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Chapter 4: Problem 120

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) \(\hat{p}=550 / 1000=0.55\)

Short Answer

Expert verified
Given the same sample proportion, larger sample sizes provide stronger evidence in favor of the alternative hypothesis, because they lead to larger test statistics and thus smaller p-values. Hence, in this scenario, a sample size of 1000 yields the strongest evidence for the alternative hypothesis.

Step by step solution

01

Understand the given hypothesis

The null hypothesis \(H_0: p=0.5\) states that there is no effect or relationship (here probability is 0.5). The alternative hypothesis \(H_a: p>0.5\) states the existence of an effect or relationship (here probability is greater than 0.5).
02

Calculate the sample proportion

The sample proportion in each case is given by \(\hat{p} = nr/n\), where \(nr\) denotes the 'number of success cases' and \(n\) is the total sample size. Here, each case gives us the same sample proportion of \(\hat{p}=0.55\).
03

Compute Test Statistic

Compute the Test statistic \(Z\) using the formula \(Z = (\hat{p} - p)/ \sqrt{(p(1-p)/n)}\). Usually, this would be done using stat software like StatKey or similar as specified in the task. The greater the absolute value of \(Z\), the stronger is the evidence against \(H_0\) in favor of \(H_a\).
04

Analysis of results

Even though all the sample cases give the same proportion of \(0.55\), the sample size changes. The larger the sample size, the smaller the standard deviation (\(\sqrt{p(1-p)/n}\)), thus leading to a larger test statistic. Consequently, the strength of evidence in favor of the alternative hypothesis increases with the sample size.

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Key Concepts

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

Understanding the P-value
In hypothesis testing, the p-value is a critical concept that indicates the probability of observing results at least as extreme as the ones obtained, under the assumption that the null hypothesis is true. The lower the p-value, the stronger the evidence that we should reject the null hypothesis. For instance, if we get a p-value less than the significance level (e.g., 0.05), this suggests that the observed effect is statistically significant, prompting us to consider the alternative hypothesis as more likely to be true.
  • A p-value of 0.05 or less typically indicates strong evidence against the null hypothesis.
  • The p-value allows us to quantify the extremeness of our data, given the null hypothesis is true.
  • It's a tool to express the strength of the evidence against the null hypothesis in favor of the alternative.
In cases where multiple sample sizes are analyzed, like in the original exercise, each p-value will likely differ due to the change in sample size, thus influencing the conclusion regarding the significance of the results.
The Influence of Sample Size
Sample size plays a significant role in hypothesis testing, impacting the precision of our estimates and the strength of our conclusions. Larger sample sizes generally provide more precise estimates of population parameters, reduce the standard error, and thus tighten the confidence intervals.
  • As sample size increases, the estimator (like sample proportion) becomes more reliable.
  • The test statistic's magnitude typically grows with larger sample sizes, enhancing the evidence strength against the null hypothesis.
  • Smaller sample sizes might lead to higher variability and less decisive evidence.
In the given exercise, while the sample proportion remains constant at 0.55, the results differ due to varying sample sizes (100, 500, 1000). Larger samples (like 1000) usually provide significantly robust evidence supporting the alternative hypothesis, as indicated by a more pronounced test statistic and often a lower p-value.
Analyzing the Alternative Hypothesis
The alternative hypothesis ( H_a ) in hypothesis testing is a statement that proposes a potential outcome from the experiment that we aim to support. It's usually in contrast with the null hypothesis, which implies no effect or status quo.
  • The alternative hypothesis can be one-sided (as used in the exercise: p > 0.5 ) or two-sided ( p 鈮� 0.5 ).
  • The goal of testing is often to provide enough evidence to reject the null, thereby supporting the alternative hypothesis.
  • A higher test statistic, influenced by larger sample sizes, can provide stronger support for the alternative hypothesis.
In the given task, the alternative hypothesis suggests the probability is greater than 0.5, meaning we're seeking evidence that exceeds this threshold. Therefore, with larger sample sizes, greater evidence becomes accessible, making it more probable to reject the null in support of the hypothesis that p is indeed greater than 0.5.

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Most popular questions from this chapter

Data 4.3 on page 265 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, \(A\) or \(B\), should the chain get chickens from in order to avoid too high a level of arsenic?

Giving a Coke/Pepsi taste test to random people in New York City to determine if there is evidence for the claim that Pepsi is preferred.

Income East and West of the Mississippi For a random sample of households in the US, we record annual household income, whether the location is east or west of the Mississippi River, and number of children. We are interested in determining whether there is a difference in average household income between those east of the Mississippi and those west of the Mississippi. (a) Define the relevant parameter(s) and state the null and alternative hypotheses. (b) What statistic(s) from the sample would we use to estimate the difference?

Exercises 4.59 to 4.64 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 \(H_{a}: p>0.5\) Sample data: \(\hat{p}=30 / 50=0.60\) with \(n=50\)

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?
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