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Chapter 10: Problem 32

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 .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
Based on the calculated p-value which is less than the commonly used significance level of \(0.05\), we reject the null hypothesis and conclude that there is statistical evidence to suggest that the proportion of adult Americans who want access to the web in their cars is less than \(50%\). Thus, the marketing manager's claim is statistically not valid.

Step by step solution

01

Define the hypotheses

First, we define our null hypothesis (H0) and alternative hypothesis (Ha). The null hypothesis represents the marketing manager's claim, and the alternative hypothesis will represent the statement we want to test against. Here, H0: \(p >= 0.50\) and Ha: \(p < 0.50\), where \(p\) is the proportion of all adult Americans who want car web access.
02

Assume the null hypothesis to calculate the test statistic

Assuming the null hypothesis is true, we have \(p = 0.50\). The test statistics for the claim is calculated using the Z-test formula for one sample proportion which is given by \(Z = (X - np_0)/sqrt(np_0q_0)\), where \(n\) is the total number of observations, in this case \(1005\), \(X\) is the number of successful observations, in this case \(46%\) of \(1005\) which equals \(462.3\), and \(p_0\) and \(q_0\) are the assumed successes and failures under the null hypothesis, \(p_0 = 0.5\) and \(q_0 = 1 - p_0 = 0.5\) respectively.
03

Calculate the test statistic

Inserting the values into the formula, we get \(Z = (462.3 - 1005*0.5)/sqrt(1005*0.5*0.5) 鈮� -1.76\)
04

Find the p-value

The Z score tells us how many standard deviations an element is from the mean. A negative Z score suggests the raw score is below the population mean. Now, we will use a Z-table (or other statistical software or calculators) to find the p-value associated with this Z score. The p-value for Z = -1.76 is approximately \(0.0392\)
05

Draw conclusions

We generally reject the null hypothesis if the p-value is less than a predetermined significance level, which is often \(0.05\). Since the p-value we calculated is approximately \(0.0392\), which is less than \(0.05\), we reject the null hypothesis. This suggests there is statistical evidence that the proportion of adult Americans who want access to the web in their cars is less than \(50%\) with a level of confidence typical for statistical analysis, thus the marketing manager is not correct in his estimation.

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Key Concepts

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

Null Hypothesis
The null hypothesis, represented as \(H_0\), is a foundational concept in hypothesis testing. It posits that there is no effect or difference, and it's what we assume to be true until we have enough evidence to suggest otherwise. In this exercise, the null hypothesis states that the true proportion \(p\) of adult Americans who want web access in their cars is greater than or equal to 0.50. This aligns with the marketing manager's claim.
Alternative Hypothesis
The alternative hypothesis, \(H_a\), challenges the null hypothesis. It's what researchers hope to support with evidence. In our scenario, \(H_a: p < 0.50\). This means we are testing whether the proportion of adult Americans wanting car web access is actually less than 50%. The aim here is to gather proof that contradicts the null hypothesis and supports this alternative claim.
P-Value
The p-value quantifies the evidence against the null hypothesis. It's the probability of observing a test statistic at least as extreme as the one observed, assuming the null hypothesis is true. A smaller p-value indicates stronger evidence against the null hypothesis. In the context of this exercise, the calculated p-value is approximately 0.0392. Since this is less than the common significance level of 0.05, we have enough statistical evidence to reject the null hypothesis.
Z-Test
The Z-test is used to determine if there is a significant difference between sample and population proportions. It helps decide if we can reject the null hypothesis. We calculate the Z score using the formula:
  • \(Z = \frac{{X - np_0}}{{\sqrt{np_0q_0}}}\)
Here, \(X\) represents actual successes, \(n\) is the sample size, and \(p_0\) is the hypothesized proportion. In this exercise, the computed Z score was approximately -1.76, indicating the sample proportion is less than the hypothesized 0.50.
Sample Proportion
Sample proportion is the fraction or percentage of a certain outcome in a sample. It's a key part of hypothesis testing for proportions. In this scenario, the sample proportion is 0.46, representing 46% of the surveyed adults. This sample proportion acts as our observed statistic, which we compare against the hypothesized proportion of 50% to determine if there is significant statistical evidence to support the marketing manager's claim or to reject it.

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Most popular questions from this chapter

The Economist collects data each year on the price of a Big Mac in various countries around the world. The price of a Big Mac for a sample of McDonald's restaurants in Europe in May 2009 resulted in the following Big Mac prices (after conversion to U.S. dollars): \(\begin{array}{llllllll}3.80 & 5.89 & 4.92 & 3.88 & 2.65 & 5.57 & 6.39 & 3.24\end{array}\) The mean price of a Big Mac in the U.S. in May 2009 was \(\$ 3.57 .\) For purposes of this exercise, assume it is reasonable to regard the sample as representative of European McDonald's restaurants. Does the sample provide convincing evidence that the mean May 2009 price of a Big Mac in Europe is greater than the reported U.S. price? Test the relevant hypotheses using \(\alpha=.05\).

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According to a survey of 1000 adult Americans conducted by Opinion Research Corporation, 210 of those surveyed said playing the lottery would be the most practical way for them to accumulate \(\$ 200,000\) in net

An article titled "Teen Boys Forget Whatever It Was" appeared in the Australian newspaper The Mercury (April 21, 1997). It described a study of academic performance and attention span and reported that the mean time to distraction for teenage boys working on an independent task was 4 minutes. Although the sample size was not given in the article, suppose that this mean was based on a random sample of 50 teenage Australian boys and that the sample standard deviation was \(1.4\) minutes. Is there convincing evidence that the average attention span for teenage boys is less than 5 minutes? Test the relevant hypotheses using \(\alpha=.01\).

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