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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

Occasionally, warning flares of the type contained in most automobile emergency kits fail to ignite. A consumer group wants to investigate a claim that the proportion of defective flares made by a particular manufacturer is higher than the advertised value of \(0.10 .\) A large number of flares will be tested, and the results will be used to decide between \(H_{0}: p=0.10\) and \(H_{a}: p>0.10,\) where \(p\) represents the actual proportion of defective flares made by this manufacturer. If \(H_{0}\) is rejected, charges of false advertising will be filed against the manufacturer. a. Explain why the alternative hypothesis was chosen to be \(H: p>0.10 .\) b. Complete the last two columns of the following table. (Hint: See Example 10.7 for an example of how this is done.)

A number of initiatives on the topic of legalized gambling have appeared on state ballots. A political candidate has decided to support legalization of casino gambling if he is convinced that more than two-thirds of American adults approve of casino gambling. Suppose that 1035 of the people in a random sample of 1523 American adults said they approved of casino gambling. Is there convincing evidence that more than two-thirds approve?

In a survey of 1000 women age 22 to 35 who work full-time, 540 indicated that they would be willing to give up some personal time in order to make more money (USA TODAY, March 4,2010 ). The sample was selected to be representative of women in the targeted age group. a. Do the sample data provide convincing evidence that a majority of women age 22 to 35 who work fulltime would be willing to give up some personal time for more money? Test the relevant hypotheses using \(\alpha=0.01\) b. Would it be reasonable to generalize the conclusion from Part (a) to all working women? Explain why or why not.

Researchers at the University of Washington and Harvard University analyzed records of breast cancer screening and diagnostic evaluations ("Mammogram Cancer Scares More Frequent Than Thought," USA TODAY, April 16,1998 ). Discussing the benefits and downsides of the screening process, the article states that although the rate of falsepositives is higher than previously thought, if radiologists were less aggressive in following up on suspicious tests, the rate of false-positives would fall, but the rate of missed cancers would rise. Suppose that such a screening test is used to decide between a null hypothesis of \(H_{0}:\) no cancer is present and an alternative hypothesis of \(H_{a}:\) cancer is present. (Although these are not hypotheses about a population characteristic, this exercise illustrates the definitions of Type I and Type II errors.) a. Would a false-positive (thinking that cancer is present when in fact it is not) be a Type I error or a Type II error? b. Describe a Type I error in the context of this problem, and discuss the consequences of making a Type I error. c. Describe a Type II error in the context of this problem, and discuss the consequences of making a Type II error. d. Recall the statement in the article that if radiologists were less aggressive in following up on suspicious tests, the rate of false-positives would fall but the rate of missed cancers would rise. What aspect of the relationship between the probability of a Type I error and the probability of a Type II error is being described here?

The paper "Teens and Distracted Driving"" (Pew Internet \& American Life Project, 2009 ) reported that in a representative sample of 283 American teens age 16 to \(17,\) there were 74 who indicated that they had sent a text message while driving. For purposes of this exercise, assume that this sample is a random sample of 16- to 17 -year-old Americans. Do these data provide convincing evidence that more than a quarter of Americans age 16 to 17 have sent a text message while driving? Test the appropriate hypotheses using a significance level of 0.01 . (Hint: See Example 10.11 .)

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