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Chapter 10: Problem 72

Many people have misconceptions about how profitable small, consistent investments can be. In a survey of 1,010 randomly selected American adults (Associated Press, October 29,1999 ), only 374 responded that they thought that an investment of \(\$ 25\) per week over 40 years with a \(7 \%\) annual return would result in a sum of over \(\$ 100,000\) (the actual amount would be \(\$ 286,640\) ). Is there convincing evidence that less than \(40 \%\) of U.S. adults think that such an investment would result in a sum of over \(\$ 100,000\) ? Test the relevant hypotheses using \(\alpha=0.05\).

Short Answer

Expert verified
There is insufficient evidence at the 0.05 level of significance to conclude that less than 40% of American adults believe that a regular $25 per week investment over 40 years with a 7% return would yield more than $100,000.

Step by step solution

01

Set up the hypothesis

The null hypothesis \(H_0\) is that 40% (or 0.4) of U.S adults think such an investment would yield over $100,000. Thus, \(H_0 : p = 0.4\). The alternative hypothesis \(H_a\) is that less than 40% of U.S. adults think so. Thus, \(H_a : p < 0.4\). p represents the population proportion.
02

Compute the sample proportion

The sample proportion \(\hat{p}\) is the number of successes (those who believe the investment yields over $100,000) divided by the total number in the sample. So, \(\hat{p} = \frac{374}{1010} = 0.37\).
03

Compute the Test Statistic

We compute the Z-score test statistic using the formula \(Z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}\), where n is the sample size. Substituting the appropriate values, \(Z = \frac{0.37 - 0.4}{\sqrt{\frac{0.4(1-0.4)}{1010}}}\), and calculating we find that the value of the Z statistic is approximately -1.36.
04

Find the P-value and make a decision

We now find the p-value corresponding to the observed value of the Z statistic. The p-value is the probability of observing a Z statistic as extreme as -1.36 or more, assuming the null hypothesis is true. From the standard normal distribution, the p-value is approximately 0.087. Since this p-value is greater than the significance level of 0.05, we fail to reject the null hypothesis.
05

Draw the Conclusion

There is not enough evidence at the 0.05 level of significance to support the claim that less than 40% of U.S. adults believe that an investment of $25 per week over 40 years with a 7% annual return would result in a sum of over $100,000.

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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 about the population that we assume to be true until proven otherwise. It's denoted as H0 and in the context of our exercise, it asserts that at least 40% of U.S. adults believe that an investment of \(25 per week over 40 years with a 7% return would yield over \)100,000. Formally, we write it as H0 : p = 0.4. Establishing a null hypothesis is critical as it provides a basis for comparison against our sample findings and helps in determining whether to support or reject it based on statistical analysis.

The null hypothesis is not a 'guess' but a presumption that there is no effect or no difference, and it's crucial to disprove it with evidence from our sample to assert that an alternative scenario is more plausible.
Alternative Hypothesis
Contrasting with the null hypothesis, we have the alternative hypothesis, represented as Ha or H1. It serves as a statement that we believe to be true if the null hypothesis is rejected. For our survey problem, the alternative hypothesis proposes that less than 40% of U.S. adults think a \(25 weekly investment would grow beyond \)100,000 after 40 years. It is expressed mathematically as Ha : p < 0.4.

Alternative hypotheses pinpoint the expected direction of the expected outcome if the null hypothesis is deemed invalid. They are often what the research or question is aiming to demonstrate or support, making them pivotal to hypothesis testing.
Sample Proportion
The sample proportion, denoted as \(\bar{p}\bar{p}\bar{p}\) ), is a statistic that measures the fraction of individuals in a sample that have a particular attribute of interest. In the given problem, this proportion is the number of people who believe the investment exceeds $100,000 divided by the total sample size: \bar{p} = \frac{374}{1010} = 0.37. It is pivotal as a point estimate of the true population proportion. Understanding sample proportions allows us to use a group's response to infer insights about the overall population's beliefs or behaviors.

When we conduct tests such as a Z-test or T-test, the sample proportion helps build the test statistic which determines how far our sample outcome is from what is stipulated by the null hypothesis.
Test Statistic
The test statistic is a calculated value that falls within a known distribution and is used to assess the evidence against the null hypothesis. In the provided exercise, a Z-test is appropriate, and we calculate the Z-score as such: Z = \frac{\(\bar{p} - p\)}{\sqrt{\frac{p(1-p)}{n}}} where \(\bar{p}\) is the sample proportion, p is the null hypothesis proportion, and n is the sample size. This statistic tells us how many standard deviations our sample proportion is from the hypothesized population proportion.

By comparing our test statistic to a critical value or using it to determine a p-value, we can make informed decisions on whether to retain or reject our null hypothesis.
P-value
The p-value is a fundamental concept in hypothesis testing. It represents the probability of observing our sample results, or more extreme, considering the null hypothesis is true. In simpler terms, it answers the question, 'If the null hypothesis were true, how strange is our data?' For our example, the p-value is approximately 0.087. If this value is lower than our chosen significance level, we have enough evidence to reject the null hypothesis.

