/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 70 Students at the Akademia Podlaka... [FREE SOLUTION] | 91Ó°ÊÓ

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Students at the Akademia Podlaka conducted an experiment to determine whether the Belgium-minted Euro coin was equally likely to land heads up or tails up. Coins were spun on a smooth surface, and in 250 spins, 140 landed with the heads side up (New Scientist, January 4,2002 ). Should the students interpret this result as convincing evidence that the proportion of the time the coin would land heads up is not. 5? Test the relevant hypotheses using \(\alpha=.01\). Would your conclusion be different if a significance level of \(.05\) had been used? Explain.

Short Answer

Expert verified
The answer depends on the calculated test statistic and its comparison to the critical value at the \(\alpha=.01\) and \(.05\) levels of significance, which will be obtained from Steps 2 to 5 in the step-by-step solution.

Step by step solution

01

Set Up the Hypotheses

The null hypothesis (\(H_0\)) is that the coin is fair (or in this case, the proportion (\(p_0\)) of the time the coin lands heads up is 0.5). The alternate hypothesis (\(H_1\)) is that the coin is not fair (or \(p \neq p_0\)). So, \(H_0: p = 0.5\) and \(H_1: p \neq 0.5\).
02

Calculate the Test Statistic

The test statistic for a proportion is given by the formula \( Z = ( \hat{p} - p_0 ) / (\sqrt{p_0*(1-p_0)/n}) \) where \(\hat{p}\) is the sample proportion, \(p_0\) is the hypothesized population proportion, and \(n\) is the sample size. Here, \(\hat{p} = 140/250 = 0.56\), \(p_0 = 0.5\) and \(n = 250\). So, calculate the test statistic value by substituting these values into the formula.
03

Compare to Critical Value

The critical value for a two-tailed test at \(\alpha=.01\) level of significance is approximately 2.58. If the calculated test statistic value is greater than 2.58, reject the null hypothesis on the ground that it is too unlikely to obtain a sample like the one observed, assuming the null hypothesis is true. If the calculated test statistic value is less than 2.58, there is insufficient evidence to reject the null hypothesis.
04

Compare to New Significance Level

Repeat the comparison done in Step 3, but now using the significance level of \(.05\). The critical value is approximately \(1.96\) for a two-tailed test at this significance level. Compare the calculated test statistic value against \(1.96\) and draw a conclusion.
05

Conclusion

Based on Steps 3 and 4, explain the conclusion and how the change in significance level can affect the decision of rejecting the null hypothesis.

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Key Concepts

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

Null and Alternate Hypothesis
When conducting hypothesis testing in statistics, determining the null and alternate hypotheses is a crucial first step. The null hypothesis, represented as \(H_0\), is a default assumption that there is no effect or no difference, which we aim to test against the evidence. In our coin example, it states that the Belgian Euro coin has a fair chance of landing heads up \(p = 0.5\).

On the other side, we have the alternate hypothesis \(H_1\), which suggests that the initial assumption of the null hypothesis is false. It is usually set up to demonstrate what we are trying to prove; in this case, that the coin is biased \(p eq 0.5\).

The objective of hypothesis testing is to determine whether we can reject the null hypothesis, thereby providing evidence that supports the alternate hypothesis.
Test Statistic Calculation
The test statistic is a value used to determine whether to reject the null hypothesis. It compares the observed data to what is expected under the null hypothesis. To calculate the test statistic for sample proportions, we use the formula:
\[ Z = ( \hat{p} - p_0 ) / (\sqrt{p_0*(1-p_0)/n}) \]
where \(\hat{p}\) is the sample proportion, \(p_0\) is the hypothesized population proportion, and \(n\) is the sample size. For our coin-tossing example, \(\hat{p}\) is 0.56, \(p_0\) is 0.5, and \(n\) is 250. Inserting these values into the calculation provides us with our test statistic, which is then compared to the critical value associated with our chosen significance level.
Significance Level
In hypothesis testing, the significance level \(\alpha\) represents the probability of rejecting the null hypothesis when it is actually true, known as a Type I error. It is a threshold for determining the statistical significance of the test. Common choices for \(\alpha\) are 0.05, 0.01, and 0.10.

