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Past experience is that when individuals are approached with a request to fill out and return a particular questionnaire in a provided stamped and addressed envelope, the response rate is \(40 \%\). An investigator believes that if the person distributing the questionnaire were stigmatized in some obvious way, potential respondents would feel sorry for the distributor and thus tend to respond at a rate higher than \(40 \%\). To test this theory, a distributor wore an eye patch. Of the 200 questionnaires distributed by this individual, 109 were returned. Does this provide evidence that the response rate in this situation is greater than the previous rate of \(40 \%\) ? State and test the appropriate hypotheses using a significance plevel of 0.05 .

Short Answer

Expert verified
In summary, using a one-sample proportion z-test, we found enough evidence to conclude that the response rate in this situation is indeed greater than 40% when the distributor is stigmatized, at a 0.05 significance level.

Step by step solution

01

State the null and alternative hypotheses

To test if the stigmatized distributor has a higher response rate (> 40%), we set up the following hypotheses: \(H_0\): The response rate is equal to 40% (p = 0.4) \(H_a\): The response rate is greater than 40% (p > 0.4)
02

Population parameter and sample estimate

The population parameter of interest is the proportion (p). The sample estimate is \(\hat{p}\), the proportion of questionnaires returned in the sample. In our case, the sample estimate (\(\hat{p}\)) is calculated as follows: \(\hat{p} = \frac{\text{Number of returned questionnaires}}{\text{Total questionnaires distributed}} = \frac{109}{200} = 0.545\)
03

Calculate the test statistic

We will perform a one-sample proportion z-test to determine if there is evidence to support that the response rate is indeed higher (> 40%) when the distributor is stigmatized. The test statistic (z) is calculated as: \(z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}\) Where \(p\) is the hypothesized response rate (0.4), \(\hat{p}\) is the calculated response rate in our sample (0.545), and \(n\) is the sample size (200). \(z = \frac{0.545 - 0.4}{\sqrt{\frac{0.4(1-0.4)}{200}}} \approx 4.46\)
04

Determine the p-value and compare with the significance level

Since this is a right-tailed test (we are testing for \(p > 0.4\)), we will calculate the p-value as: P(Z > 4.46) Using a Z-table or statistical software, we find that: P(Z > 4.46) ≈ 0.000004 Now, we compare this p-value to our chosen significance level (0.05): 0.000004 < 0.05
05

Make a decision and interpret the result

Since the p-value (0.000004) is smaller than the significance level (0.05), we reject the null hypothesis (\(H_0\)) in favor of the alternative hypothesis (\(H_a\)). This means that there is enough evidence to conclude that the response rate in this situation is indeed greater than 40% when the distributor is stigmatized, at a 0.05 significance level.

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Key Concepts

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

Response Rate
The response rate is a vital concept in statistical studies, particularly when measuring how many participants in a sample return feedback or data, such as questionnaires or surveys. It is given as a percentage, signifying the proportion of returned questionnaires compared to those distributed. In our original exercise, the previous known response rate is 40%. This percentage indicates that out of every 100 people approached, 40 are expected to return the completed questionnaire.

Response rates are crucial for ensuring the reliability of data in research. A higher response rate typically suggests more robust findings because it reduces the potential for non-response bias. Non-response bias can occur when the views of respondents differ significantly from those who did not respond, potentially skewing results.

In this exercise, the investigator is testing if altering conditions, such as the distributor wearing an eye patch, impacts the response rate significantly. Specifically, they were examining if the response rate would increase beyond the established 40% norm.
One-Sample Proportion Z-Test
The one-sample proportion z-test is a statistical method used to determine if a sample proportion is significantly different from a known population proportion. In this context, it is employed to see if the response rate of 54.5% (from the sample of distributed questionnaires) is significantly greater than the known rate of 40%.

