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Symposium is part of a larger work referred to as Plato's Dialogues. Wishart and Leach (see source in Problem 15\()\) found that about \(21.4 \%\) of five- syllable sequences in Symposium are of the type in which four are short and one is long. Suppose an antiquities store in Athens has a very old manuscript that the owner claims is part of Plato's Dialogues. A random sample of 493 five-syllable sequences from this manuscript showed that 136 were of the type four short and one long. Do the data indicate that the population proportion of this type of five-syllable sequence is higher than that found in Plato's Symposium? Use \(\alpha=0.01\).

Short Answer

Expert verified
The manuscript's proportion is significantly higher than 21.4%.

Step by step solution

01

Define the Hypotheses

We need to set up our null and alternative hypotheses. The null hypothesis \( H_0 \) states that the population proportion of the sequences in the manuscript is equal to 21.4%, while the alternative hypothesis \( H_a \) states that the population proportion is greater than 21.4%. Thus, the hypotheses are:- \( H_0: p = 0.214 \)- \( H_a: p > 0.214 \)
02

Determine the Test Statistic

We will use a z-test for a single proportion since we are testing if one proportion is greater than the hypothesized proportion. The test statistic is calculated as follows:\[z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}\]where \( \hat{p} = \frac{136}{493} \) is the sample proportion, \( p = 0.214 \) is the population proportion from the hypothesis, and \( n = 493 \) is the sample size. Calculate \( \hat{p} \):\[\hat{p} = \frac{136}{493} \approx 0.276\].
03

Calculate the Standard Error

The standard error (SE) is calculated using the formula:\[SE = \sqrt{\frac{p(1-p)}{n}}\]Substitute \( p = 0.214 \) and \( n = 493 \):\[SE = \sqrt{\frac{0.214(1-0.214)}{493}} \approx 0.019\].
04

Compute the z-Statistic

Substitute \( \hat{p} \), \( p \), and the SE into the z-test formula:\[z = \frac{0.276 - 0.214}{0.019} \approx 3.263\]
05

Decide Using the Significance Level

For a right-tailed test at \( \alpha = 0.01 \), we compare the calculated z-statistic against the z-critical value. The z-critical value for \( \alpha = 0.01 \) is approximately 2.33. Since \( z = 3.263 \) is greater than 2.33, we reject the null hypothesis.
06

Conclude the Hypothesis Test

Since we rejected the null hypothesis, the data suggest that the proportion of sequences of type "four short and one long" in the manuscript is significantly higher than 21.4% at a \( 1\% \) significance level.

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Key Concepts

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

Z-test
The Z-test is a powerful tool in statistical hypothesis testing used when we're dealing with large sample sizes or known population variances. In this context, we use a Z-test to compare the sample proportion of sequences found in the manuscript with the known proportion from Plato's "Symposium." It helps us determine if the difference between these two proportions is statistically significant.

One key reason to choose a Z-test in this scenario is that our sample size, 493 sequences, is large enough to assume normal distribution according to the Central Limit Theorem.
This allows us to use the Z-test formula effectively.

The computed result, known as the Z-statistic, allows us to compare it against a critical value which lets us decide whether to reject or fail to reject the null hypothesis.
  • When the Z-statistic exceeds a certain critical value, it suggests that the observed effect is real, not due to random sampling error.
Proportion
Proportion refers to the fraction or percentage of a total that has a particular attribute. In the context of hypothesis testing, we are interested in comparing the sample proportion of a specific trait with a known or hypothesized proportion.

In our problem, we calculate the sample proportion, denoted as \( \hat{p} \), to find the proportion of the five-syllable sequences that match the type being studied. By computing \( \hat{p} = \frac{136}{493} \), we find that approximately 27.6% of sequences match the four short, one long pattern.

This sample proportion is crucial as it serves as our test statistic's basis, allowing us to perform a Z-test. Calculating the effect of this proportion and comparing it against the hypothesized population proportion of 21.4% gives insight into whether the manuscript could potentially belong to "Symposium."
  • Proportions like these help in making inferences about a population based on sample data.
  • Understanding the proportion facilitates interpretation of findings in practical terms, such as determining authenticity in this exercise.
Standard Error
The standard error (SE) is a measure of how much we expect the sample proportion \( \hat{p} \) to fluctuate from the true population proportion \( p \) due to random sampling variation. It essentially provides an estimate of the variability around the sample mean.

