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Chapter 4: Problem 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. \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1} \neq p_{2}\)

Short Answer

Expert verified
The notation for a sample statistic in this case is \((\hat{p_1} - \hat{p_2})\). This represents the difference in sample proportions, which will be used to create the randomization distribution for the hypothesis test.

Step by step solution

01

Understand Hypothesis Testing

The first step is to understand that in a hypothesis test, we compare a null hypothesis (which is usually a claim of no effect, or no difference) against an alternative hypothesis (which usually claims an effect or difference). In this particular case, our null hypothesis is that both populations have the same proportion \(p_1 = p_2\), and our alternative hypothesis is that the proportions are not equal \(p_1 \neq p_2\).
02

Identify the Sample Statistic

We usually use sample statistics as an estimate of the corresponding population parameter. This lets us compare our observed data to what we would expect under the null hypothesis. In a test for proportions, this will usually be the sample proportion \(\hat{p}\). When comparing two proportions, we would usually use the difference in sample proportions \(\hat{p_1} - \hat{p_2}\).
03

Specify the Notation for the Sample Statistic

We now need to specify a notation for the sample statistic. In the case of comparing two proportions, we decided to use the difference in sample proportions, \(\hat{p_1} - \hat{p_2}\). So, the required notation for the sample statistic in this exercise is \((\hat{p_1} - \hat{p_2})\).

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Key Concepts

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

Understanding the Null Hypothesis
The null hypothesis plays a crucial role in hypothesis testing. It is the initial claim made about a population parameter. Often, it asserts that there is no effect or no difference between groups. In the context of the exercise, the null hypothesis is that the proportions of two populations are equal, denoted as
  • \( H_{0}: p_{1} = p_{2} \)
This means we start by assuming any observed difference between the groups is due to random chance. Hypothesis testing then determines if the data provides sufficient evidence to reject this claim. When you conduct an experiment, the null hypothesis serves as the baseline statement to measure other claims against.
Exploring the Alternative Hypothesis
While the null hypothesis suggests no change, the alternative hypothesis is what researchers want to prove. It's the statement reflecting a new effect or a difference between groups. For our exercise, the alternative hypothesis states that the two population proportions are not equal:
  • \( H_{a}: p_{1} eq p_{2} \)
This opposing claim challenges the status quo. Researchers look for evidence to support the alternative hypothesis if they suspect a noteworthy effect exists beyond random chance. By gathering and analyzing data, we assess if the findings support or reject the null hypothesis in favor of this alternative scenario.
Defining a Sample Statistic
A sample statistic is a calculated data value that provides an estimate of a corresponding population parameter. In hypothesis tests, sample statistics play a pivotal role in evaluating the null hypothesis. Our focus in the exercise is on comparing two sample proportions. This involves exploring the:
  • Difference in sample proportions, noted as \( \hat{p_1} - \hat{p_2} \)
By calculating this difference from a sample, we get a clear numerically expressed idea of how the sample data behaves compared to the assumption of no difference. We use this figure to make inferences about the populations and to see if such observed differences are in line with what is expected under the null hypothesis. The sample statistic is essential for building the randomization distribution and helps in assessing whether the observed effect holds any genuine significance.

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Most popular questions from this chapter

Approval Rating for Congress In a Gallup poll \(^{51}\) conducted in December 2015 , a random sample of \(n=824\) American adults were asked "Do you approve or disapprove of the way Congress is handling its job?" The proportion who said they approve is \(\hat{p}=0.13,\) and a \(95 \%\) confidence interval for Congressional job approval is 0.107 to 0.153 . If we use a 5\% significance level, what is the conclusion if we are: (a) Testing to see if there is evidence that the job approval rating is different than \(14 \%\). (This happens to be the average sample approval rating from the six months prior to this poll.) (b) Testing to see if there is evidence that the job approval rating is different than \(9 \%\). (This happens to be the lowest sample Congressional approval rating Gallup ever recorded through the time of the poll.)

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.

In this exercise, we see that it is possible to use counts instead of proportions in testing a categorical variable. Data 4.7 describes an experiment to investigate the effectiveness of the two drugs desipramine and lithium in the treatment of cocaine addiction. The results of the study are summarized in Table 4.14 on page \(323 .\) The comparison of lithium to the placebo is the subject of Example 4.34 . In this exercise, we test the success of desipramine against a placebo using a different statistic than that used in Example 4.34. Let \(p_{d}\) and \(p_{c}\) be the proportion of patients who relapse in the desipramine group and the control group, respectively. We are testing whether desipramine has a lower relapse rate then a placebo. (a) What are the null and alternative hypotheses? (b) From Table 4.14 we see that 20 of the 24 placebo patients relapsed, while 10 of the 24 desipramine patients relapsed. The observed difference in relapses for our sample is $$\begin{aligned}D &=\text { desipramine relapses }-\text { placebo relapses } \\\&=10-20=-10\end{aligned}$$ If we use this difference in number of relapses as our sample statistic, where will the randomization distribution be centered? Why? (c) If the null hypothesis is true (and desipramine has no effect beyond a placebo), we imagine that the 48 patients have the same relapse behavior regardless of which group they are in. We create the randomization distribution by simulating lots of random assignments of patients to the two groups and computing the difference in number of desipramine minus placebo relapses for each assignment. Describe how you could use index cards to create one simulated sample. How many cards do you need? What will you put on them? What will you do with them?

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: \(H_{0}: \mu=15\) vs \(H_{a}: \mu \neq 15\) (a) \(95 \%\) confidence interval for \(\mu: \quad 13.9\) to 16.2 (b) \(95 \%\) confidence interval for \(\mu: \quad 12.7\) to 14.8 (c) \(90 \%\) confidence interval for \(\mu: \quad 13.5\) to 16.5

The same sample statistic is used to test a hypothesis, using different sample sizes. In each case, use StatKey or other technology to find the p-value and indicate whether the results are significant at a \(5 \%\) level. Which sample size provides the strongest evidence for the alternative hypothesis? Testing \(H_{0}: p=0.5\) vs \(H_{a}: p>0.5\) using \(\hat{p}=0.55\) with each of the following sample sizes: (a) \(\hat{p}=55 / 100=0.55\) (b) \(\hat{p}=275 / 500=0.55\) (c) \(\hat{p}=550 / 1000=0.55\)
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