A common mistake is to misinterpret the p-value as the probability that the null hypothesis is true. Instead, it quantifies how extreme the observed data is under the null hypothesis - a small p-value indicates a higher incompatibility of the null hypothesis with the observed data.
Significance Level
The significance level, denoted by 伪, is a threshold chosen by the researcher to decide whether to reject the null hypothesis. It is the probability of making a Type I error, which is rejecting the null hypothesis when it is actually true. In the example provided, a significance level of 伪 = 0.05 is used. This means there is a 5% risk of concluding that a difference exists when there is none. It is a predefined parameter that researchers choose based on the desired confidence of their conclusions.

A smaller 伪 value indicates a more stringent criterion for claiming a result to be statistically significant, thus reducing the risk of a Type I error. When the p-value exceeds the significance level, we fail to reject the null hypothesis, suggesting insufficient evidence to support the alternative hypothesis.

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Most popular questions from this chapter

Assuming a random sample from a large population, for which of the following null hypotheses and sample sizes is the large-sample \(z\) test appropriate? a. \(H_{0}: p=0.8, n=40\) b. \(H_{0}: p=0.4, n=100\) c. \(H_{0}: p=0.1, n=50\) d. \(H_{0}: p=0.05, n=750\)

The article "The Benefits of Facebook Friends: Social Capital and College Students' Use of Online Social Network Sites" (Journal of Computer-Mediated Communication [2007]: \(1143-1168\) ) describes a study of \(n=286\) undergraduate students at Michigan State University. Suppose that it is reasonable to regard this sample as a random sample of undergraduates at Michigan State. You want to use the survey data to decide if there is evidence that more than \(75 \%\) of the students at this university have a Facebook page that includes a photo of themselves. Let \(p\) denote the proportion of all Michigan State undergraduates who have such a page. (Hint: See Example 10.10\()\) a. Describe the shape, center, and spread of the sampling distribution of \(\hat{p}\) for random samples of size 286 if the null hypothesis \(H_{0}: p=0.75\) is true. b. Would you be surprised to observe a sample proportion as large as \(\hat{p}=0.83\) for a sample of size 286 if the null hypothesis \(H_{0}: p=0.75\) were true? Explain why or why not. c. Would you be surprised to observe a sample proportion as large as \(\hat{p}=0.79\) for a sample of size 286 if the null hypothesis \(H_{0}: p=0.75\) were true? Explain why or why not. d. The actual sample proportion observed in the study was \(\hat{p}=0.80 .\) Based on this sample proportion, is there convincing evidence that the null hypothesis \(H_{0}: p=\) 0.75 is not true, or is \(\hat{p}\) consistent with what you would expect to see when the null hypothesis is true? Support your answer with a probability calculation.

Describe the two types of errors that might be made when a hypothesis test is carried out.

The article "Cops Get Screened for Digital Dirt" (USA Today, Nov. 12,2010 ) summarizes a report on law enforcement agency use of social media to screen applicants for employment. The report was based on a survey of 728 law enforcement agencies. One question on the survey asked if the agency routinely reviewed applicants' social media activity during background checks. For purposes of this exercise, suppose that the 728 agencies were selected at random and that you want to use the survey data to decide if there is convincing evidence that more than \(25 \%\) of law enforcement agencies review applicants' social media activity as part of routine background checks. a. Describe the shape, center, and spread of the sampling distribution of \(\hat{p}\) for samples of size 728 if the null hypothesis \(H_{0}: p=0.25\) is true. b. Would you be surprised to observe a sample proportion as large as \(\hat{p}=0.27\) for a sample of size 728 if the null hypothesis \(H_{0}: p=0.25\) were true? Explain why or why not. c. Would you be surprised to observe a sample proportion as large as \(\hat{p}=0.31\) for a sample of size 728 if the null hypothesis \(H_{0}: p=0.25\) were true? Explain why or why not. d. The actual sample proportion observed in the study was \(\hat{p}=0.33 .\) Based on this sample proportion, is there convincing evidence that more than \(25 \%\) of law enforcement agencies review social media activity as part of background checks, or is this sample proportion consistent with what you would expect to see when the null hypothesis is true?

The article "Poll Finds Most Oppose Return to Draft, Wouldn't Encourage Children to Enlist" (Associated Press, December 18,2005 ) reports that in a random sample of 1,000 American adults, 700 indicated that they oppose the reinstatement of a military draft. Suppose you want to use this information to decide if there is convincing evidence that the proportion of American adults who oppose reinstatement of the draft is greater than two-thirds. a. What hypotheses should be tested in order to answer this question? b. The \(P\) -value for this test is \(0.013 .\) What conclusion would you reach if \(\alpha=0.05 ?\)
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