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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: \(H_{0}: p=0.6\) vs \(H_{a}: p>0.6\) Sample data: \(\hat{p}=52 / 80=0.65\) with \(n=80\)

Short Answer

Expert verified
The null and alternative hypotheses are \(H_{0}: p=0.6\) and \(H_{a}: p>0.6\), respectively. The sample proportion is \(\hat{p}=52 / 80=0.65\) from \(n=80\) observations. A randomization distribution needs to be generated using software followed by the calculation of the p-value, which gives the chance of observing what we did if the null hypothesis was true.

Step by step solution

01

State the Hypotheses

The null hypothesis (\(H_{0}\)) is that the population proportion \(p\) equals 0.6, while the alternative hypothesis (\(H_{a}\)) is that the population proportion \(p\) is greater than 0.6. In other words, \(H_{0}: p=0.6\) and \(H_{a}: p>0.6\).
02

Analyze Sample Data

The sample data are given as a proportion: \(\hat{p} = 52/80 = 0.65\). This sample data will be used to conduct the hypothesis test.
03

Generate Randomization Distribution

Using StatKey or similar software, generate a randomization distribution. In StatKey, select 'Test for a Single Proportion', then 'Edit Data' to enter the sample information. Then, use the software to create the randomization distribution, which models what might happen if the null hypothesis is true.
04

Calculate p-value

Calculate the p-value, which is the probability under the null hypothesis of obtaining a result as extreme as, or more extreme than, the observed result. This p-value can be obtained by counting what proportion of the simulated randomization statistics are greater than or equal to the observed sample proportion of 0.65. The p-value is then this proportion.

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Key Concepts

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

Population Proportion
In hypothesis testing, the concept of a population proportion is fundamental. A population proportion is the ratio of members in a population that have a particular characteristic. It's often denoted by the symbol \( p \). For example, if you are interested in determining whether more than 60% of a population prefers a certain brand, 0.6 would be your comparative value for \( p \).
Given the null hypothesis in our exercise, we assume \( p = 0.6 \), which means that 60% of the entire population would hold a certain trait if the null hypothesis held true. Under the alternative hypothesis \( p > 0.6 \), we speculate that the proportion is greater than 60%.
Understanding the population proportion helps us frame the hypothesis and decide the focus of our analysis. Our main job in testing these hypotheses is to obtain evidence through sampling to either support or refute our initial assumption about the population.
Sample Data Analysis
Sample data analysis is crucial for performing hypothesis tests. The process essentially entails analyzing a smaller subset or sample from the broader population to draw conclusions about the entire group. Here, the sample proportion, commonly denoted as \( \hat{p} \), is calculated as the ratio of observed successes to the total number in the sample.
In the given exercise, we observe a sample proportion \( \hat{p} = 0.65 \) from a sample size \( n = 80 \), derived from 52 successes out of 80 attempts. This sample proportion is key because it gives us a tangible statistic to compare against the population proportion suggested by our hypothesis.
By comparing the sample data \( \hat{p} \) to the hypothesized population proportion \( p \), we can use statistical methods to assess the likelihood of our sample result occurring if the null hypothesis were true. Essentially, this step bridges the theoretical aspects of hypothesis testing with real-world data.
Randomization Distribution
Randomization distribution is a powerful tool in hypothesis testing. It helps us understand the distribution of a test statistic under the assumption that the null hypothesis is true. This involves simulating data that reflects how the sample statistic might behave if it were centered around the null hypothesis value.
Tools like StatKey facilitate the creation of a randomization distribution, where you can "Test for a Single Proportion" and "Edit Data" to input sample information. This gives you a visual and theoretical foundation to explore what results might be expected purely by chance.
In our exercise, we use randomization distribution to generate numerous hypothetical sample proportions under the null assumption that \( p = 0.6 \). Observing how frequently our sample proportion \( \hat{p} = 0.65 \) occurs within this distribution, we can derive the p-value. This measures how extreme our sample statistic is in comparison to the population parameter as assumed by the null hypothesis.

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Most popular questions from this chapter

The Ignorance Surveys were conducted in 2013 using random sampling methods in four different countries under the leadership of Hans Rosling, a Swedish statistician and international health advocate. The survey questions were designed to assess the ignorance of the public to global population trends. The survey was not just designed to measure ignorance (no information), but if preconceived notions can lead to more wrong answers than would be expected by random guessing. One question asked, "In the last 20 years the proportion of the world population living in extreme poverty has \(\ldots, "\) and three choices were provided: 1) "almost doubled" 2) "remained more or less the same," and 3) "almost halved." Of 1005 US respondents, just \(5 \%\) gave the correct answer: "almost halved." 34 We would like to test if the percent of correct choices is significantly different than what would be expected if the participants were just randomly guessing between the three choices. (a) What are the null and alternative hypotheses? (b) Using StatKey or other technology, construct a randomization distribution and compute the p-value. (c) State the conclusion in context.

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After exercise, massage is often used to relieve pain, and a recent study 33 shows that it also may relieve inflammation and help muscles heal. In the study, 11 male participants who had just strenuously exercised had 10 minutes of massage on one quadricep and no treatment on the other, with treatment randomly assigned. After 2.5 hours, muscle biopsies were taken and production of the inflammatory cytokine interleukin-6 was measured relative to the resting level. The differences (control minus massage) are given in Table 4.11 . $$ \begin{array}{lllllllllll} 0.6 & 4.7 & 3.8 & 0.4 & 1.5 & -1.2 & 2.8 & -0.4 & 1.4 & 3.5 & -2.8 \end{array} $$ (a) Is this an experiment or an observational study? Why is it not double blind? (b) What is the sample mean difference in inflammation between no massage and massage? (c) We want to test to see if the population mean difference \(\mu_{D}\) is greater than zero, meaning muscle with no treatment has more inflammation than muscle that has been massaged. State the null and alternative hypotheses. (d) Use Statkey or other technology to find the p-value from a randomization distribution. (e) Are the results significant at a \(5 \%\) level? At a \(1 \%\) level? State the conclusion of the test if we assume a \(5 \%\) significance level (as the authors of the study did).

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