The test involves several steps: formulating the null and alternative hypotheses, calculating the test statistic, and then interpreting the findings.
  • In step one, the null hypothesis (\( H_0 \)) states that the response rate is equal to the population proportion (40%), while the alternative hypothesis (\( H_a \)) proposes that it is greater.
  • The test statistic (\( z \)) is calculated using the formula: \( z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \) , where \( \hat{p} \) is the sample proportion (0.545), \( p \) is the population proportion (0.4), and \( n \) is the sample size (200).
The outcome of this calculation was a z-value of approximately 4.46, which is then compared against a standard normal distribution to infer statistical significance.
Significance Level
The significance level, often denoted as \( \alpha \), is a threshold set by the researcher to determine when to reject the null hypothesis. It reflects the degree of risk one is willing to take for a Type I error, which occurs when the null hypothesis is true, but we incorrectly reject it.

In hypothesis testing, if the p-value calculated from the test statistic is smaller than the significance level, it suggests that the observed effect is unlikely due to sampling variability alone. This prompts the rejection of the null hypothesis in favor of the alternative.

In our specific exercise, the significance level is set at 0.05, meaning there is a 5% risk of making a Type I error. Given the calculated p-value of approximately 0.000004, which is well below 0.05, the results are deemed statistically significant. Thus, the findings support the hypothesis that the response rate exceeds the previously known rate of 40% when the distributor is stigmatized. This level of significance provides strong evidence that the observed increase is not just by chance.

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Most popular questions from this chapter

In a survey of 1005 adult Americans, \(46 \%\) indicated that they were somewhat interested or very interested in having web access in their cars (USA TODAY, May 1,2009 ). Suppose that the marketing manager of a car manufacturer claims that the \(46 \%\) is based only on a sample and that \(46 \%\) is close to half, so there is no reason to believe that the proportion of all adult Americans who want car web access is less than \(0.50 .\) Is the marketing manager correct in his claim? Provide statistical evidence to support your answer. For purposes of this exercise, assume that the sample can be considered representative of adult Americans.

USA TODAY, (February 17, 2011) described a survey of 1008 American adults. One question on the survey asked people if they had ever sent a love letter using e-mail. Suppose that this survey used a random sample of adults and that you want to decide if there is evidence that more than \(20 \%\) of American adults have written a love letter using e-mail. a. Describe the shape, center, and variability of the sampling distribution of \(\hat{p}\) for random samples of size 1008 if the null hypothesis \(H_{0}: p=0.20\) is true. b. Based on your answer to Part (a), what sample proportion values would convince you that more than \(20 \%\) of adults have sent a love letter via e-mail?

The article titled "13\% of Americans Don't Use the Internet. Who Are They?" describes a study conducted by the Pew Research Center (pewrearch.org, September 7,2016 , retrieved December 1,2016 ). Suppose that the title of this article is based on a representative sample of 600 adult Americans. Does this support the claim that the proportion of adult Americans who do not use the Internet is greater than \(0.10(10 \%) ?\)

For which of the following \(P\) -values will the null hypothesis be rejected when performing a test with a significance level of \(0.05 ?\) a. 0.001 d. 0.047 b. 0.021 e. 0.148 c. 0.078

Medical personnel are required to report suspected cases of child abuse. Because some diseases have symptoms that are similar to those of child abuse, doctors who see a child with these symptoms must decide between two competing hypotheses: \(H_{0}:\) symptoms are due to child abuse \(H:\) symptoms are not due to child abuse (Although these are not hypotheses about a population characteristic, this exercise illustrates the definitions of Type I and Type II errors.) The article "Blurred Line Between Illness, Abuse Creates Problem for Authorities" (Macon Telegraph, February \(28,\) 2000 ) included the following quote from a doctor in Atlanta regarding the consequences of making an incorrect decision: "If it's disease, the worst you have is an angry family. If it is abuse, the other kids (in the family) are in deadly danger." a. For the given hypotheses, describe Type I and Type II errors. b. Based on the quote regarding consequences of the two kinds of error, which type of error is considered more serious by the doctor quoted? Explain.

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