The formula for standard error in a proportion test is given by: \[ SE = \sqrt{\frac{p(1-p)}{n}} \] where \( p \) is the hypothesized population proportion, and \( n \) is the sample size.

In our exercise, the calculated SE was approximately 0.019. This small standard error suggests that the sample proportion should be close to the population proportion, assuming the null hypothesis is true.
  • The standard error helps gauge the precision of our sample statistic as an estimate of the population parameter.
  • By taking into account the SE when computing the Z-statistic, we are considering the potential sampling error in our inference.
Null Hypothesis
In statistical hypothesis testing, the null hypothesis \( H_0 \) stands as a default assumption that there is no effect or difference. It's what we test against to determine the presence of a significant change or difference.

For our scenario, the null hypothesis posits that the proportion of sequences in the manuscript matches the known proportion from the "Symposium," which is 21.4%. We officially state this as \( H_0: p = 0.214 \).

The whole purpose of setting a null hypothesis is to establish a baseline for comparison. When we conduct our Z-test, we assess whether to reject this null hypothesis based on the evidence from our sample data. If we reject it, as we do in this instance where the calculated Z-statistic is greater than the critical value, it suggests that the proportion in the manuscript is significantly different and indeed higher than 21.4%.
  • The null hypothesis helps in structuring the hypothesis test and provides a baseline for significance testing.
  • It’s important because it represents the absence of an effect or relationship, serving as a starting point in statistical testing.

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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 \(\alpha=0.05\)

When conducting a test for the difference of means for two independent populations \(x_{1}\) and \(x_{2}\), what alternate hypothesis would indicate that the mean of the \(x_{2}\) population is smaller than that of the \(x_{1}\) population? Express the alternate hypothesis in two ways.

Women athletes at the University of Colorado, Boulder, have a long-term graduation rate of \(67 \%\) (Source: Cbronicle of Higher Education). Over the past several years, a random sample of 38 women athletes at the school showed that 21 eventually graduated. Does this indicate that the population proportion of women athletes who graduate from the University of Colorado, Boulder, is now less than \(67 \%\) ? Use a \(5 \%\) level of significance.

Please read the Focus Problem at the beginning of this chapter. Recall that Benford's Law claims that numbers chosen from very large data files tend to have " \(1 "\) as the first nonzero digit disproportionately often. In fact, research has shown that if you randomly draw a number from a very large data file, the probability of getting a number with " 1 " as the leading digit is about \(0.301\) (see the reference in this chapter's Focus Problem). Now suppose you are an auditor for a very large corporation. The revenue report involves millions of numbers in a large computer file. Let us say you took a random sample of \(n=215\) numerical entries from the file and \(r=46\) of the entries had a first nonzero digit of \(1 .\) Let \(p\) represent the population proportion of all numbers in the corporate file that have a first nonzero digit of \(1 .\) j. Test the claim that \(p\) is less than \(0.301\). Use \(\alpha=0.01\). ii. If \(p\) is in fact less than \(0.301\), would it make you suspect that there are not enough numbers in the data file with leading 1 's? Could this indicate that the books have been "cooked" by "pumping up" or inflating the numbers? Comment from the viewpoint of a stockholder. Comment from the perspective of the Federal Bureau of Investigation as it looks for money laundering in the form of false profits. iii. Comment on the following statement: "If we reject the null hypothesis at level of significance \(\alpha\), we have not proved \(H_{0}\) to be false. We can say that the probability is \(\alpha\) that we made a mistake in rejecting \(H_{0} . "\) Based on the outcome of the test, would you recommend further investigation before accusing the company of fraud?

Consider a test for \(\mu\). If the \(P\) -value is such that you can reject \(H_{0}\) at the \(5 \%\) level of significance, can you always reject \(H_{0}\) at the \(1 \%\) level of significance? Explain.

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