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In Exercises 4.112 to \(4.116,\) the null and alternative hypotheses for a test are given as well as some information about the actual sample(s) and the statistic that is computed for each randomization sample. Indicate where the randomization distribution will be centered. In addition, indicate whether the test is a left-tail test, a right-tail test, or a twotailed test. Hypotheses: \(H_{0}: p=0.5\) vs \(H_{a}: p<0.5\) Sample: \(\hat{p}=0.4, n=30\) Randomization statistic \(=\hat{p}\)

Short Answer

Expert verified
The center of the randomization distribution will be at p = 0.5. The test is a left-tail test.

Step by step solution

01

Understanding the Types of Tests

Left-tail test is when the testing parameter is less than the given value, Right-tail test is when the testing parameter is greater than the given value, and a Two-tailed test is when the testing parameter is not equal to the given value. In this case, the alternative hypothesis \(H_a: p<0.5\) indicates a left-tail test.
02

Determining the Center of Randomization Distribution

The center of the randomization distribution under the null hypothesis is equal to the claimed population proportion in the null hypothesis. Here, the null hypothesis \(H_0: p = 0.5\), so the center of the randomization distribution will be at p = 0.5.
03

Deciding the Type of the Test

In this case, the alternative hypothesis is \(H_a: p<0.5\), which means, if the observed value of p is significantly less than 0.5, then we would reject the null hypothesis in favor of the alternative hypothesis. So, this is a left-tail test.

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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 is a fundamental concept in hypothesis testing. It is the statement that there is no effect or no difference, and it serves as the starting point for statistical testing. In any test, the null hypothesis, denoted as \(H_0\), is the claim about the population that we wish to test. For example, the null hypothesis \(H_0: p = 0.5\) suggests that the population proportion is 0.5. This is a neutral position, asserting that any observed difference is due to random variation rather than a real effect.

When we perform hypothesis testing, our goal is to see if there is enough statistical evidence to reject the null hypothesis in favor of the alternative hypothesis. A failure to reject \(H_0\) means that there isn't sufficient evidence to support a change or difference and, conversely, rejecting it implies that the sample provides strong enough evidence against it.

Always remember that an outcome of a hypothesis test doesn't "prove" \(H_0\) true or false. Instead, we make inferences based on the data, which are subject to errors and variability.
Randomization Distribution
The randomization distribution is a key part of understanding how data behaves under the null hypothesis. It's essentially a model of what the dataset would look like if the null hypothesis were true, and it acts as a reference for determining statistical significance. The idea is to mimic the process that could have produced your data, if \(H_0\) were true, multiple times to understand the variability of your statistic.

In a randomized sample model, each sample proportion can be plotted to form a distribution. For example, with \(H_0: p = 0.5\), if you collected enough samples, the randomization distribution should be centered around 0.5. This centering around the null value helps assess how extreme the observed sample statistic is when compared to what is expected by chance alone.

  • Randomization approximates sampling variability
  • It is helpful in determining the probability of observing a sample statistic as extreme as or more extreme than the one obtained
Left-Tail Test
A left-tail test is a type of hypothesis test where the region of rejection is on the left side of the sampling distribution. This means our concern is with deviations that are less than what is stated in the null hypothesis. It's used when you suspect that the observed statistic is smaller than the hypothesized parameter as stated in the null hypothesis.

Consider \(H_a: p < 0.5\). This indicates a left-tail test because you are interested in knowing whether the true proportion is less than 0.5. Consequently, the left tail of the randomization distribution becomes critical because it represents those sample statistics more extreme than our observed value, under the assumption that the null hypothesis holds.

In practice, to determine if the left-tail test reveals a significant result, you would look at a test statistic and compare it against a critical value or calculate the p-value. If the measures are extreme enough, it suggests the observed sample has a low probability of occurring if \(H_0\) is true, allowing us to confidently reject the null hypothesis.

  • Focuses on detecting changes towards lower values than hypothesized
  • Important in contexts where a decrease from a specified standard is of interest

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Most popular questions from this chapter

Arsenic in Chicken Data 4.5 on page 228 discusses a test to determine if the mean level of arsenic in chicken meat is above 80 ppb. If a restaurant chain finds significant evidence that the mean arsenic level is above \(80,\) the chain will stop using that supplier of chicken meat. The hypotheses are $$ \begin{array}{ll} H_{0}: & \mu=80 \\ H_{a}: & \mu>80 \end{array} $$ where \(\mu\) represents the mean arsenic level in all chicken meat from that supplier. Samples from two different suppliers are analyzed, and the resulting p-values are given: Sample from Supplier A: p-value is 0.0003 Sample from Supplier B: p-value is 0.3500 (a) Interpret each p-value in terms of the probability of the results happening by random chance. (b) Which p-value shows stronger evidence for the alternative hypothesis? What does this mean in terms of arsenic and chickens? (c) Which supplier, \(\mathrm{A}\) or \(\mathrm{B}\), should the chain get chickens from in order to avoid too high a level of arsenic?

Testing for a Gender Difference in Compassionate Rats In Exercise 3.80 on page 203 , we found a \(95 \%\) confidence interval for the difference in proportion of rats showing compassion, using the proportion of female rats minus the proportion of male rats, to be 0.104 to \(0.480 .\) In testing whether there is a difference in these two proportions: (a) What are the null and alternative hypotheses? (b) Using the confidence interval, what is the conclusion of the test? Include an indication of the significance level. (c) Based on this study would you say that female rats or male rats are more likely to show compassion (or are the results inconclusive)?

In Exercises 4.14 and \(4.15,\) determine whether the sets of hypotheses given are valid hypotheses. State whether each set of hypotheses is valid for a statistical test. If not valid, explain why not. (a) \(H_{0}: \mu=15 \quad\) vs \(\quad H_{a}: \mu \neq 15\) (b) \(H_{0}: p \neq 0.5 \quad\) vs \(\quad H_{a}: p=0.5\) (c) \(H_{0}: p_{1}p_{2}\) (d) \(H_{0}: \bar{x}_{1}=\bar{x}_{2} \quad\) vs \(\quad H_{a}: \bar{x}_{1} \neq \bar{x}_{2}\)

Beer and Mosquitoes Does consuming beer attract mosquitoes? Exercise 4.17 on page 232 discusses an experiment done in Africa testing possible ways to reduce the spread of malaria by mosquitoes. In the experiment, 43 volunteers were randomly assigned to consume either a liter of beer or a liter of water, and the attractiveness to mosquitoes of each volunteer was measured. The experiment was designed to test whether beer consumption increases mosquito attraction. The report \(^{27}\) states that "Beer consumption, as opposed to water consumption, significantly increased the activation... of An. gambiae [mosquitoes]... \((P<0.001)\) (a) Is this convincing evidence that consuming beer is associated with higher mosquito attraction? Why or why not? (b) How strong is the evidence for the result? Explain. (c) Based on these results, is it reasonable to conclude that consuming beer causes an increase in mosquito attraction? Why or why not?

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

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