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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 I. 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 \(.50 .\) 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 as representative of adult Americans.

Short Answer

Expert verified
The final answer to this problem will be based on the calculated p-value. Since the calculation of this p-value requires access to statistical tables or software, it cannot be precisely determined within this context. However, if the p-value is less than 0.05, the marketing manager's claim will be considered incorrect. If the p-value is greater than 0.05, there is insufficient evidence to refute the marketing manager's claim.

Step by step solution

01

Define Hypotheses

The first step in testing the marketing manager's claim is to define the null and alternative hypotheses. In this case, the null hypothesis (H0) would be that the population proportion is equal to 0.50, i.e., H0: p= 0.50. The alternative hypothesis (H1) would be that the population proportion is less than 0.50, i.e., H1: p < 0.50.
02

Test Statistic

The test statistic can be calculated using the formula: Z = (p̂ - p0) / sqrt((p0(1 - p0))/n), where p̂ is the sample proportion, p0 is the population proportion given by H0, and n is the sample size. Substituting the given values, we get Z = (0.46-0.50) / sqrt((0.50*(1 - 0.50))/1005).
03

Determine the p-value

The p-value associated with a given Z value reports the likelihood of finding a statistic as extreme as, or more extreme than, the observed statistic under the null hypothesis. It can be found using Z-tables or software. If the p-value is less than the significance level (usually chosen as 0.05), we reject the null hypothesis.
04

Conclusion

If the calculated p-value is less than the chosen significance level, we reject the null hypothesis and assert that the marketing manager's claim is not correct. Alternatively, if the p-value is greater than the significance level, we fail to reject the null hypothesis, suggesting that the marketing manager's claim could be correct.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis is a statement that implies no effect or no difference in the population. It is essentially a baseline against which we measure our results. For our exercise, the null hypothesis (denoted as \(H_0\)) suggests that the true population proportion of adult Americans who want car web access is \(0.50\).
This represents the claim that the proportion is similar enough to half, showing no significant difference from the proverbial midpoint.
In mathematical terms:
  • \(H_0: p = 0.50\)
Our goal is to test this claim against the evidence gathered from the sample.
Alternative Hypothesis
The alternative hypothesis provides the opposite stance to the null hypothesis. It proposes that there is a significant effect or difference. When conducting a hypothesis test, we wish to see if the data provides enough evidence to support this alternative perspective.
In our given example, the alternative hypothesis (denoted as \(H_1\) or \(H_a\)) suggests that the true population proportion is less than \(0.50\), contradicting the marketing manager's statement.
Mathematically, this is expressed as:
  • \(H_1: p < 0.50\)
This suggests that less than half of adult Americans want web access in their cars.
P-value
The p-value plays a crucial role in hypothesis testing, as it helps quantify the evidence against the null hypothesis. It provides the probability of observing the test results under the assumption that the null hypothesis is true.
A smaller p-value indicates stronger evidence in favor of the alternative hypothesis. In decision-making, we compare the p-value to a chosen significance level (usually \(0.05\)):
  • If \(\text{p-value} < 0.05\), we reject the null hypothesis.
  • If \(\text{p-value} \geq 0.05\), we fail to reject the null hypothesis.
In this exercise, the p-value will help us determine if the difference between the surveyed and claimed proportions is statistically significant.
Test Statistic
The test statistic is a standard value we calculate from our sample data, which follows a certain distribution under the null hypothesis. For proportion tests, like in this exercise, we often use the Z-statistic, which tells us how far our sample proportion is from the assumed population proportion under the null hypothesis.
The formula for the Z-statistic in testing proportions is:
  • \[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]
where:
  • \(\hat{p}\) is the sample proportion (\(0.46\) in this case).
  • \(p_0\) is the population proportion under the null hypothesis (\(0.50\)).
  • \(n\) is the sample size (1005).
Computing this will give us our test statistic, which we use to find our p-value.
Population Proportion
Population proportion is a parameter that indicates the fraction of the population that shares a certain characteristic. In hypothesis testing, we often compare the observed sample proportion (\(\hat{p}\)) with an expected population proportion (\(p\)) based on past claims or analysis.
In our example, the survey suggests that \(46\%\) of adult Americans are interested in car web access, while the hypothesized population proportion (from the null hypothesis) is \(50\%\).
Understanding and comparing these proportions through hypothesis tests helps in verifying claims about the population and provides insights into whether the observed sample accurately reflects the broader population trend.

