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Chapter 9: Problem 12

Athabasca Fishing Lodge is located on Lake Athabasca in northern Canada. In one of its recent brochures, the lodge advertises that \(75 \%\) of its guests catch northern pike over 20 pounds. Suppose that last summer 64 out of a random sample of 83 guests did, in fact, catch northern pike weighing over 20 pounds. Does this indicate that the population proportion of guests who catch pike over 20 pounds is different from \(75 \%\) (either higher or lower)? Use \(\alpha=0.05\).

Short Answer

Expert verified
No significant difference from 75% (p-value > 0.05, fail to reject).

Step by step solution

01

Define the Hypotheses

First, establish the null and alternative hypotheses. The null hypothesis is that the population proportion of guests catching northern pike over 20 pounds is 75%, denoted as \( H_0: p = 0.75 \). The alternative hypothesis is that the proportion is different from 75%, which can be expressed as \( H_a: p eq 0.75 \).
02

Set the Significance Level

We're given a significance level of \( \alpha = 0.05 \). This indicates that there's a 5% risk of concluding that a difference exists when there is no actual difference.
03

Calculate the Sample Proportion

Determine the sample proportion (\( \hat{p} \)) of guests who caught northern pike over 20 pounds. This is calculated as the number of successes divided by the sample size: \( \hat{p} = \frac{64}{83} \approx 0.7711 \).
04

Compute the Standard Error

The standard error of the sample proportion is computed using the formula:\[ SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.75(1-0.75)}{83}} \approx 0.0484 \]
05

Calculate the Test Statistic

Use the z-test for proportions to calculate the test statistic:\[Z = \frac{\hat{p} - p}{SE} = \frac{0.7711 - 0.75}{0.0484} \approx 0.437 \]
06

Determine the p-Value

Using a standard normal distribution table, determine the p-value that corresponds to the calculated test statistic. Because we have a two-tailed test, we look for the probability outside the range 鈥� for \( Z \approx 0.437 \), the p-value is approximately 0.662 (as it is a two-tailed test, we multiply by 2).
07

Compare p-Value and Significance Level

Since the p-value (0.662) is greater than the significance level (0.05), we fail to reject the null hypothesis.
08

Conclusion

There is not enough evidence to conclude that the actual proportion of guests catching pike over 20 pounds is different from 75%.

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Key Concepts

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

Understanding the Null Hypothesis
When conducting a hypothesis test, the first step involves defining your null hypothesis. The null hypothesis is a statement of no effect or no difference. It's where you assume that any observed difference is due to chance alone. In our fishing lodge example, the null hypothesis (\( H_0 \)) is that the true population proportion of guests who catch northern pike over 20 pounds is 75%. This means we start by assuming there's no change or deviation from what the brochure claims.
The null hypothesis is essential because it provides a benchmark against which we measure our test statistic. If the test shows that the results would be rare if the null hypothesis were true, we might reject this hypothesis in favor of an alternative.
Deciphering the Alternative Hypothesis
The alternative hypothesis is what you look to see if the data supports. It represents a new effect or difference that we suspect might exist based on our sample. In our scenario, the alternative hypothesis (\( H_a \)) suggests the proportion of guests catching pike over 20 pounds is different from 75%. This hypothesis is not just about verifying a difference; it's a reflection of curious inquiry into what could be true beyond the status quo. The alternative hypothesis is what we hope to support through our evidence, providing a meaningful narrative when our results show a significant difference.
Exploring the Z-Test for Proportions
The z-test for proportions is a statistical method used to determine if there is a significant difference between a sample proportion and a known or hypothesized population proportion. It is particularly useful when you have a large sample size. In this context, it helps to evaluate if the sample proportion of guests catching large pike is significantly different from the claimed 75%.To perform this test, we calculate a test statistic using the formula:\[ Z = \frac{\hat{p} - p}{SE} \]Where:
  • \( \hat{p} \) is the sample proportion
  • \( p \) is the hypothesized population proportion
  • \( SE \) is the standard error
The z-test helps determine whether the observed sample proportion is there by chance or represents a true population difference.
The Role of Sample Proportion
The sample proportion is a fundamental statistic used in hypothesis testing for proportions. It provides an estimate of the true population proportion based on sample data. In our exercise, the sample proportion (\( \hat{p} \)) is calculated by dividing the number of guests who caught pike over 20 pounds by the total number of guests sampled. For the fishing lodge data, this calculation is:\[ \hat{p} = \frac{64}{83} \approx 0.7711 \]Understanding the sample proportion is crucial as it serves as a basis for comparison against the claimed population proportion. It directly impacts the calculation of the z-test statistic. A sample proportion close to the hypothesized proportion might suggest that the null hypothesis is true, whereas a significant deviation could indicate a real difference in the population.

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Most popular questions from this chapter

Harper's Index reported that \(80 \%\) of all supermarket prices end in the digit 9 or \(5 .\) Suppose you check a random sample of 115 items in a supermarket and find that 88 have prices that end in 9 or \(5 .\) Does this indicate that less than \(80 \%\) of the prices in the store end in the digits 9 or 5 ? Use \(\alpha=0.05\).

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 manager's hours per week lost due to hot tempers, flaming e-mails, and general unproductive tensions: $$\begin{array}{llllllll} x_{1}: & 1 & 5 & 8 & 4 & 2 & 4 & 10 \end{array}$$ The variable \(x_{2}\) measures manager's hours per week lost due to disputes regarding technical workers' superior attitudes that their colleagues are "dumb and dispensable": $$\begin{array}{lllllllll} x_{2}: & 10 & 5 & 4 & 7 & 9 & 4 & 10 & 3 \end{array}$$ 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.

Would you favor spending more federal tax money on the arts? This question was asked by a research group on behalf of The National Institute (Reference: Painting by Numbers, J. Wypijewski, University of California Press). Of a random sample of \(n_{1}=93\) politically conservative voters, \(r_{1}=21\) responded yes. Another random sample of \(n_{2}=83\) politically moderate voters showed that \(r_{2}=22\) responded yes. Does this information indicate that the population proportion of conservative voters inclined to spend more federal tax money on funding the arts is less than the proportion of moderate voters so inclined? Use \(\alpha=0.05\).

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. Gentle Ben is a Morgan horse at a Colorado dude ranch. Over the past 8 weeks, a veterinarian took the following glucose readings from this horse (in \(\mathrm{mg} / 100 \mathrm{ml}\) ). \(\begin{array}{llllllll}93 & 88 & 82 & 105 & 99 & 110 & 84 & 89\end{array}\) The sample mean is \(\bar{x} \approx 93.8\). Let \(x\) be a random variable representing glucose readings taken from Gentle Ben. We may assume that \(x\) has a normal distribution, and we know from past experience that \(\sigma=12.5 .\) The mean glucose level for horses should be \(\mu=85 \mathrm{mg} / 100 \mathrm{ml}\) (Reference: Merck Veterinary Manual). Do these data indicate that Gentle Ben has an overall average glucose level higher than 85? Use \(\alpha=0.05\).

In a statistical test, we have a choice of a left-tailed test, a right-tailed test, or a two-tailed test. Is it the null hypothesis or the alternate hypothesis that determines which type of test is used? Explain your answer.
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