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In a representative sample of adult Americans ages 26 to 32 years, \(27 \%\) indicated that they owned a fitness band that kept track of the number of steps walked each day and their daily activity levels ("Digital Democracy Survey", Deloitte Development LLC, 2016, www2, deloitte.com/us/en.html, retrieved November 30 , 2016). Suppose that the sample size was 500 . Is there convincing evidence that more than one-quarter of all adult Americans in this age group own a fitness band?

Short Answer

Expert verified
In this problem, we conducted a one-sample proportion hypothesis test to determine if more than one-quarter of adult Americans ages 26 to 32 years own a fitness band. The null hypothesis was that the proportion of adult Americans who own a fitness band is 0.25, with the alternative hypothesis being that the proportion is greater than 0.25. We calculated a test statistic of Z ≈ 1.13, which gave us a p-value of 0.129. Since the p-value is greater than the significance level of 0.05, we fail to reject the null hypothesis. This means there is not enough evidence to support the claim that more than one-quarter of all adult Americans in this age group own a fitness band.

Step by step solution

01

State the hypotheses

We will first state the null and alternative hypotheses. Null hypothesis (H0): p = 0.25 Alternative hypothesis (H1): p > 0.25 Here, p represents the population proportion of adult Americans in this age group who own a fitness band.
02

Calculate the test statistic

To calculate the test statistic, we will use the formula: Test statistic (Z) = \(\frac{(\hat{p} - p_0)}{\sqrt{\frac{p_0(1-p_0)}{n}}}\) where \(\hat{p}\) is the sample proportion, \(p_0\) is the proportion under the null hypothesis, and n is the sample size. In this case, \(\hat{p} = \frac{0.27 \times 500}{500} = 0.27\) \(p_0 = 0.25\) \(n = 500\) Now, we will calculate the test statistic (Z): Z = \(\frac{(0.27 - 0.25)}{\sqrt{\frac{0.25(1-0.25)}{500}}}\) Z ≈ 1.13
03

Determine the p-value

Since our alternative hypothesis is p > 0.25, we are performing a one-tailed test. To determine the p-value, we will find the area under the standard normal curve to the right of the Z value. Using a Z-table or calculator, we find the p-value: p-value ≈ 0.129
04

Compare the p-value to the significance level

In order to determine if there is convincing evidence to reject the null hypothesis, we will compare the p-value to the significance level. In this problem, we are not given a significance level, so we will use the common value of 0.05. Since the p-value (0.129) is greater than the significance level (0.05), we fail to reject the null hypothesis.
05

Draw conclusions

We fail to reject the null hypothesis, meaning that there is not enough evidence to support the claim that more than one-quarter of all adult Americans in this age group (26 to 32 years old) own a fitness band.

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Key Concepts

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

Null Hypothesis
When diving into the world of hypothesis testing, one fundamental concept to grasp is the null hypothesis, symbolized as \(H_0\). The null hypothesis is a statement of no effect or no difference and is presumed true until evidence suggests otherwise. It establishes a benchmark for testing if there is significant evidence to support an alternative view.

In our exercise, the null hypothesis claims that the true proportion (denoted as \(p\)) of adult Americans aged 26-32 who own a fitness band is 25%, expressed as \(H_0: p = 0.25\). It's the status quo or the assumption that no change has occurred. When calculating the test statistic and determining the p-value, it's the null hypothesis that we're essentially putting on trial, looking for evidence strong enough to reject it in favor of the alternative hypothesis.
Alternative Hypothesis
Contrasting the null hypothesis, the alternative hypothesis \(H_1\) or \(H_a\) represents what we're trying to provide evidence for. It is a statement that indicates a new effect or a difference from what has been traditionally accepted. It's our research hypothesis - the claim we hope the data will support.

