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Throwing darts at the stock pages to decide which companies to invest in could be a successful stock-picking strategy. Suppose a researcher decides to test this theory and randomly chooses 100 companies to invest in. After 1 year, 53 of the companies were considered winners; that is, they outperformed other companies in the same investment class. To assess whether the dart-picking strategy resulted in a majority of winners, the researcher tested \(H_{0}: p=0.5\) versus \(H_{1}: p>0.5\) and obtained a \(P\) -value of \(0.2743 .\) Explain what this \(P\) -value means and write a conclusion for the researcher.

Short Answer

Expert verified
The P-value of 0.2743 indicates there is not enough evidence to reject the null hypothesis. The dart-picking strategy does not result in a majority of winners according to this test.

Step by step solution

01

- State the Null and Alternative Hypotheses

The null hypothesis ( H_0) is that the proportion of companies considered winners is 0.5 ( p = 0.5). The alternative hypothesis ( H_1) is that the proportion of companies considered winners is greater than 0.5 ( p > 0.5).
02

- Interpret the P-Value

The P -value represents the probability of observing the test results, or more extreme results, under the null hypothesis. In this case, a P -value of 0.2743 indicates that there is a 27.43% chance of obtaining a sample proportion of winners at least as extreme as 0.53, assuming the null hypothesis is true.
03

- Evaluate the P-Value Against the Significance Level

Typically, a significance level ( α) of 0.05 is used to decide whether to reject the null hypothesis. If the P -value is greater than α, we do not reject the null hypothesis. Here, the P -value (0.2743) is greater than 0.05.
04

- Draw a Conclusion

Since the P -value is greater than the significance level of 0.05, we do not reject the null hypothesis. This means there is not enough evidence to support the claim that the dart-picking strategy results in a majority of winners.

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Key Concepts

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

P-Value Interpretation
The P-value is a key aspect of hypothesis testing. It tells us the probability of observing our data, or something more extreme, if the null hypothesis is true.

In our example, the P-value is 0.2743. This means that if the dart-picking strategy had no effect (i.e., the true proportion of winners is 0.5), there is a 27.43% chance of randomly selecting a sample with 53 or more winners out of 100.

When we evaluate a P-value, we compare it to our chosen significance level, denoted as \( \alpha \). If the P-value is lower than this threshold, we have enough evidence to reject the null hypothesis. Otherwise, we do not reject it.
Null and Alternative Hypotheses
Hypothesis testing begins with formulating two contrasting hypotheses: the null hypothesis (\( H_0 \)) and the alternative hypothesis (\( H_1 \)).

In this example,
the null hypothesis is \( H_0: p = 0.5 \), indicating that the proportion of winning companies is 0.5;
while the alternative hypothesis is \( H_1: p > 0.5 \), suggesting that the strategy results in more than 50% winners.

These hypotheses set the stage for the test. The null hypothesis represents a default position that there is no effect or no difference. The alternative hypothesis, on the other hand, reflects the claim we are testing for.

By conducting a hypothesis test and using the observed P-value, we can determine whether the data provides sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.
Significance Level (Alpha)
The significance level is a critical value that helps us decide if a result is statistically significant. It is denoted as \( \alpha \) and typically set at 0.05.

In our dart-picking example, the significance level \( \alpha \) is 0.05. This means we are willing to accept a 5% chance of concluding that the strategy results in a majority of winners when it actually does not.

We compare our P-value to \( \alpha \) to make a decision. If the P-value is less than \( \alpha \), we reject the null hypothesis, suggesting that the observed effect is statistically significant.
In this case, the P-value of 0.2743 is greater than 0.05, so we do not reject the null hypothesis.

This indicates that the observed proportion of 53 winners is not significantly higher than 50%, meaning there's not enough evidence to prove that the dart-picking strategy is effective.

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Most popular questions from this chapter

The manufacturer of Prolong Engine Treatment claims that if you add one 12 -ounce bottle of its \(\$ 20\) product, your engine will be protected from excessive wear. An infomercial claims that a woman drove 4 hours without oil, thanks to Prolong. Consumer Reports magazine tested engines in which they added Prolong to the motor oil, ran the engines, drained the oil, and then determined the time until the engines seized. (a) Determine the null and alternative hypotheses that Consumer Reports will test. (b) Both engines took exactly 13 minutes to seize. What conclusion might Consumer Reports draw based on this evidence?

To test \(H_{0}: p=0.40\) versus \(H_{1}: p>0.40,\) a simple random sample of \(n=200\) individuals is obtained and \(x=84\) successes are observed. (a) What does it mean to make a Type II error for this test? (b) If the researcher decides to test this hypothesis at the \(\alpha=0.05\) level of significance, compute the probability of making a Type II error if the true population proportion is 0.44. What is the power of the test? (c) Redo part (b) if the true population proportion is 0.47

In December \(2001,38 \%\) of adults with children under the age of 18 reported that their family ate dinner together seven nights a week. In a recent poll, 403 of 1122 adults with children under the age of 18 reported that their family ate dinner together seven nights a week. Has the proportion of families with children under the age of 18 who eat dinner together seven nights a week decreased? Use the \(\alpha=0.05\) significance level.

To test \(H_{0}: \mu=50\) versus \(H_{1}: \mu<50,\) a simple random sample of size \(n=24\) is obtained from a population that is known to be normally distributed, and the sample standard deviation is found to be 6 . (a) A researcher decides to test the hypothesis at the \(\alpha=0.05\) level of significance. Determine the sample mean that separates the rejection region from the nonrejection region. [Hint: Follow the same approach as that laid out on page \(514,\) but use Student's \(t\) -distribution to find the critical value.] (b) Suppose the true population mean is \(\mu=48.9 .\) Use technology to find the area under the \(t\) -distribution to the right of the sample mean found in part (a) assuming \(\mu=48.9\). [Hint: This can be accomplished by performing a one-sample \(t\) -test.] This represents the probability of making a Type II error, \(\beta\). What is the power of the test?

A simple random sample of size \(n=20\) is drawn from a population that is normally distributed. The sample variance is found to be \(49.3 .\) Determine whether the population variance is less than 95 at the \(\alpha=0.1\) level of significance.

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