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Chapter 4: Problem 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 \(H_{a}: p \neq 0.5\) Sample data: \(\hat{p}=28 / 40=0.70\) with \(n=40\)

Short Answer

Expert verified
Using StatKey and the given sample data, a p-value is obtained through a Single Proportion Test. Depending on the p-value result, we either reject or fail to reject the null hypothesis of the population proportion being 0.5. The exact p-value depends on the specific randomization distribution generated by StatKey.

Step by step solution

01

State the Null and Alternative Hypotheses

Null hypothesis (H0): The population proportion, p, equals 0.5. Alternative hypothesis (Ha): The population proportion, p, does not equal 0.5.
02

Present the Sample Data

The sample data includes 40 observations, where the sample proportion, \(\hat{p}\), is measured as 0.7 (or 28 out of 40).
03

Use the Single Proportion Test in StatKey

With the sample data and the hypotheses, we can use a Single Proportion Test in StatKey to analyze. For this, 'Edit Data' is used to enter the sample data.
04

Generate a Randomization Distribution and Calculate the p-value

With the sample data entered, StatKey is used to generate a randomization distribution, following the model under the null hypothesis. The p-value is then calculated, indicating the probability of observing the sample proportion given the null hypothesis is true. If the p-value is smaller than the significance level (often 0.05), we reject the null hypothesis in favor of the alternative.

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Key Concepts

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

Population Proportion
Let's start by talking about population proportion. In statistics, the population proportion (\( p \) ) is the fraction of the population that has a particular characteristic. Imagine a large bowl of marbles where some are blue and others are red; if you want to find out the proportion of blue marbles, you would look at the fraction referring to the blue ones.
In practice, it's often infeasible to observe the entire population. Therefore, a sample proportion (\( \hat{p} \)) from a smaller, representative sample is used to make inferences about the population. In our exercise, the sample proportion is 0.70, based on 28 favorable observations out of 40 trials. This sample proportion is used to test against our hypotheses concerning the population proportion.
Randomization Distribution
A randomization distribution is an essential step when performing hypothesis tests, especially in determining the validity of our null hypothesis. This distribution shows us what possible sample proportions can look like if the null hypothesis is indeed true.
To build a randomization distribution, we simulate many samples assuming the null hypothesis is correct (\(H_0: p = 0.5\)). These samples help visualize the variability we could expect in the proportion if chance alone dictated the results. The spread of these samples forms the randomization distribution. Each point on this distribution corresponds to a simulated outcome assuming the null hypothesis holds true.
  • This approach allows us to compare our actual sample result (here \(\hat{p} = 0.70\)) to this distribution.

  • It helps determine how unusual or common our sample result is under the assumption that the null hypothesis is true.
p-value Calculation
After creating a randomization distribution, we calculate the p-value, which is crucial to making decisions in hypothesis testing. The p-value represents the probability of observing a sample result as extreme as, or more extreme than, the observed sample proportion, assuming the null hypothesis is true.
In simpler terms, it helps answer: "How surprised should we be to see our sample data if the null hypothesis was true?"
A small p-value indicates that such a result would be rare under the null hypothesis, suggesting that our observed sample proportion (\(\hat{p} = 0.70\)) may not just have happened by random chance.
  • In our context, if the p-value is less than 0.05, we usually conclude that the null hypothesis is unlikely, favoring the alternative hypothesis that the population proportion is not 0.5.

  • The p-value thus helps in determining whether there's enough evidence to reject the null hypothesis.
Being aware of the significance threshold in advance (often 0.05) helps in setting the benchmark for such decisions.

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Most popular questions from this chapter

A study suggests that exposure to UV rays through the car window may increase the risk of skin cancer. \(^{52}\) The study reviewed the records of all 1,050 skin cancer patients referred to the St. Louis University Cancer Center in 2004\. Of the 42 patients with melanoma, the cancer occurred on the left side of the body in 31 patients and on the right side in the other 11 . (a) Is this an experiment or an observational study? (b) Of the patients with melanoma, what proportion had the cancer on the left side? (c) A bootstrap \(95 \%\) confidence interval for the proportion of melanomas occurring on the left is 0.579 to \(0.861 .\) Clearly interpret the confidence interval in the context of the problem. (d) Suppose the question of interest is whether melanomas are more likely to occur on the left side than on the right. State the null and alternative hypotheses. (e) Is this a one-tailed or two-tailed test? (f) Use the confidence interval given in part (c) to predict the results of the hypothesis test in part (d). Explain your reasoning. (g) A randomization distribution gives the p-value as 0.003 for testing the hypotheses given in part (d). What is the conclusion of the test in the context of this study? (h) The authors hypothesize that skin cancers are more prevalent on the left because of the sunlight coming in through car windows. (Windows protect against UVB rays but not UVA rays.) Do the data in this study support a conclusion that more melanomas occur on the left side because of increased exposure to sunlight on that side for drivers?

4.150 Approval from the FDA for Antidepressants The FDA (US Food and Drug Administration) is responsible for approving all new drugs sold in the US. In order to approve a new drug for use as an antidepressant, the FDA requires two results from randomized double-blind experiments showing the drug is more effective than a placebo at a \(5 \%\) level. The FDA does not put a limit on the number of times a drug company can try such experiments. Explain, using the problem of multiple tests, why the FDA might want to rethink its guidelines. 4.151 Does Massage Really Help Reduce Inflammation in Muscles? In Exercise 4.112 on page \(301,\) 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?

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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