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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 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.

Short Answer

Expert verified
Our hypothesis test results show that the proportion of adult Americans wanting car web access is likely to be less than 50%. The marketing manager's claim that it is at least 0.50 is not supported by the data, as our p-value of 0.0049 is less than the common significance level of 0.05, providing enough evidence to reject the null hypothesis and accept the alternative hypothesis.

Step by step solution

01

Set up the null and alternative hypotheses

Our null hypothesis is that the proportion of adult Americans who want web access in their cars is at least 50%, and the alternative hypothesis is that the proportion is less than 50%. \( H_0: p \geq 0.50 \) \( H_a: p < 0.50 \)
02

Find the sample proportion

From the survey, we know that 46% of the 1005 participants are interested in car web access. To find the sample proportion, divide the number of people interested by the total number of people in the sample. \( \hat{p} = \frac{46}{100} = 0.46 \)
03

Calculate the test statistic

To test the hypothesis, we will use a one-proportion z-test. The formula for the z-test statistic is: \( z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \) Here, \(p_0\) is the null proportion (0.50), and n is the sample size (1005). Plugging in the values we have in the formula, we get: \( z = \frac{0.46 - 0.50}{\sqrt{\frac{0.50(1-0.50)}{1005}}} = -2.58 \)
04

Find the p-value

Since we're performing a one-tailed test (with the alternative hypothesis being \(p < 0.50\)), we will find the area to the left of the test statistic on a standard normal distribution table (z-table). A z-score of -2.58 corresponds to a p-value of 0.0049.
05

Interpret the results

Our p-value is 0.0049, which is less than the common significance level of 0.05. This means that there is enough evidence to reject the null hypothesis and accept the alternative hypothesis. Therefore, our statistical evidence shows that the proportion of adult Americans wanting car web access is likely to be less than 50%. The marketing manager's claim that it is at least 0.50 is not supported by the data.

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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 represents a statement that there is no effect or no difference and generally reflects the current belief or default position about a population parameter. It is denoted as \( H_0 \). In our context, the null hypothesis suggests that the actual proportion of adult Americans who want web access in their cars is at least 50%. This means that the survey result of 46% could simply be random fluctuation in the data without indicating true preference change.

The null hypothesis is essential because it provides a baseline against which we measure the strength of evidence. If our statistical test finds enough evidence against the null hypothesis, we might conclude that the alternative explanation is more likely to be true. However, if the evidence isn't strong enough, we don't reject the null hypothesis. Here, our null hypothesis is \( H_0: p \geq 0.50 \)."},{
Alternative Hypothesis
The alternative hypothesis posits a different perspective from the null hypothesis and indicates the presence of an effect or difference. In our example, the alternative hypothesis is that the proportion of adult Americans who desire web access in their cars is less than 50%, represented by \( H_a: p < 0.50 \).

This hypothesis is typically what the researcher wants to prove. If we find sufficient evidence to reject the null hypothesis, we accept the alternative hypothesis. In statistical testing, the alternative hypothesis usually encompasses the new theory or claim that researchers wish to validate, relying on the data provided. The strength of hypothesis tests lies in determining whether the alternative hypothesis is more believable based on the sample data."},{
One-Proportion Z-Test
A one-proportion z-test is a statistical test used to determine whether there is evidence that a sample proportion is significantly different from a specified population proportion. In this case, it tests whether the sample proportion of 46%, from 1005 surveyed individuals, differs significantly from the 50% benchmark suggested by the null hypothesis.

To perform the test, a z-test statistic is calculated. This involves the observed sample proportion \( \hat{p} \), population proportion \( p_0 \) under the null hypothesis, and the sample size \( n \). The formula for the z-test statistic is:\[z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\]
The test statistic allows us to locate our observed value within the standard normal distribution, helping to determine the p-value. A smaller p-value suggests stronger evidence against the null hypothesis, supporting the claim that the true population proportion is less than hypothesized."},{
Sample Proportion
Sample proportion, denoted by \( \hat{p} \), is a critical metric in hypothesis testing with proportions. It represents the part of the sample that has a particular characteristic. In our scenario, the sample proportion is derived from people interested in car web access. With 46% of the 1005 respondents expressing interest, our sample proportion for this survey is \( \hat{p} = 0.46 \).

The sample proportion acts as an estimate of the corresponding population proportion. It is central in forming statistics due to its role in calculating the z-test statistic, measuring the deviation from the assumed population proportion. Accurate interpretation of the sample proportion is essential as it directly influences the hypothesis test's conclusions. Recognizing its value can aid in determining whether the observed data possibly represents a broader trend within the population."}]} patrick.stackexchange.com.addRowNextSteps.com.addInfo.com.addDetails..addStep.wikistack.com.addTapSteps.com.add patrick.stackexchange.com.addNextSteps.com.add.com.unwrap@editor.crm.marketing.slimpicker.vendor.com/addANTDAUT/pro-admin_settings=*/comsetCoalesce-apply-supportLink.com.GetDataLink.com鞐 蟺位畏蟻慰蠁慰蟻委蔚蟼 纬喂伪 蟿畏谓 蠀蟺畏蟻蔚蟽委伪 魏伪喂 蟿伪 蟽蠂蔚蟿喂魏维 未喂魏伪喂蠋渭伪蟿伪. 渭慰喂蟻维蟽慰蠀 渭委伪 蟽蠀谓蟿伪纬萎 魏伪喂 苇谓伪 渭萎谓蠀渭伪 蟽蠀谓蟿伪纬萎 纬喂伪 维位位慰蠀蟼- 蟿畏蟼 蟺委蔚蟽畏蟼 蠈蟿喂 畏 Associazione 蠈蟺蠅蟼 蔚委谓伪喂 伪位位喂蠋蟼, 蟿慰蠀 Swearealignment" 魏伪喂 蔚尉伪蟻蟿萎渭伪蟿慰蟼 伪蠀蟿萎蟼, 纬喂伪 蟽伪蟼, 蟿畏 未蔚蠉蟿蔚蟻畏 谓苇伪 蟽蠀谓蟿伪纬萎 蟽蟿慰 魏慰蠀蟿委 纬喂伪 谓伪 蟺蟻慰蟽胃苇蟽蔚喂 蟿慰 "蟺伪喂蠂谓委未喂"-serveLogin 伪蟺蠈 蟿慰 蠀蟺慰渭慰谓蔚蟿喂魏蠈 魏慰谓蟿蟻蠈位. 蟽蠀谓未苇蟽蔚喂蟼 伪蟺蠈 蟿慰 蟺蟻慰蟽蠅蟺喂魏蠈InfrastructureConnector-CompliSocialVisaPremium.com via.flowdok.gr/viewAllFeeds.cms牍勱祼鞖╇矔 於旍旮办偓霌れ潃 攴鸽Μ瓿 鞖半Μ鞚 鞁滌儊鞁 氇╈埁瓿 瓿勳暯鞚 於╈嫟旮 氚橂偐頃 雼轨嫚鞚 毹鸽ΜtogetherAltoMAX鞐怰elationshipTechie.getString(

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