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In Exercises 4.40 to 4.44 , null and alternative hypotheses for a test are given. Give the notation \((\bar{x},\) for example) for a sample statistic we might record for each simulated sample to create the randomization distribution. \(H_{0}: p=0.5\) vs \(H_{a}: p \neq 0.5\)

Short Answer

Expert verified
The sample statistic that might be recorded for each simulated sample to create a randomization distribution for the given null and alternative hypotheses is the sample proportion \(\hat{p}\).

Step by step solution

01

Understanding the Null and Alternative Hypotheses

The null hypothesis \(H_{0}: p=0.5\) suggests that the population proportion is hypothesized to be 0.5. The alternative hypothesis \(H_{a}: p \neq 0.5\) states that the population proportion is something other than 0.5.
02

Selecting the Sample Statistic

One might record a sample statistic for each simulated sample to create the randomization distribution. In this problem with binomial distribution, the appropriate statistic would be the sample proportion \(\hat{p}\), which estimates the true population proportion p.
03

Formulate the Sample Statistic

\(\hat{p}\) can be formulated as \(\hat{p} = X/n\), where X represents the number of 'successes' in our sample and n is the total sample size. It gives us the proportion of successes in our sample.
04

Using the Sample Statistic

This statistic \(\hat{p}\) can then be used to perform the hypothesis test by comparing its value to the hypothesized population proportion in \(H_{0}\).

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Key Concepts

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

Null hypothesis
The **null hypothesis** represents a statement that there is no effect or no difference, and it is usually denoted as \(H_0\). In hypothesis testing context, it serves as the default or starting assumption for a test. For example, in a situation where we want to know if a coin is fair, the null hypothesis would be that the probability of getting heads \((p)\) is 0.5. This is because we are assuming that the coin is fair unless we have evidence to prove otherwise.
  • The null hypothesis assumes that any kind of difference or significance you see in a set of data is due to chance.
  • It is important to remember that the null hypothesis is always expressed as an equality.
Testing a null hypothesis involves using a sample statistic, like the sample proportion \(\hat{p}\), to determine if there is sufficient evidence to reject it in favor of an alternative hypothesis.
Alternative hypothesis
The **alternative hypothesis**, denoted as \(H_a\), posits that there is an effect or a difference. It challenges the null hypothesis and is what you want to prove in hypothesis testing. Continuing with our coin example, if you suspect that the coin is not fair, the alternative hypothesis would assert that \(p eq 0.5\). This indicates that the probability of heads is different from 0.5, suggesting either a bias towards heads or tails.
  • The alternative hypothesis is essentially what the researcher is trying to establish.
  • Unlike the null hypothesis, the alternative hypothesis is always expressed as a non-equality \((eq, >, <)\).
It is crucial to define the alternative hypothesis correctly, as it guides the research and analysis direction. If statistical analysis provides enough evidence against \(H_0\), then \(H_a\) is accepted.
Randomization distribution
A **randomization distribution** is a distribution of a sample statistic, built under the assumption that the null hypothesis is true. This method involves repeatedly simulating samples from the null hypothesis to generate a distribution of the sample statistic (like \(\hat{p}\), the sample proportion in this context). - **Purpose:**The main purpose of a randomization distribution is to help us determine how the observed data compare to what we would expect if the null hypothesis were true.- **Construction:**Creating a randomization distribution involves using the known null value (\(p = 0.5\) in this case) and simulating many samples from it. For each sample, compute the sample proportion \(\hat{p}\) and plot these values to form the distribution.- **Interpretation:**Once the distribution is available, you can compare the actual sample statistic to this randomization distribution. A significant deviation of \(\hat{p}\) from 0.5, taking into account the variability in the distribution, would lead to rejecting the null hypothesis in favor of the alternative hypothesis.Using a randomization distribution is a powerful approach because it does not assume a particular theoretical distribution, making it versatile for various hypothesis testing scenarios.

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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 \(H_{a}: p<0.5\) Sample data: \(\hat{p}=38 / 100=0.38\) with \(n=100\)

Figure 4.25 shows a scatterplot of the acidity (pH) for a sample of \(n=53\) Florida lakes vs the average mercury level (ppm) found in fish taken from each lake. The full dataset is introduced in Data 2.4 on page 71 and is available in FloridaLakes. There appears to be a negative trend in the scatterplot, and we wish to test whether there is significant evidence of a negative association between \(\mathrm{pH}\) and mercury levels. (a) What are the null and alternative hypotheses? (b) For these data, a statistical software package produces the following output: $$ r=-0.575 \quad p \text { -value }=0.000017 $$ Use the p-value to give the conclusion of the test. Include an assessment of the strength of the evidence and state your result in terms of rejecting or failing to reject \(H_{0}\) and in terms of \(\mathrm{pH}\) and mercury. (c) Is this convincing evidence that low \(\mathrm{pH}\) causes the average mercury level in fish to increase? Why or why not?

Mating Choice and Offspring Fitness: MiniExperiments Exercise 4.153 explores the question of whether mate choice improves offspring fitness in fruit flies, and describes two seemingly identical experiments yielding conflicting results (one significant, one insignificant). In fact, the second source was actually a series of three different experiments, and each full experiment was comprised of 50 different mini-experiments (runs), 10 each on five different days. (a) Suppose each of the 50 mini-experiments from the first study were analyzed individually. If mating choice has no impact on offspring fitness, about how many of these \(50 \mathrm{p}\) -values would you expect to yield significant results at \(\alpha=0.05 ?\) (b) The 50 p-values, testing the alternative \(H_{a}\) : \(p_{C}>p_{N C}\) (proportion of flies surviving is higher in the mate choice group) are given below: $$ \begin{array}{lllllllllll} \text { Day 1: } & 0.96 & 0.85 & 0.14 & 0.54 & 0.76 & 0.98 & 0.33 & 0.84 & 0.21 & 0.89 \\ \text { Day 2: } & 0.89 & 0.66 & 0.67 & 0.88 & 1.00 & 0.01 & 1.00 & 0.77 & 0.95 & 0.27 \\ \text { Day 3: } & 0.58 & 0.11 & 0.02 & 0.00 & 0.62 & 0.01 & 0.79 & 0.08 & 0.96 & 0.00 \\ \text { Day 4: } & 0.89 & 0.13 & 0.34 & 0.18 & 0.11 & 0.66 & 0.01 & 0.31 & 0.69 & 0.19 \\ \text { Day 5: } & 0.42 & 0.06 & 0.31 & 0.24 & 0.24 & 0.16 & 0.17 & 0.03 & 0.02 & 0.11 \end{array} $$ How many are actually significant using \(\alpha=0.05 ?\) (c) You may notice that two p-values (the fourth and last run on day 3 ) are 0.00 when rounded to two decimal places. The second of these is actually 0.0001 if we report more decimal places. This is very significant! Would it be appropriate and/or ethical to just report this one run, yielding highly statistically significant evidence that mate choice improves offspring fitness? Explain. (d) You may also notice that two of the p-values on day 2 are 1 (rounded to two decimal places). If we had been testing the opposite alternative, \(H_{a}:\) \(p_{C}

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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