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91Ó°ÊÓ

GRE performance again Instead of advertising the percentage of customers who improve by at least 10 points, a manager suggests testing whether the mean score improves at all. For each customer they record the difference in score before and after taking the course (After \(-\) Before). a. State the null and alternative hypotheses. b. The P-value from the test is 0.65 . Does this provide any evidence that their course works? c. From part \(b,\) what can you tell, if anything, about the mean difference in the sample scores?

Short Answer

Expert verified
Null Hypothesis: The mean score after the course minus the mean score before the course equals zero. Alternative Hypothesis: The mean score after the course minus the mean score before the course does not equal zero. The P-value of 0.65 is high, which means there is not enough evidence to reject the null hypothesis, implying the course might not have significantly improved the scores. If anything can be said about the mean difference in sample scores, it is likely close to zero.

Step by step solution

01

Formulating the Null and Alternative Hypotheses

Null Hypothesis (\(H_0\)): The mean score difference (After - Before) is 0. (Mean difference = 0). In other words, there's no difference in the mean scores before and after taking the course. \n Alternative Hypothesis (\(H_a\)): The mean score difference (After - Before) is not 0. (Mean difference ≠ 0). So there's a difference in the mean scores before and after taking the course.
02

Interpret the P-value

A P-value of 0.65 is quite large (greater than 0.05), which means that we fail to reject the null hypothesis. This P-value indicates that there is not enough evidence to suggest that the course results in an improvement in scores.
03

Understanding the mean difference based on the P-value

Given that the P-value is quite high, it suggests that the sample mean difference of score is therefore close to zero, assuming that other conditions for the test hold. Since the test has failed to provide strong evidence against the null hypothesis, the mean difference in before and after the course scores is likely to be zero or very small.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis plays a crucial role. It's essentially a statement that suggests there is no effect or difference in the scenario you're studying. This hypothesis sets the baseline for the testing.

- In the context of the given GRE performance scenario, the null hypothesis (\(H_0\)) claims that the mean score difference between before and after the course is zero.- By assuming the mean difference is zero, it implies there's no impact of the course on the scores.

Under this assumption, we gather data and perform statistical tests to either find evidence against this assumption or fail to find such evidence. If evidence exists against the null hypothesis, we might consider rejecting it in favor of an alternative explanation.
Alternative Hypothesis
Once you’ve understood the null hypothesis, the alternative hypothesis comes into play. Unlike the null hypothesis, the alternative hypothesis suggests there is an effect or a difference.

- For the GRE scenario, the alternative hypothesis (\(H_a\)) claims the mean score difference is not zero. This means the scores have either increased or decreased after the course.- It opposes the null hypothesis and provides the framework to show the course had an impact on the scores.

The goal of hypothesis testing is to analyze data to see if there is enough evidence to support this alternative hypothesis. It seeks to show that something different or unexpected is occurring compared to what the null hypothesis predicts.
P-value
The P-value is a critical component in hypothesis testing, providing a means to make decisions. It quantifies the probability of observing the data, assuming that the null hypothesis is true.
- In the example provided about GRE scores, a P-value of 0.65 indicates a high probability under the assumption of the null hypothesis being true. - Since 0.65 is greater than the typical threshold of 0.05, it suggests that the observed results aren't unusual assuming the null is correct.

This high P-value tells us that there's a lack of evidence to reject the null hypothesis. In authentic testing scenarios, it's essential to remember that a high P-value doesn't verify the null hypothesis, but rather indicates insufficient evidence to dismiss it.
Mean Difference
The concept of mean difference helps in quantifying the change or effect in a research study. Essentially, it's the average amount by which scores vary from one group to another.
- Within the context of the test on GRE scores, the mean difference refers to the change in scores after subtracting the scores before from those after the course. - A high P-value indicates that this mean difference is likely near zero, suggesting either no significant improvement or a small effect from the course.

By focusing on the mean difference, researchers can better understand the magnitude of changes and make informed decisions based on the statistical significance demonstrated through hypothesis testing.

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Most popular questions from this chapter

Expensive medicine Developing a new drug can be an expensive process, resulting in high costs to patients. A pharmaceutical company has developed a new drug to reduce cholesterol, and it will conduct a clinical trial to compare the effectiveness to the most widely used current treatment. The results will be analyzed using a hypothesis test. a. If the test yields a low P-value and the researcher rejects the null hypothesis that the new drug is not more effective, but it actually is not better, what are the consequences of such an error? b. If the test yields a high \(\mathrm{P}\) -value and the researcher fails to reject the null hypothesis, but the new drug is more effective, what are the consequences of such an error?

WebZine A magazine is considering the launch of an online edition. The magazine plans to go ahead only if it's convinced that more than \(25 \%\) of current readers would subscribe. The magazine contacted a Simple Random Sample of 500 current subscribers, and 137 of those surveyed expressed interest. What should the company do? Test an appropriate hypothesis and state your conclusion. Be sure the appropriate assumptions and conditions are satisfied before you proceed.

Bad medicine Occasionally, a report comes out that a drug that cures some disease turns out to have a nasty side effect. For example, some antidepressant drugs may cause suicidal thoughts in younger patients. A researcher wants to study such a drug and look for evidence of a side effect. a. If the test yields a low P-value and the researcher rejects the null hypothesis, but there is actually no ill side effect of the drug, what are the consequences of such an error? b. If the test yields a high \(\mathrm{P}\) -value and the researcher fails to reject the null hypothesis, but there is a bad side effect of the drug, what are the consequences of such an error?

Pollution A company with a fleet of 150 cars found that the emissions systems of 7 out of the 22 they tested failed to meet pollution control guidelines. Is this strong evidence that more than \(20 \%\) of the fleet might be out of compliance? Test an appropriate hypothesis and state your conclusion. Be sure the appropriate assumptions and conditions are satisfied before you proceed.

Obesity 2016 In \(2016,\) the Centers for Disease Control and Prevention reported that \(36.5 \%\) of adults in the United States are obese. A county health service planning a new awareness campaign polls a random sample of 750 adults living there. In this sample, 228 people were found to be obese based on their answers to a health questionnaire. Do these responses provide strong evidence that the \(36.5 \%\) figure is not accurate for this region? Correct the mistakes you find in a student's attempt to test an appropriate hypothesis. $$ \begin{array}{l} \mathrm{H}_{0}: \hat{p}=0.365 \\ \mathrm{H}_{\mathrm{A}}: \hat{p}<0.365 \\ \mathrm{SRS}, 750 \geq 10 \\ \frac{228}{750}=0.304 ; S D(\hat{p})=\sqrt{\frac{(0.304)(0.696)}{750}}=0.017 \end{array} $$ \(z=\frac{0.304-0.365}{0.017}=-3.588\) $$ \mathrm{P}=P(z>-3.588)=0.9998 $$ There is more than a \(99.98 \%\) chance that the stated percentage is correct for this region.

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