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Bill Alther is a zoologist who studies Anna's hummingbird (Calypte anna) (Reference: Hummingbirds by K. Long and W. Alther). Suppose that in a remote part of the Grand Canyon, a random sample of six of these birds was caught, weighed, and released. The weights (in grams) were \(\begin{array}{llllll}3.7 & 2.9 & 3.8 & 4.2 & 4.8 & 3.1\end{array}\) The sample mean is \(\bar{x}=3.75\) grams. Let \(x\) be a random variable representing weights of Anna's hummingbirds in this part of the Grand Canyon. We assume that \(x\) has a normal distribution and \(\sigma=0.70\) gram. It is known that for the population of all Anna's hummingbirds, the mean weight is \(\mu=4.55\) grams. Do the data indicate that the mean weight of these birds in this part of the Grand Canyon is less than \(4.55\) grams? Use \(\alpha=0.01\).

Short Answer

Expert verified
Yes, the data indicate the mean weight is less than 4.55 grams at the 0.01 significance level.

Step by step solution

01

State the Hypotheses

We begin by setting up our null and alternative hypotheses. The null hypothesis, denoted by \( H_0 \), is that the mean weight \( \mu \) of the hummingbirds in this part of the Grand Canyon is equal to the general population mean weight of 4.55 grams. Mathematically, it can be stated as \( H_0: \mu = 4.55 \). The alternative hypothesis, denoted by \( H_a \), is that the mean weight is less than 4.55 grams, which can be stated as \( H_a: \mu < 4.55 \).
02

Compute the Test Statistic

Using the formula for the test statistic for one-sample mean when the population standard deviation is known:\[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]Substitute the given values: \( \bar{x} = 3.75 \), \( \mu = 4.55 \), \( \sigma = 0.70 \), and \( n = 6 \):\[ z = \frac{3.75 - 4.55}{0.70 / \sqrt{6}} \approx \frac{-0.8}{0.285} \approx -2.81 \]
03

Determine the Critical Value

For a one-tailed test at \( \alpha = 0.01 \), we look up the critical value in the standard normal distribution table. The critical value \( z_{\alpha} \) for \( \alpha = 0.01 \) in the left tail is approximately \( -2.33 \).
04

Make the Decision

Compare the calculated test statistic \( z = -2.81 \) with the critical value \( z_{\alpha} = -2.33 \). Since \( z = -2.81 \) is less than \( -2.33 \), it falls in the rejection region.
05

Conclusion

Based on our decision, we have sufficient evidence at the \( \alpha = 0.01 \) significance level to reject the null hypothesis. This suggests that the mean weight of Anna's hummingbirds in this part of the Grand Canyon is indeed less than 4.55 grams.

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Key Concepts

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

One-Sample Z-Test
The One-Sample Z-Test is a statistical method used to determine whether there is a significant difference between the mean of a sample and a known population mean. In this case, the zoologist wants to compare the average weight of Anna's hummingbirds in a specific region with the known average weight for the entire population. It's perfect for situations where:
  • The sample size is small (typically less than 30).
  • The population standard deviation is known.
  • The data is normally distributed.
To perform this test, you calculate a Z-score using the formula:\[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]where:
  • \(\bar{x}\) is the sample mean.
  • \(\mu\) is the population mean.
  • \(\sigma\) is the population standard deviation.
  • \(n\) is the sample size.
This formula helps us understand how many standard deviations the sample mean is from the population mean. In our hummingbird example, the calculated Z-score was -2.81, which indicates the sample mean weight is significantly different, and lower, compared to the population mean. This method allows scientists like Bill Alther to objectively assess differences in their observations.
Statistical Significance
Statistical significance is a way to quantify whether a result from data analysis likely reflects a true effect rather than random chance. In hypothesis testing, we use a significance level, represented by \(\alpha\), to determine if our results are significant. Common \(\alpha\) levels are 0.05 or 0.01, which correspond to a 5% or 1% threshold for what is considered significant.

In the hummingbird exercise, the significance level \(\alpha\) was set at 0.01, meaning we are seeking a result that has less than a 1% probability of occurring if the null hypothesis is true. A hypothesis test compares the p-value (a probability score derived from the test statistic) to the \(\alpha\). If the p-value is less than \(\alpha\), the null hypothesis is rejected.

This threshold helps researchers understand if their findings, as in the hummingbird weight data, are likely due to a real difference in the sample—such as environmental factors in the Grand Canyon—rather than random variation.
Normal Distribution
The normal distribution, often called a bell curve, is critical in hypothesis testing because many statistical methods, including the Z-test, assume data that is normally distributed. It's characterized by having:
  • A symmetric shape around the mean.
  • Most data falling close to the mean.
  • Decreasing probabilities as you move away from the mean.
In the context of the hummingbird weights, assuming a normal distribution means that these weights, if visualized on a graph, would form a bell-shaped curve. This assumption justifies using the Z-test since the hummingbird data is said to follow a normal distribution.

Why this assumption? Normal distribution is common in nature and measurements, making it a useful model for biological data like weight. It helps provide a backbone for statistical inference, allowing conclusions about population parameters based on sample data. This is why scientists, even like our zoologist, value it for its ability to simplify complex analyses.

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