If the absolute value of the test statistic is greater than the critical value associated with \(\alpha\), we reject the null hypothesis, suggesting the sample provides sufficient evidence that the null hypothesis may not be true. With our coin example, we initially used \(\alpha=.01\) but also considered the commonly used level of \(\alpha=.05\). The choice of \(\alpha\) reflects how stringent we are in our decision-making process; a smaller \(\alpha\), such as 0.01, requires stronger evidence to reject the null hypothesis than a larger \(\alpha\) would.
Sample Proportion Analysis
Sample proportion analysis deals with summarizing the proportion of items in a sample that have a particular attribute. In our scenario, we want to know the proportion of times the coin lands heads up. We use the formula: \(\hat{p} = x/n\) where \(x\) is the number of heads and \(n\) is the total number of spins.

In our calculation, there were 140 heads in 250 spins, thus \(\hat{p} = 140/250 = 0.56\). We analyze this sample proportion in the context of the null hypothesis and using the test statistic to determine if the observed proportion \(\hat{p}\) is significantly different from the hypothesized proportion under \(H_0\).

This analysis informs us whether the deviation observed (from \(p_0 = 0.5\)) is simply due to the randomness of drawing a sample or if it's statistically significant enough to suggest a true bias in the coin.

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Most popular questions from this chapter

Researchers at the University of Washington and Harvard University analyzed records of breast cancer screening and diagnostic evaluations ("Mammogram Cancer Scares More Frequent than Thought," USA Today. April 16, 1998). Discussing the benefits and downsides of the screening process, the article states that, although the rate of false-positives is higher than previously thought, if radiologists were less aggressive in following up on suspicious tests, the rate of false-positives would fall but the rate of missed cancers would rise. Suppose that such a screening test is used to decide between a null hypothesis of \(H_{0}:\) no cancer is present and an alternative hypothesis of \(H_{a}\) : cancer is present. (Although these are not hypotheses about a population characteristic, this exercise illustrates the definitions of Type I and Type II errors.) a. Would a false-positive (thinking that cancer is present when in fact it is not) be a Type I error or a Type II error? b. Describe a Type I error in the context of this problem, and discuss the consequences of making a Type I error. c. Describe a Type II error in the context of this problem, and discuss the consequences of making a Type II error. d. What aspect of the relationship between the probability of Type I and Type II errors is being described by the statement in the article that if radiologists were less aggressive in following up on suspicious tests, the rate of false-positives would fall but the rate of missed cancers would rise?

A comprehensive study conducted by the National Institute of Child Health and Human Development tracked more than 1000 children from an early age through elementary school (New york Times, November 1, 2005). The study concluded that children who spent more than 30 hours a week in child care before entering school tended to score higher in math and reading when they were in the third grade. The researchers cautioned that the findings should not be a cause for alarm because the effects of child care were found to be small. Explain how the difference between the sample mean math score for third graders who spent long hours in child care and the known overall mean for third graders could be small but the researchers could still reach the conclusion that the mean for the child care group is significantly higher than the overall mean for third graders.

Explain why the statement \(\bar{x}=50\) is not a legitimate hypothesis.

The power of a test is influenced by the sample size and the choice of significance level. a. Explain how increasing the sample size affects the power (when significance level is held fixed). b. Explain how increasing the significance level affects the power (when sample size is held fixed).

Consider the following quote from the article "Review Finds No Link Between Vaccine and Autism" (San Luis Obispo Tribune, October 19,2005\()\) : "'We found no evidence that giving MMR causes Crohn's disease and/or autism in the children that get the MMR,' said Tom Jefferson, one of the authors of The Cochrane Review. 'That does not mean it doesn't cause it. It means we could find no evidence of it.'" (MMR is a measlesmumps-rubella vaccine.) In the context of a hypothesis test with the null hypothesis being that MMR does not cause autism, explain why the author could not conclude that the MMR vaccine does not cause autism.

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