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Chapter 9: Problem 12

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 no sufficient evidence to conclude that the average length of trout is less than 19 inches.

Step by step solution

01

Define Hypotheses

We start by defining the null and alternative hypotheses. The null hypothesis is\( H_0: \mu = 19 \) inches, suggesting no difference from the average reported by your friend. The alternative hypothesis is \( H_a: \mu < 19 \) inches, indicating that the average length is less than 19 inches.
02

Determine the Test Statistic

To decide which statistical test to use, we will use a t-test since we have sample data rather than the entire population data and the population standard deviation is unknown. The test statistic is calculated using the formula: \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]Where \( \bar{x} = 18.5 \), \( \mu = 19 \), \( s = 3.2 \), and \( n = 51 \). Substitute these values to find:\[ t = \frac{18.5 - 19}{3.2/\sqrt{51}} \]
03

Calculate the Test Statistic

Now, calculate the test statistic:\[ t = \frac{-0.5}{3.2/\sqrt{51}} = \frac{-0.5}{3.2 / 7.141} \approx \frac{-0.5}{0.448} \approx -1.116 \]
04

Determine Critical Value

Using the t-distribution table, we find the critical value for a one-tailed test with \( \alpha = 0.05 \) and \( df = n - 1 = 50 \) degrees of freedom. The critical value is approximately -1.676.
05

Make a Decision

Compare the calculated test statistic to the critical value: \( t \approx -1.116 \) and \( t_{critical} = -1.676 \). Since \(-1.116 > -1.676\), we do not reject the null hypothesis.
06

Conclusion

There is not enough statistical evidence to support the claim that the average length of a trout caught in Pyramid Lake is less than 19 inches at the \( \alpha = 0.05 \) significance level.

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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. In our exercise, we are using it to compare the average length of trout caught at Pyramid Lake to the hypothesized mean. It is particularly useful when the sample size is small or when the population standard deviation is unknown.

The formula for the t-test is:
  • \( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \)
Where:
  • \( \bar{x} \) is the sample mean
  • \( \mu \) is the population mean
  • \( s \) is the sample standard deviation
  • \( n \) is the sample size
Understanding these terms and the formula is essential, as they will guide you in calculating the test statistic, which we compare against a critical value found in the t-distribution table.
Explaining the Null Hypothesis
The null hypothesis is a critical part of hypothesis testing and serves as a statement of no effect or no difference. In our case, the null hypothesis \( H_0 \) states that the average length of trout in Pyramid Lake is exactly 19 inches. It acts as a starting point for our statistical test.

We assume the null hypothesis to be true and use statistical testing to determine whether there is enough evidence to reject it. This does not mean we prove the null true; rather, failing to reject it means there is insufficient evidence to support a difference.
Exploring the Alternative Hypothesis
The alternative hypothesis offers a contradictory statement to the null hypothesis. It presents what we are testing for, the possibility of a new effect of result that is different from the current theory or belief. For this exercise, the alternative hypothesis \( H_a \) suggests that the average length of trout in Pyramid Lake is less than 19 inches.

Developing a clear alternative hypothesis is crucial as it guides the direction of the hypothesis test. It specifically represents the researcher's suspected result or claim they are trying to validate or find evidence for.
Defining the Significance Level
The significance level, denoted by \( \alpha \), is the threshold that determines when we reject the null hypothesis. It represents the probability of making a Type I error - incorrectly rejecting a true null hypothesis. In most research, a significance level of \( \alpha = 0.05 \) is standard.

In the current exercise:
  • A significance level of 0.05 means there is a 5% risk of concluding that a difference exists when there is none.
  • It is set before data collection or analysis begins, maintaining the objectivity and integrity of the test.
The significance level sets the critical value against which our test statistic is compared. If the test statistic falls beyond this critical value, the null hypothesis is rejected in favor of the alternative hypothesis.

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Most popular questions from this chapter

A random sample of \(n_{1}=288\) voters registered in the state of California showed that 141 voted in the last general election. A random sample of \(n_{2}=216\) registered voters in the state of Colorado showed that 125 voted in the most recent general election. (See reference in Problem 25.) Do these data indicate that the population proportion of voter turnout in Colorado is higher than that in California? Use a \(5 \%\) level of significance.

When using a Student's \(t\) distribution for a paired differences test with \(n\) data pairs, what value do you use for the degrees of freedom?

How much customers buy is a direct result of how much time they spend in the store. A study of average shopping times in a large national houseware store gave the following information (Source: Why We Buy: The Science of Shopping by P. Underhill): Women with female companion: \(8.3 \mathrm{~min}\). Women with male companion: \(4.5 \mathrm{~min}\). Suppose you want to set up a statistical test to challenge the claim that a woman with a female friend spends an average of \(8.3\) minutes shopping in such a store. (a) What would you use for the null and alternate hypotheses if you believe the average shopping time is less than \(8.3\) minutes? Is this a right-tailed, left-tailed, or two-tailed test? (b) What would you use for the null and alternate hypotheses if you believe the average shopping time is different from \(8.3\) minutes? Is this a right- tailed, left-tailed, or two-tailed test? Stores that sell mainly to women should figure out a way to engage the interest of men! Perhaps comfortable seats and a big TV with sports programs. Suppose such an entertainment center was installed and you now wish to challenge the claim that a woman with a male friend spends only \(4.5\) minutes shopping in a houseware store. (c) What would you use for the null and alternate hypotheses if you believe the average shopping time is more than \(4.5\) minutes? Is this a right-tailed, lefttailed, or two-tailed test? (d) What would you use for the null and alternate hypotheses if you believe the average shopping time is different from \(4.5\) minutes? Is this a right- tailed, left-tailed, or two-tailed test?

Let \(x\) be a random variable that represents the \(\mathrm{pH}\) of arterial plasma (i.e., acidity of the blood). For healthy adults, the mean of the \(x\) distribution is \(\mu=7.4\) (Reference: Merck Manual, a commonly used reference in medical schools and nursing programs). A new drug for arthritis has been developed. However, it is thought that this drug may change blood \(\mathrm{pH} . \mathrm{A}\) random sample of 31 patients with arthritis took the drug for 3 months. Blood tests showed that \(\bar{x}=8.1\) with sample standard deviation \(s=1.9 .\) Use a \(5 \%\) level of significance to test the claim that the drug has changed (either way) the mean \(\mathrm{pH}\) level of the blood.

Consider a test for \(\mu\). If the \(P\) -value is such that you can reject \(H_{0}\) for \(\alpha=0.01\), can you always reject \(H_{0}\) for \(\alpha=0.05\) ? Explain.
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