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Chapter 4: Problem 4

A situation is described for a statistical test and some hypothetical sample results are given. In each case: (a) State which of the possible sample results provides the most evidence for the claim. (b) State which (if any) of the possible results provide no evidence for the claim. Testing to see if there is evidence that the proportion of US citizens who can name the capital city of Canada is greater than \(0.75 .\) Use the following possible sample results: Sample A: 31 successes out of 40 Sample B: \(\quad 34\) successes out of 40 Sample C: 27 successes out of 40 Sample \(\mathrm{D}: \quad 38\) successes out of 40

Short Answer

Expert verified
The sample with the most evidence for the claim is sample D, while sample C provides no evidence for the claim.

Step by step solution

01

Calculate the Proportion

The proportion is simply the division of 'successes' by the total. So, for each of the given samples, compute the proportion as follows: \n\nSample A: \(\frac{31}{40} = 0.775\) \n\nSample B: \(\frac{34}{40} = 0.85\) \n\nSample C: \(\frac{27}{40} = 0.675\) \n\nSample D: \(\frac{38}{40} = 0.95\)
02

Identify the Sample with the Most Evidence

The claim is that the proportion of US citizens who can name the capital city of Canada is greater than 0.75. So, the closer the proportion to 1, the stronger the evidence for the claim. Comparing those values, Sample D with a proportion of 0.95 gives the most evidence.
03

Identify the Sample with the Least Evidence

Now, inspecting which sample provides no evidence or contradict the claim, it's noticed that only Sample C has a proportion less than 0.75. It means all other sample results provide some evidence supporting the claim. However, Sample C with 0.675 contradicts the claim, thus it would provide no evidence in support of the claim.

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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: \(H_{0}: p=0.6\) vs \(H_{a}: p>0.6\) Sample data: \(\hat{p}=52 / 80=0.65\) with \(n=80\)

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 \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1}>p_{2}\) using \(\hat{p}_{1}-\hat{p}_{2}=0.45-0.30=0.15\) with each of the following sample sizes: (a) \(\hat{p}_{1}=9 / 20=0.45\) and \(\hat{p}_{2}=6 / 20=0.30\) (b) \(\hat{p}_{1}=90 / 200=0.45\) and \(\hat{p}_{2}=60 / 200=0.30\) (c) \(\hat{p}_{1}=900 / 2000=0.45\) and \(\hat{p}_{2}=600 / 2000=0.30\)

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.

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 if a well-known company is lying in its advertising. If there is evidence that the company is lying, the Federal Trade Commission will file a lawsuit against them.

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?
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