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According to the Washington Post, nearly 45\% of all Americans are born with brown eyes, although their eyes don't necessarily stay brown. \(^{\star}\) A random sample of 80 adults found 32 with brown eyes. Is there sufficient evidence at the .01 level to indicate that the proportion of brown eyed adults differs from the proportion of Americans who are born with brown eyes?

Short Answer

Expert verified
There is no sufficient evidence to reject the null hypothesis at the 0.01 level.

Step by step solution

01

Set Hypotheses

Define the null and alternative hypotheses for the problem. The null hypothesis (H0) states that the proportion of brown-eyed adults is the same as the proportion of Americans born with brown eyes. The alternative hypothesis (H1) states that the proportion of brown-eyed adults differs. Mathematically, these hypotheses can be expressed as: \( H_0: p = 0.45 \) \( H_1: p eq 0.45 \)
02

Calculate the Test Statistic

First, determine the sample proportion \( \hat{p} \) by dividing the number of brown-eyed adults by the sample size. Here, \( \hat{p} = \frac{32}{80} = 0.4 \). The test statistic for a proportion can be calculated using the formula:\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \]Substitute \( \hat{p} = 0.4 \), \( p_0 = 0.45 \), and \( n = 80 \), to get: \[ z = \frac{0.4 - 0.45}{\sqrt{\frac{0.45 \times 0.55}{80}}} \]
03

Simplify the Test Statistic

Calculate the standard error: \[ SE = \sqrt{\frac{0.45 \times 0.55}{80}} \approx 0.0541 \]Then use it to simplify the test statistic:\[ z = \frac{-0.05}{0.0541} \approx -0.924 \]
04

Determine the Critical Value

Since this is a two-tailed test at the 0.01 significance level, you need to find the critical z-values that correspond to \( \alpha/2 = 0.005 \) in each tail. Using a standard normal distribution table, the critical values are approximately \( z = \pm 2.576 \).
05

Compare Test Statistic and Critical Value

Compare the calculated test statistic with the critical value. The test statistic \( z = -0.924 \) is between \(-2.576\) and \(+2.576\).
06

Make a Decision

The calculated test statistic is not in the critical region, so we do not reject the null hypothesis.
07

Conclusion

There is not enough evidence at the 0.01 significance level to indicate that the proportion of brown-eyed adults differs from the proportion of Americans who are born with brown eyes.

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Key Concepts

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

Proportion Testing
Proportion Testing is a statistical method used to compare a sample proportion to a population proportion to determine if there is a significant difference. In this case, we are comparing the proportion of brown-eyed adults found in a sample to the known proportion of 0.45 of Americans born with brown eyes. This test helps us decide if the observed sample proportion is significantly different from the expected population proportion, or if it could have occurred by random chance. In our exercise, we start by considering the null hypothesis, which assumes no difference between the sample and population proportions. The alternative hypothesis, on the other hand, considers that a difference exists. We calculate a test statistic, specifically a Z-test statistic, to determine how far away the sample proportion is from the hypothesized population proportion.
Significance Level
The Significance Level, often denoted by \(\alpha\), is a threshold used in hypothesis testing to decide whether to reject the null hypothesis. It represents the probability of rejecting the null hypothesis when it is actually true, known as a Type I error. Common significance levels are 0.05, 0.01, and 0.10. In this problem, the significance level is set at 0.01. This means that there is a 1% risk of concluding that there is a difference in the proportions when, in fact, there isn't. A lower significance level such as this requires stronger evidence to reject the null hypothesis, making it less likely to observe a false positive, i.e., claiming a difference when none exists. In scenarios where the calculated test statistic does not fall beyond the critical value set by the significance level, we do not reject the null hypothesis.
Z-Test
A Z-Test is a type of hypothesis test that helps us determine if there is a significant difference between sample data and a known population parameter. It is specifically used when the population variance is known or the sample size is large, which is typically considered to be more than 30. In proportion testing, the Z-test statistic is calculated using the formula:\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \]where \( \hat{p} \) is the sample proportion, \( p_0 \) is the population proportion, and \( n \) is the sample size. In our exercise, using a sample of 80 individuals, we found a sample proportion of 0.4. After computing the Z-test statistic, we compared it to the critical Z-values determined by our significance level to decide whether the sample proportion significantly differs from the population proportion.
Sample Proportion
Sample Proportion, represented as \( \hat{p} \), is the proportion of individuals in a sample who have a particular attribute. It is calculated by dividing the number of individuals with the attribute by the total number of individuals in the sample. In our exercise, the sample proportion calculated was \( \hat{p} = \frac{32}{80} = 0.4 \). This proportion is an estimate of the true population proportion. We use the sample proportion, together with the population proportion, to compute the test statistic in the Z-test. By evaluating this sample measure against the population proportion, we can infer if our sample proportion significantly deviates from the hypothesized population value, under the framework of a hypothesis test. When the sample proportion differs from the assumed population proportion beyond what would be expected due to random sampling variability, a significant result can be inferred.

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Most popular questions from this chapter

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