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Chapter 10: Problem 77

Many people have misconceptions about how profitable small, consistent investments can be. In a survey of 1010 randomly selected U.S. 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 correct amount is \(\$ 286,640\) ). Is there sufficient evidence to conclude that less than \(40 \%\) of U.S. adults are aware that such an investment would result in a sum of over \(\$ 100,000\) ? Test the relevant hypotheses using \(\alpha=.05\).

Short Answer

Expert verified
Yes, there is enough evidence to support the claim that less than 40% of U.S. adults know that investing $25 per week for 40 years at 7% annual return would yield more than $100,000.

Step by step solution

01

Formulate the null and alternative hypotheses

The null hypothesis \(H_0\) would be that 40 percent (.40) of people knows that the regular small investment over the 40 years would result in > $100,000. Formally, \(H_0: p = 0.40\). The alternative hypothesis \(H_A\) is that less than 40% of people know this, so formally, \(H_A: p < 0.40\).
02

Compute the sample proportion

To find the sample proportion, divide the number who responded affirmatively by the total number of respondents. It鈥檚 \(p' = 374/1010 = 0.37\). It shows that 37% (less than 40%) of surveyed population have this knowledge.
03

Check the sampling distribution

The sampling distribution should be approximately normally distributed. This can be tested using these inequalities: \(np_0 > 5\) and \(n(1 - p_0) > 5\). Both \(1010 * 0.40 = 404\) and \(1010 * 0.60 = 606\) are more than 5. So, the approximation of a normal distribution can be used.
04

Calculate Standard Error and Z-Score

We calculate the standard deviation (Standard Error - SE) of the sampling distribution by the formula: SE = \(\sqrt{[p_0 * (1 - p_0)]/n}\) where \(p_0\) is the population proportion under \(H_0\) and n is the total survey count. So, SE = \(\sqrt{ [(0.40 * 0.60) / 1010] } = 0.0156\). Then, we find the Z-score which is the number of standard deviations the observed sample proportion \(p'\) is from the expected population proportion \(p_0\). So, Z = \((p' - p_0) / SE = (0.37 - 0.40) / 0.0156 = -1.923\).
05

Making a decision

Then, check Z-table for the probability associated with -1.923, which is 0.0274. We see that the p-value is 0.0274 which is less than our chosen alpha level of 0.05. This means we reject the null hypothesis. That is, we have sufficient evidence to support the claim that less than 40% of U.S. adults are aware that such an investment 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.

Sample Proportion
The sample proportion is a crucial concept in hypothesis testing. It represents the fraction of the sample that exhibits a particular characteristic. In our exercise, the sample proportion, denoted as \( p' \), is calculated to determine how many among the surveyed individuals knew about the investment potential.
To find the sample proportion, divide the number of favorable responses by the total sample size. Here, 374 out of 1010 adults said that a small weekly investment could yield over $100,000. Mathematically, it is expressed as:
  • \( p' = \frac{374}{1010} = 0.37 \) or 37%.
This result shows us that only 37% of the surveyed group is aware of this investment outcome, which is less than the 40% benchmark we are testing against.
Understanding the concept of sample proportion is essential as it becomes a pivotal point in comparing to the hypothesized population proportion during statistical testing.
Standard Error
The standard error (SE) is a measure of the variability or spread of the sampling distribution. It plays a key role in hypothesis testing because it helps us understand the range within which our sample proportion might vary if we were to take another sample from the same population.
In a hypothesis test involving proportions, we calculate the standard error using the formula:
  • \[ SE = \sqrt{ \frac{p_0 (1 - p_0)}{n} } \]
  • where \( p_0 \) is the population proportion specified in the null hypothesis, and \( n \) is the sample size.
For our case, with \( p_0 = 0.40 \) and \( n = 1010 \), the standard error is:
  • \( SE = \sqrt{ \frac{0.40(1-0.40)}{1010} } = 0.0156 \)
This standard error tells us how much variability exists around the sample proportion, assisting us in calculating the Z-score for the test.
Z-Score
The Z-score is a statistic that describes how far a sample proportion deviates from the population proportion under the null hypothesis, in terms of standard errors. It helps us understand whether our observed sample proportion is significantly different from what is expected if the null hypothesis were true.
To find the Z-score, use the formula:
  • \[ Z = \frac{p' - p_0}{SE} \]
  • where \( p' \) is the sample proportion, \( p_0 \) is the population proportion, and SE is the standard error calculated previously.
In our survey, with \( p' = 0.37 \), \( p_0 = 0.40 \), and \( SE = 0.0156 \), the Z-score is:
  • \( Z = \frac{0.37 - 0.40}{0.0156} = -1.923 \)
A Z-score indicates how many standard errors the sample proportion is away from the hypothesized population proportion. A negative Z-score suggests that the sample proportion is smaller than the expected population proportion. By comparing this Z-score with critical values or using a p-value, we can make an informed decision regarding the null hypothesis.

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Most popular questions from this chapter

In a survey conducted by Yahoo Small Business. 1432 of 1813 adults surveyed said that they would alter their shopping habits if gas prices remain high (Associated Press, November 30,2005 ). The article did not say how the sample was selected, but for purposes of this exercise, assume that it is reasonable to regard this sample as representative of adult Americans. Based on these survey data, is it reasonable to conclude that more than three-quarters of adult Americans plan to alter their shopping habits if gas prices remain high?

A television manufacturer claims that (at least) \(90 \%\) of its TV sets will need no service during the first 3 years of operation. A consumer agency wishes to check this claim, so it obtains a random sample of \(n=100\) purchasers and asks each whether the set purchased needed repair during the first 3 years after purchase. Let \(p\) be the sample proportion of responses indicating no repair (so that no repair is identified with a success). Let \(\pi\) denote the true proportion of successes for all sets made by this manufacturer. The agency does not want to claim false advertising unless sample evidence strongly suggests that \(\pi<.9 .\) The appropriate hypotheses are then \(H_{0}: \pi=.9\) versus \(H_{a}: \pi<.9\). a. In the context of this problem, describe Type \(I\) and Type II errors, and discuss the possible consequences of each. b. Would you recommend a test procedure that uses \(\alpha=.10\) or one that uses \(\alpha=.01 ?\) Explain.

The article referenced in Exercise \(10.34\) also reported that 470 of 1000 randomly selected adult Americans thought that the quality of movies being produced was getting worse. a. Is there convincing evidence that fewer than half of adult Americans believe that movie quality is getting worse? Use a significance level of \(.05\). b. Suppose that the sample size had been 100 instead of 1000 , and that 47 thought that the movie quality was getting worse (so that the sample proportion is still . 47 ). Based on this sample of 100 , is there convincing evidence that fewer than half of adult Americans believe that movie quality is getting worse? Use a significance level of \(.05\). c. Write a few sentences explaining why different conclusions were reached in the hypothesis tests of Parts (a) and (b).

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.

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 false-positives 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. What aspect of the relationship between the probability of Type I and Type II errors is being described by 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?
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