deviation \(s=3.2\) inches. Do these data indicate that the average length of a trout caught in Pyramid Lake is less than \(\mu=19\) inches? Use \(\alpha=0.05.\)

Short Answer

Expert verified
There is not enough evidence to conclude that the average trout length is less than 19 inches.

Step by step solution

01

State the Hypotheses

We will start by setting up the null and alternative hypotheses. The null hypothesis \(H_0\) is that the average length of trout is equal to 19 inches: \(H_0: \mu = 19\). The alternative hypothesis \(H_a\) is that the average length of trout is less than 19 inches: \(H_a: \mu < 19\).
02

Determine the Significance Level

The significance level \(\alpha\) is given as 0.05. This will be used to determine the critical value and to compare against the p-value.
03

Calculate the Test Statistic

We'll use a t-test for the sample since the population standard deviation is unknown. The test statistic \(t\) is calculated using the formula: \[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \] where \(\bar{x} = 18.5\), \(\mu = 19\), \(s = 3.2\), \(n = 51\). Substitute these values in: \[ t = \frac{18.5 - 19}{\frac{3.2}{\sqrt{51}}} = \frac{-0.5}{0.4486} \approx -1.114 \]
04

Determine the Critical Value

Since \(n = 51\), degrees of freedom \(df = n - 1 = 50\). For a left-tailed test at \(\alpha = 0.05\), locate the critical t-value from the t-distribution table. The critical value \(t_{0.05, 50}\) is approximately -1.676.
05

Make a Decision

Compare the calculated t-statistic with the critical value. Since \(t = -1.114\) is greater than \(-1.676\), we fail to reject the null hypothesis \(H_0\).
06

Conclusion

At the 0.05 significance level, the data do not provide sufficient evidence to reject the claim that the average length of trout is 19 inches. This means there is not enough evidence to support the claim that the average length of trout is less than 19 inches.

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Key Concepts

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

Significance Level
Before diving deeper into hypothesis testing, it's important to understand the concept of the significance level, denoted as \( \alpha \). This is a threshold that researchers set to determine whether the results of a statistical test are significant. In essence, the significance level represents the probability of rejecting the null hypothesis when it is true. In our example, the significance level is set at 0.05.

This means if the calculated p-value is less than 0.05, we reject the null hypothesis, indicating that the sample data provide enough evidence against the null hypothesis. On the other hand, if the p-value is greater than 0.05, we fail to reject it. Choosing a significance level is crucial because it controls how confidently we decide to reject the null hypothesis.

Commonly used significance levels are 0.05, 0.01, and 0.10, with smaller values indicating stricter criteria to reject the null hypothesis.
t-test
The t-test is a statistical test used to compare the mean from a sample to a known value or to another sample. In this case, the one-sample t-test is used because we're testing whether the sample mean differs from the known population mean. Unlike some other statistical tests, the t-test is used when the population standard deviation is unknown, which is why it fits well for our example where only the sample standard deviation is available.

The formula for the t-test statistic is:
  • \( t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \)
where:
  • \( \bar{x} \) is the sample mean,
  • \( \mu \) is the population mean under the null hypothesis,
  • \( s \) is the sample standard deviation, and
  • \( n \) is the sample size.
The calculated t-statistic helps determine how far our sample mean is from the hypothesized population mean in terms of the sample's standard error.
Critical Value
The critical value is pivotal in hypothesis testing. It acts as a boundary that determines whether the test statistic falls in the rejection region or not. For our exercise, we are dealing with a left-tailed test, due to the alternative hypothesis suggesting a mean less than 19 inches.

To find the critical value, you need both the significance level and the degrees of freedom, which is the sample size minus one (\( n - 1 \)). For our example, with \( n = 51 \), degrees of freedom are 50. By consulting the t-distribution table with this information, we find the critical value for \( \alpha = 0.05 \), which is approximately -1.676.

The critical value divides the t-distribution into regions where the null hypothesis is either rejected or not rejected based on where the test statistic falls relative to this boundary. It is the cut-off point beyond which the evidence against the null hypothesis is considered strong enough to reject it.
Null Hypothesis
In hypothesis testing, the null hypothesis, denoted as \( H_0 \), represents a statement of no effect or no difference. It's the default assumption that there is no relationship or that a condition does not hold. Within the context of our exercise, the null hypothesis is that the average length of trout in Pyramid Lake is equal to 19 inches (\( H_0: \mu = 19 \)).

The purpose of the null hypothesis is to provide a basis for statistical testing. It serves as a benchmark against which the sample data are compared. If the data provide sufficient evidence, the null hypothesis might be rejected, suggesting something else may be true. It's essential to note that failing to reject the null hypothesis does not prove it true; it merely suggests insufficient evidence to support the alternative.
Alternative Hypothesis
The alternative hypothesis, denoted as \( H_a \) or \( H_1 \), proposes a competing statement to the null hypothesis. It is what the researcher aims to support through statistical testing. For our exercise, the alternative hypothesis indicates that the average length of trout is less than 19 inches (\( H_a: \mu < 19 \)). This suggests that any deviation from the null hypothesis in this direction supports the alternative.

