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In Exercises 4.107 to \(4.111,\) 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. $$ H_{0}: \mu=15 \text { vs } H_{a}: \mu<15 $$

Short Answer

Expert verified
The appropriate sample statistic to be recorded for each simulated sample to create a randomization distribution in this case would be the sample mean, denoted by \(\bar{x}\).

Step by step solution

01

Understand the hypothesis

First, we need to identify the null and alternative hypotheses and understand what they are implying. The null hypothesis \(H_{0}: \mu=15\) is claiming that the population mean is equal to 15. The alternative hypothesis \(H_{a}: \mu<15\) is claiming that the population mean is less than 15.
02

Identify the sample statistic

To determine the population mean, \(\mu\), we collect sample data and calculate a sample mean, \(\bar{x}\). This \(\bar{x}\) is the notation for the sample statistic that we would record in each simulated sample to create a randomization distribution.
03

Summary

To test the null hypothesis \(H_{0}: \mu=15\) against the alternative hypothesis \(H_{a}: \mu<15\), we would use the sample mean \(\bar{x}\) as our sample statistic for the randomization distribution. It’s an estimate of the population mean \(\mu\), capable of taking different values for different samples. If the \(\bar{x}\) from our simulated samples falls below 15 in a statistically significant way, this will lead us to reject the null hypothesis in favor of the alternative hypothesis. It reflects our confidence that the actual population mean is indeed less than 15.

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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, often symbolized as \(H_0\), is a foundation of any hypothesis test in statistics. It represents the default position that there is no effect or no difference, and it sets the stage for statistical analysis. In the context of the given exercise, the null hypothesis \(H_0: \mu=15\) posits that the population mean \(\mu\) is equal to 15. Think of it as our starting point, where we assume there is no deviation from this value. It's the status quo that we're challenging with our sample data.

Why do we need a null hypothesis? By suggesting that nothing has changed or no difference exists, we create a benchmark against which we can measure the evidence obtained from our sample data. In essence, it creates a measurable statement that can either be refuted or not by the data we collect. If our sample provides sufficient evidence against \(H_0\), we can consider rejecting the null hypothesis.
Alternative Hypothesis
While the null hypothesis serves as our starting assumption, the alternative hypothesis, symbolized as \(H_a\) or \(H_1\), represents a contrast to the null hypothesis. It is a statement we hope or suspect to be true instead of the null hypothesis. In our exercise, the alternative hypothesis \(H_a: \mu<15\) asserts that the population mean \(\mu\) is less than 15.

This hypothesis is the researcher's initial claim, which they aim to support with sample data. As opposed to the precise statement of the null, the alternative hypothesis can be directional or non-directional (e.g., \(\mueq15\)). Its formulation depends on what the researcher wants to prove or explore. If the evidence leans towards \(H_a\), then we consider this an indication that the null hypothesis may not be an accurate reflection of the population.
Sample Mean
The sample mean, denoted by \(\bar{x}\), is a critical statistic in hypothesis testing. It represents the average value of a sample, and it's used as an estimate for the population mean \(\mu\). In our specific question, the sample mean would be calculated from the collected data to determine if it supports the null or alternative hypothesis.

Here's the importance of the sample mean: it serves as a bridge between the sample data and what we infer about the population as a whole. It can fluctuate from one sample to another due to natural sampling variability, so researchers use it to assess whether observed differences are likely to be real or just due to chance. When evaluating the null hypothesis \(H_0: \mu=15\), we compare \(\bar{x}\) from our sample with the hypothesized population mean. A significant difference between \(\bar{x}\) and \(15\) might suggest that we reject the null hypothesis.
Randomization Distribution
The randomization distribution is a construct we use to understand the behavior of a sample statistic, like the sample mean \(\bar{x}\), under numerous re-samples or simulations of a study. By recording \(\bar{x}\) for each of these simulated samples, as outlined in the exercise, we form a distribution that shows the possible values \(\bar{x}\) could take if the null hypothesis \(H_0\) were true.

Why is this distribution so valuable? It allows us to assess the likelihood of observing a sample statistic as extreme as, or more extreme than, the one we calculated from our actual data. If our actual \(\bar{x}\) lies in the tail of the randomization distribution (especially if it's in the direction that supports \(H_a\)), we might conclude that such an extreme result is rare under the null hypothesis. This lends support to the alternative hypothesis, giving us grounds to question \(H_0\) and suggesting that our observed data may not be a product of random chance alone.

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Most popular questions from this chapter

In Exercises 4.107 to \(4.111,\) 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. $$ H_{0}: \mu_{1}=\mu_{2} \operatorname{vs} H_{a}: \mu_{1}>\mu_{2} $$

For each situation described in Exercises 4.93 to 4.98 , 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 \(\alpha=0.01)\) A pharmaceutical company is testing to see whether its new drug is significantly better than the existing drug on the market. It is more expensive than the existing drug. Which significance level would the company prefer? Which significance level would the consumer prefer?

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.

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: \(\hat{p}=38 / 100=0.38\) with \(n=100\)

Exercise and the Brain Exercise 4.19 on page 232 describes a study investigating the effects of exercise on cognitive function. \(^{28}\) Separate groups of mice were exposed to running wheels for \(0,2,4,7,\) or 10 days. Cognitive function was measured by \(Y\) maze performance. The study was testing whether exercise improves brain function, whether exercise reduces levels of BMP (a protein which makes the brain slower and less nimble), and whether exercise increases the levels of noggin (which improves the brain's ability). For each of the results quoted in parts (a), (b), and (c), interpret the information about the p-value in terms of evidence for the effect. (a) "Exercise improved Y-maze performance in most mice by the 7 th day of exposure, with further increases after 10 days for all mice tested \((p<.01)^{*}\) (b) "After only two days of running, BMP ... was reduced ... and it remained decreased for all subsequent time-points \((p<.01) . "\) (c) "Levels of noggin ... did not change until 4 days, but had increased 1.5 -fold by \(7-10\) days of exercise \((p<.001)\)." (d) Which of the tests appears to show the strongest statistical effect? (e) What (if anything) can we conclude about the effects of exercise on mice?

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