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Most popular questions from this chapter

Water samples are taken from water used for cooling as it is being discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharged water is at most \(150^{\circ} \mathrm{F}\), there will be no negative effects on the river's ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge water temperature above \(150^{\circ} \mathrm{F}\), a scientist will take 50 water samples at randomly selected times and will record the water temperature of each sample. She will then use a \(z\) statistic $$ z=\frac{\bar{x}-150}{\frac{\sigma}{\sqrt{n}}} $$ to decide between the hypotheses \(H_{0}: \mu=150\) and \(H_{a}: \mu>150,\) where \(\mu\) is the mean temperature of discharged water. Assume that \(\sigma\) is known to be 10 . a. Explain why use of the \(z\) statistic is appropriate in this setting. b. Describe Type I and Type II errors in this context. \(c\). The rejection of \(H_{0}\) when \(z \geq 1.8\) corresponds to what value of \(\alpha\) ? (That is, what is the area under the \(z\) curve to the right of \(1.8 ?\) ) d. Suppose that the actual value for \(\mu\) is 153 and that \(H_{0}\) is to be rejected if \(z \geq 1.8 .\) Draw a sketch (similar to that of Figure 10.5 ) of the sampling distribution of \(\bar{x},\) and shade the region that would represent \(\beta\), the probability of making a Type II error. e. For the hypotheses and test procedure described, compute the value of \(\beta\) when \(\mu=153\). f. For the hypotheses and test procedure described, what is the value of \(\beta\) if \(\mu=160\) ? g. What would be the conclusion of the test if \(H_{0}\) is rejected when \(z \geq 1.8\) and \(\bar{x}=152.4\) ? What type of error might have been made in reaching this conclusion?

The paper titled "Music for Pain Relief" (The Cochrane Database of Systematic Reviews, April 19 , 2006 ) concluded, based on a review of 51 studies of the effect of music on pain intensity, that "Listening to music reduces pain intensity levels ... However, the magnitude of these positive effects is small, the clinical relevance of music for pain relief in clinical practice is unclear." Are the authors of this paper claiming that the pain reduction attributable to listening to music is not statistically significant, not practically significant, or neither statistically nor practically significant? Explain.

The article "Boy or Girl: Which Gender Baby Would You Pick?" (LiveScience, March 23. 2005 , www.livescience.com) summarized the findings of a study that was published in Fertility and Sterility. The LiveScience article makes the following statements: "When given the opportunity to choose the sex of their baby, women are just as likely to choose pink socks as blue, a new study shows" and "Of the 561 women who participated in the study, 229 said they would like to choose the sex of a future child. Among these 229 , there was no greater demand for boys or girls." These statements are equivalent to the claim that for women who would like to choose the baby's sex, the proportion who would choose a girl is 0.50 or \(50 \%\). a. The journal article on which the LiveScience summary was based ("Preimplantation Sex-Selection Demand and Preferences in an Infertility Population," Fertility and Sterility [2005]: \(649-658\) ) states that of the 229 women who wanted to select the baby's sex, 89 wanted a boy and 140 wanted a girl. Does this provide convincing evidence against the statement of no preference in the LiveScience summary? Test the relevant hypotheses using \(\alpha=\) .05. Be sure to state any assumptions you must make about the way the sample was selected in order for your test to be appropriate. b. The journal article also provided the following information about the study: \- A survey with 19 questions was mailed to 1385 women who had visited the Center for Reproductive Medicine at Brigham and Women's Hospital. \- 561 women returned the survey. Do you think it is reasonable to generalize the results from this survey to a larger population? Do you have any concerns about the way the sample was selected or about potential sources of bias? Explain.

The international polling organization Ipsos reported data from a survey of 2000 randomly selected Canadians who carry debit cards (Canadian Account Habits Survey, July 24, 2006 ). Participants in this survey were asked what they considered the minimum purchase amount for which it would be acceptable to use a debit card. Suppose that the sample mean and standard deviation were \(\$ 9.15\) and \(\$ 7.60\), respectively. (These values are consistent with a histogram of the sample data that appears in the report.) Do these data provide convincing evidence that the mean minimum purchase amount for which Canadians consider the use of a debit card to be appropriate is less than \(\$ 10\) ? Carry out a hypothesis test with a significance level of .01 .

For the following pairs, indicate which do not comply with the rules for setting up hypotheses, and explain why: a. \(H_{0}: \mu=15, H_{a}: \mu=15\) b. \(H_{0}: p=.4, H_{a}: p>.6\) c. \(H_{0}: \mu=123, H_{a}: \mu<123\) d. \(H_{0}: \mu=123, H_{d}: \mu=125\) e. \(\quad H_{0}: \hat{p}=.1, H_{a}: \hat{p} \neq .1\)

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