An alternative hypothesis can be one-tailed or two-tailed, depending on whether the scope of the test is to detect an effect in one specific direction or in either direction. In this case, we have a one-tailed test since we're only interested in proving if the mean length is less than 19 inches. By focusing on the alternative hypothesis, hypothesis testing seeks to establish the plausibility that the sample has a different mean than the hypothesized population mean.

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Most popular questions from this chapter

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. Compute the \(z\) 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) Interpret your conclusion in the context of the application. Total blood volume (in ml) per body weight (in \(\mathrm{kg}\) ) is important in medical research. For healthy adults, the red blood cell volume mean is about \(\mu=28 \mathrm{ml} / \mathrm{kg}\) (Reference: Laboratory and Diagnostic Tests by F. Fischbach). Red blood cell volume that is too low or too high can indicate a medical problem (see reference). Suppose that Roger has had seven blood tests, and the red blood cell volumes were $$\begin{array}{rrrrrr} 32 & 25 & 41 & 35 & 30 & 37 & 29 \end{array}$$ The sample mean is \(\bar{x} \approx 32.7 \mathrm{ml} / \mathrm{kg} .\) Let \(x\) be a random variable that represents Roger's red blood cell volume. Assume that \(x\) has a normal distribution and \(\sigma=4.75 .\) Do the data indicate that Roger's red blood cell volume is different (either way) from \(\mu=28 \mathrm{ml} / \mathrm{kg} ?\) Use a 0.01 level of significance.

Please provide the following information for Problems. (a) What is the level of significance? State the null and alternate hypotheses. (b) Check Requirements What sampling distribution will you use? What assumptions are you making? Compute the sample test statistic and corresponding \(z\) or \(t\) value as appropriate. (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) Interpret your conclusion in the context of the application. Note: For degrees of freedom \(d . f .\) not in the Student's \(t\) table, use the closest \(d . f .\) that is smaller. In some situations, this choice of \(d . f .\) may increase the \(P\) -value a small amount and therefore produce a slightly more "conservative" answer. Management: Lost Time In her book Red Ink Behaviors, Jean Hollands reports on the assessment of leading Silicon Valley companies regarding a manager's lost time due to inappropriate behavior of employees. Consider the following independent random variables. The first variable \(x_{1}\) measures a manager's hours per week lost due to hot tempers, flaming e-mails, and general unproductive tensions: The variable \(x_{2}\) measures a manager's hours per week lost due to disputes regarding technical workers' superior attitudes that their colleagues are "dumb and dispensable": i. Use a calculator with sample mean and standard deviation keys to verify that \(\bar{x}_{1} \approx 4.86, s_{1} \approx 3.18, \bar{x}_{2}=6.5,\) and \(s_{2} \approx 2.88\) ii. Does the information indicate that the population mean time lost due to hot tempers is different (either way) from population mean time lost due to disputes arising from technical workers' superior attitudes? Use \(\alpha=0.05 .\) Assume that the two lost-time population distributions are mound-shaped and symmetric.

For a Student's \(t\) distribution with \(d . f .=10\) and \(t=2.930\), (a) find an interval containing the corresponding \(P\) -value for a two-tailed test. (b) find an interval containing the corresponding \(P\) -value for a right- tailed test.

To test \(\mu\) for an \(x\) distribution that is mound-shaped using sample size \(n \geq 30,\) how do you decide whether to use the normal or the Student's \(t\) distribution?

Tree-ring dating from archaeological excavation sites is used in conjunction with other chronologic evidence to estimate occupation dates of prehistoric Indian ruins in the southwestern United States. It is thought that Burnt Mesa Pueblo was occupied around 1300 A.D. (based on evidence from potsherds and stone tools). The following data give tree-ring dates (A.D.) from adjacent archaeological sites (Bandelier Archaeological Excavation Project: Summer 1990 Excavations at Burnt Mesa Pueblo, edited by T. Kohler, Washington State University Department of Anthropology): $$\begin{aligned} &\begin{array}{ccccc} 1189 & 1267 & 1268 & 1275 & 1275 \end{array}\\\ &1271 \quad 1272 \quad 1316 \quad 1317 \quad1230 \end{aligned}$$ i. Use a calculator with mean and standard deviation keys to verify that \(\bar{x}=1268\) and \(s \approx 37.29\) years. ii. Assuming the tree-ring dates in this excavation area follow a distribution that is approximately normal, does this information indicate that the population mean of tree-ring dates in the area is different from (either higher or lower than) that in 1300 A.D.? Use a \(1 \%\) level of significance.

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