In the context of our provided exercise, the alternative hypothesis proposes that more than 25% of adult Americans in the specified age group own a fitness band (expressed as \(H_1: p > 0.25\)). In hypothesis testing, if the evidence is strong enough to reject the null hypothesis, we can then embrace the alternative hypothesis.
Test Statistic
The test statistic is a crucial part of hypothesis testing as it enables us to measure the distance of the sample statistic from the hypothesized parameter assuming the null hypothesis is true. It's essentially the standardized value that signals how far off our sample result is if the null hypothesis holds.

For example, in our exercise scenario, we calculated the test statistic (Z) using the formula: \[Z = \frac{(\hat{p} - p_0)}{\sqrt{\frac{p_0(1-p_0)}{n}}}\] This value helps us assess whether the observed sample proportion (\(\hat{p}\)=27%) is significantly different from the null hypothesis value of \(p_0\)=25% given the sample size (\(n=500\)). The resultant Z-value then guides us in determining the p-value.
P-value
One of the most pivotal concepts in hypothesis testing is the p-value. It serves as a bridge between the test statistic and our decision on the hypothesis. The p-value represents the probability of obtaining results at least as extreme as the observed results, given that the null hypothesis is true.

In the exercise, we seek the probability of finding a sample proportion of fitness band owners as high as we did (or higher) if, in reality, only 25% of the population own them. We find that the p-value for our test statistic (Z ≈ 1.13) is approximately 0.129. This means that there's a 12.9% chance of observing such data if the null hypothesis is correct. P-values are compared to the significance level to make the final decision in hypothesis testing.
Significance Level
The significance level, noted as \(\alpha\), is a threshold used to decide whether the p-value provides strong enough evidence against the null hypothesis. It's pre-determined by the researcher and represents the risk we are willing to take of incorrectly rejecting a true null hypothesis - known as a type I error.

Often, a significance level of 0.05 is used, meaning there's a 5% risk of wrongly rejecting the null hypothesis. In the exercise, if the p-value is less than the significance level, we would have sufficient evidence to reject \(H_0\) and accept \(H_1\). However, with a p-value of 0.129, which is greater than the typical significance level of 0.05, we do not have enough evidence to reject the null hypothesis. The outcome is that we fail to reject \(H_0\), suggesting that we cannot confidently claim that more than a quarter of the specified population owns a fitness band.

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Most popular questions from this chapter

The paper "Teens and Distracted Driving"" (Pew Internet \& American Life Project, 2009 ) reported that in a representative sample of 283 American teens age 16 to \(17,\) there were 74 who indicated that they had sent a text message while driving. For purposes of this exercise, assume that this sample is a random sample of 16- to 17 -year-old Americans. Do these data provide convincing evidence that more than a quarter of Americans age 16 to 17 have sent a text message while driving? Test the appropriate hypotheses using a significance level of 0.01 . (Hint: See Example 10.11 .)

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 falsepositives 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. Recall 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. What aspect of the relationship between the probability of a Type I error and the probability of a Type II error is being described here?

In a survey of 1000 women age 22 to 35 who work full-time, 540 indicated that they would be willing to give up some personal time in order to make more money (USA TODAY, March 4,2010 ). The sample was selected to be representative of women in the targeted age group. a. Do the sample data provide convincing evidence that a majority of women age 22 to 35 who work fulltime would be willing to give up some personal time for more money? Test the relevant hypotheses using \(\alpha=0.01\) b. Would it be reasonable to generalize the conclusion from Part (a) to all working women? Explain why or why not.

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. \(H_{0}: p=0.05, n=750\)

Occasionally, warning flares of the type contained in most automobile emergency kits fail to ignite. A consumer group wants to investigate a claim that the proportion of defective flares made by a particular manufacturer is higher than the advertised value of \(0.10 .\) A large number of flares will be tested, and the results will be used to decide between \(H_{0}: p=0.10\) and \(H_{a}: p>0.10,\) where \(p\) represents the actual proportion of defective flares made by this manufacturer. If \(H_{0}\) is rejected, charges of false advertising will be filed against the manufacturer. a. Explain why the alternative hypothesis was chosen to be \(H: p>0.10 .\) b. Complete the last two columns of the following table. (Hint: See Example 10.7 for an example of how this is done.)

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