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Chapter 4: Problem 117

Exercises 4.117 to 4.122 give null and alternative hypotheses for a population proportion, as well as sample results. Use StatKey or other technology to generate a randomization distribution and calculate a p-value. StatKey tip: Use "Test for a Single Proportion" and then "Edit Data" to enter the sample information. Hypotheses: \(H_{0}: p=0.5\) vs \(H_{a}: p>0.5\) Sample data: \(\hat{p}=30 / 50=0.60\) with \(n=50\)

Short Answer

Expert verified
The implementation of these steps in StatKey or another technology would provide the p-value for the given hypotheses and sample data. The p-value helps determine if the observed data is statistically significant for the proportions and would lead to accepting or rejecting the null hypothesis.

Step by step solution

01

Set up the Hypotheses

Set up the null hypothesis and the alternative hypothesis. The null hypothesis is \(H_{0}: p=0.5\). This means that the proportion of the population is 0.5. The alternative hypothesis is \(H_{a}: p>0.5\). This means that the proportion of the population is more than 0.5.
02

Input the Sample Information

The sample proportion \(\hat{p}\), is calculated using the given data, \(\hat{p} = \frac{Number\:of\:successes}{Total\:number\:of\:trials} = \frac{30}{50}=0.6\). The sample size, \(n\), is given as 50. This information needs to be entered in the 'Test for a Single Proportion' and 'Edit Data' section in StatKey or another technology being used.
03

Generate a Randomization Distribution and Calculate a P-value

With the sample information entered, run the simulation to generate a randomization distribution. From this distribution, calculate the p-value. The p-value is the probability that, given the null hypothesis is true, we would obtain a sample as extreme or more extreme than our current sample. This is computed by finding the proportion of simulated samples that were greater than or equal to our observed statistic, in this case, the sample proportion of 0.6.
04

Interpreting P-value

If the p-value is very small, it means, assuming the null hypothesis is true, the likelihood of getting a sample as extreme as the observed sample is relatively low, leading us to reject the null hypothesis in favor of the alternative hypothesis. If the p-value is large, this means that our sample result is relatively common, and we would not reject 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 Alternative Hypotheses
Understanding the null and alternative hypotheses is fundamental to hypothesis testing in statistics. The null hypothesis, denoted as ull hypothesis is a statement of no effect or no difference. In the context of the given exercise, the null hypothesis ull hypothesis asserts that the population proportion is equal to 0.5, which often represents the status quo or a baseline value to compare against.

The alternative hypothesis, noted by H_{a}, represents what the researcher is trying to find evidence for; it is the counterclaim to the null hypothesis. In this case, the alternative hypothesis suggests that the population proportion is greater than 0.5. Essentially, it indicates that there's an effect or a difference that the researcher believes may be present.
Population Proportion
The population proportion, often denoted as p, is a parameter that represents the ratio of individuals in a population who have a particular attribute to the total population size. Estimating the population proportion is a common objective in statistical studies. For the given exercise, the sample proportion, for computing the population proportion. However, one should be cautious and remember that the sample should be random and representative to make valid inferences about the entire population.
P-value
The p-value is a critical concept in hypothesis testing. It provides a measure of the evidence against the null hypothesis provided by the sample. Found by calculating the likelihood of obtaining a sample statistic as extreme or more extreme than the one measured, the p-value assumes the null hypothesis is true. A low p-value indicates that such an extreme result is rare under the null hypothesis, leading to consideration of the alternative hypothesis.

In our exercise, the p-value would reveal how likely it is to get a sample proportion of 0.6 or more if the actual population proportion is 0.5. Deciding on a significance level (commonly 0.05) beforehand allows us to determine if the p-value is sufficiently small to reject the null hypothesis in favor of the alternative.
Randomization Distribution
The concept of randomization distribution comes into play when you want to assess how likely your observed sample statistic is under the null hypothesis. Created by simulating many samples assuming the null hypothesis is true, the randomization distribution is a collection of sample statistics used to observe the variability and pattern of the data. This information is crucial for calculating the p-value.

In practice, we look at this distribution to see where the observed sample statistic fits, which informs the extremeness of the observed result. The further away from the center of this distribution, the less likely is the result under the null hypothesis, thus strengthening the evidence for the alternative hypothesis.
StatKey
StatKey is a software tool designed to help users with the visualization and computation needed for statistical analyses, including hypothesis testing. When utilizing StatKey for the exercise, one would use the 'Test for a Single Proportion' feature to enter the specific sample information and then utilize the simulation capabilities to generate the randomization distribution.

By engaging with tools like StatKey, students and practitioners can focus on understanding and interpreting the results rather than being bogged down by the manual calculation process. It provides a visual way of understanding statistical concepts and enhances learning by providing an interactive experience for users.

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Most popular questions from this chapter

In Exercises 4.146 to \(4.149,\) hypotheses for a statistical test are given, followed by several possible confidence intervals for different samples. In each case, use the confidence interval to state a conclusion of the test for that sample and give the significance level used. Hypotheses: \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2} .\) In addition, in each case for which the results are significant, state which group ( 1 or 2 ) has the larger mean. (a) \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}:\) $$ 0.12 \text { to } 0.54 $$ (b) \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) : -2.1 to 5.4 (c) \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) : -10.8 to -3.7

4.20 Taste Test A taste test is conducted between two brands of diet cola, Brand \(A\) and Brand \(B\), to determine if there is evidence that more people prefer Brand A. A total of 100 people participate in the taste test. (a) Define the relevant parameter(s) and state the null and alternative hypotheses. (b) Give an example of possible sample results that would provide strong evidence that more people prefer Brand A. (Give your results as number choosing Brand \(\mathrm{A}\) and number choosing Brand B.) (c) Give an example of possible sample results that would provide no evidence to support the claim that more people prefer Brand A. (d) Give an example of possible sample results for which the results would be inconclusive: The sample provides some evidence that Brand \(\mathrm{A}\) is preferred but the evidence is not strong.

In Exercises 4.112 to \(4.116,\) the null and alternative hypotheses for a test are given as well as some information about the actual sample(s) and the statistic that is computed for each randomization sample. Indicate where the randomization distribution will be centered. In addition, indicate whether the test is a left-tail test, a right-tail test, or a twotailed test. Hypotheses: \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1}>p_{2}\) Sample: \(\hat{p}_{1}=0.3, n_{1}=20\) and \(\hat{p}_{2}=0.167, n_{2}=12\) Randomization statistic \(=\hat{p}_{1}-\hat{p}_{2}\)

Flipping Coins We flip a coin 150 times and get 90 heads, so the sample proportion of heads is \(\hat{p}=90 / 150=0.6 .\) To test whether this provides evidence that the coin is biased, we create a randomization distribution. Where will the distribution be centered? Why?

Does Massage Really Help Reduce Inflammation in Muscles? In Exercise 4.132 on page \(279,\) we learn that massage helps reduce levels of the inflammatory cytokine interleukin-6 in muscles when muscle tissue is tested 2.5 hours after massage. The results were significant at the \(5 \%\) level. However, the authors of the study actually performed 42 different tests: They tested for significance with 21 different compounds in muscles and at two different times (right after the massage and 2.5 hours after). (a) Given this new information, should we have less confidence in the one result described in the earlier exercise? Why? (b) Sixteen of the tests done by the authors involved measuring the effects of massage on muscle metabolites. None of these tests were significant. Do you think massage affects muscle metabolites? (c) Eight of the tests done by the authors (including the one described in the earlier exercise) involved measuring the effects of massage on inflammation in the muscle. Four of these tests were significant. Do you think it is safe to conclude that massage really does reduce inflammation?
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