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Exercises 4.59 to 4.64 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
In this task, you would generate a randomization distribution using a statistical software or technology and calculate a p-value based on that. The p-value is the likelihood of obtaining a sample proportion as extreme as, or more extreme than, the observed sample proportion, under the assumption that the null hypothesis is true. The smaller the p-value, the stronger the evidence against the null hypothesis.

Step by step solution

01

Define null and alternative hypothesis

The null hypothesis (\(H_{0}\)) suggests that the population proportion \(p\) is 0.5. The alternative hypothesis (\(H_{a}\)) suggests that the population proportion \(p\) is greater than 0.5.
02

Calculate test statistic

We need to calculate the test statistic. Under \(H_{0}\), we expect the proportion of successes to be 0.5. But in our sample, the proportion of successes is \(\hat{p}=0.60\). This is our test statistic.
03

Generate randomization distribution

We now generate a randomization distribution under the assumption that \(H_{0}\) is true, i.e. \(p=0.5\). We do this via a simulation. For each trial, we generate a sample of size \(n=50\) and we calculate the proportion of successes. We repeat this process many times (e.g. 10000 times) to obtain the randomization distribution.
04

Calculate p-value

The p-value is the probability of obtaining a result as extreme as, or more extreme than, the observed data, under the assumption that the null hypothesis is true. In this case, p-value is the proportion of trials in the randomization distribution that had a proportion of successes of 0.60 or more.

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Key Concepts

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

Population Proportion
Understanding the population proportion is crucial when conducting hypothesis tests in statistics. This concept refers to the true proportion of individuals or items within a larger, defined grouping that exhibit a certain trait or outcome. In hypothesis testing, we're often trying to determine if our sample data provides enough evidence to infer something about the population proportion.

For instance, if a coin is flipped 50 times, the population proportion we're interested in might be the proportion of flips that land heads. If we assume that the coin is fair, we would expect the population proportion, denoted as 'p', to be 0.5, meaning that half of the flips should land heads. However, if our sample data show a higher proportion of heads, say \( \hat{p} = 0.60 \), we may begin to question our assumption of fairness and consider if the true population proportion is actually greater than 0.5.
P-value Calculation
The p-value is a fundamental concept in hypothesis testing, serving as a bridge between observed data and the hypotheses we're evaluating. To put it simply, the p-value measures the strength of evidence against the null hypothesis. When we calculate a p-value, we're determining the probability of obtaining a sample statistic as extreme as the one observed, given that the null hypothesis is true.

Utilizing our example where a sample proportion \( \hat{p} = 0.60 \) is observed, the p-value calculation will help us understand the likelihood of seeing such a result if, in reality, the true population proportion is 0.5 (as stated in the null hypothesis, \( H_{0}: p = 0.5 \) ). A low p-value indicates that the observed result is unlikely under the null hypothesis, suggesting that the alternative hypothesis (in this case, \( H_{a}: p > 0.5 \)) may be true. The smaller the p-value, the stronger the evidence against the null hypothesis.
Randomization Distribution
The concept of a randomization distribution is vital for understanding how unusual our sample data is within the context of the null hypothesis. It's essentially a sampling distribution that shows us what kinds of results we might see if the null hypothesis were true and we collected many, many samples. By simulating numerous random samples, we create a distribution of statistics (like sample proportions) that could happen just by random chance.

In our coin example, where we're testing if the true population proportion of heads is greater than 0.5, we'd use randomization to create many sets of 50 coin flips. We'd simulate this under the null hypothesis, where each flip has a 50% chance of being heads. After many trials, we create a randomization distribution that gives us a visual and numerical sense of how likely different sample proportions are. This distribution is then used to assess the rarity of our observed sample statistic (\( \hat{p} = 0.60 \) ) and thus to calculate the p-value.

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Most popular questions from this chapter

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 \(H_{a}: p \neq 0.5\) Sample data: \(\hat{p}=42 / 100=0.42\) with \(n=100\)

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: \(H_{0}: p=0.5\) vs \(H_{a}: p \neq 0.5\) (a) 95\% confidence interval for \(p: \quad 0.53\) to 0.57 (b) \(95 \%\) confidence interval for \(p: \quad 0.41\) to 0.52 (c) 99\% confidence interval for \(p: \quad 0.35\) to 0.55

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 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

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.

For each situation described, indicate whether it makes more sense to use a relatively large significance level (such as \(\alpha=0.10\) ) or a relatively small significance level (such as \(\alpha=0.01\) ). Testing to see whether taking a vitamin supplement each day has significant health benefits. There are no (known) harmful side effects of the supplement.

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