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In 2016 , the National Foundation for Credit Counseling released a report titled "The 2016 Consumer \(\begin{array}{lll}\text { Financial Literacy } & \text { Survey" } & \text { (www.nfcc.org, retrieved }\end{array}\) December 1,2016\() .\) In a nationally representative sample of 1668 adult Americans, 965 indicated that they had checked their credit score within the last 12 months. Is there convincing evidence that a majority of adult Americans have checked their credit scores within the last 12 months? Test the relevant hypotheses using \(\alpha=0.05\).

Short Answer

Expert verified
In conclusion, after performing a hypothesis test with a significance level of 伪 = 0.05, we found convincing evidence that a majority of adult Americans have checked their credit scores within the last 12 months. We rejected the null hypothesis that stated the majority of adult Americans have NOT checked their credit scores within the last 12 months in favor of the alternative hypothesis.

Step by step solution

01

State the null and alternative hypothesis

Let p represent the proportion of adult Americans who have checked their credit scores within the last 12 months. The null hypothesis (H鈧) is that the majority (more than 50%) of adult Americans have NOT checked their credit scores within the last 12 months, and the alternative hypothesis (H鈧) is that the majority of adult Americans have checked their credit scores within the last 12 months. H鈧: p 鈮 0.5 H鈧: p > 0.5
02

Test statistic

Calculate the test statistic for the sample proportion. Let p-hat be the sample proportion (number of people who checked their credit score within the last 12 months divided by the total sample size). p-hat = 965/1668 = 0.5784 The test statistic, z, is given by: z = (p-hat - p鈧) / sqrt((p鈧 * (1 - p鈧)) / n) where p鈧 is the null hypothesis proportion (0.5) and n is the sample size (1668). Plug the values into the formula: z = (0.5784 - 0.5) / sqrt((0.5 * (1 - 0.5)) / 1668) = 5.831
03

Calculate the p-value

In this case, we have a right-tailed (greater than) test, so we need to find the area to the right of the test statistic, z = 5.831, using the standard normal distribution table or a calculator. The p-value is given by: P(Z > 5.831) = 1 - P(Z 鈮 5.831) 鈮 0.
04

Make a decision

Compare the p-value to the significance level (伪 = 0.05). If the p-value is less than or equal to 伪, reject the null hypothesis. If the p-value is greater than 伪, fail to reject the null hypothesis. In this case, the p-value (0) is less than the significance level (0.05), so we reject the null hypothesis.
05

State the conclusion

Since we rejected the null hypothesis, there is convincing evidence that a majority of adult Americans have checked their credit scores within the last 12 months.

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Key Concepts

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

Proportion Test
A proportion test is a statistical method used to determine if a specific proportion of a population meets a stated hypothesis. In our exercise, the focus is on whether the majority of adult Americans have checked their credit scores in the last year. This involves examining the sample data (1,668 participants in total) to estimate the proportion who did check their credit scores (965 participants), which results in a sample proportion of approximately 0.5784.

Key aspects of a proportion test include:
  • Calculating the sample proportion (\( \hat{p}\ = \frac{965}{1668} \approx 0.5784 \)
  • Establishing clear hypotheses (null and alternative)
  • Assessing the evidence through a test statistic and p-value
  • Making conclusions about the population based on sample data
This structured approach helps in determining whether observed data supports the hypothesis that a certain proportion of a population exhibits a particular behavior.
Null Hypothesis
The null hypothesis is a starting assumption in hypothesis testing that suggests no effect or no difference exists. It's a proposition that researchers aim to test against. In this context, the null hypothesis (\(H_0\) ) claims that 50% or less of adult Americans have checked their credit score in the past 12 months. This assumption serves as the default position, waiting to be challenged by data.

It's vital to frame this hypothesis clearly, as it underpins the statistical test. The goal is to see if observed data provides sufficient evidence to reject it. If evidence is inadequate to reject the null hypothesis, it remains as a plausible explanation of the population behavior.

In mathematical terms for our exercise, this is represented as:
  • \(H_0: p \leq 0.5\)
Where \(p\) is the true proportion of all adult Americans that checked their credit score.
Alternative Hypothesis
The alternative hypothesis offers the contrasting claim to the null, suggesting that a particular effect or difference does exist. In this exercise, the alternative hypothesis (\(H_1\) ) asserts that more than 50% of adult Americans have checked their credit score in the last year. This is what researchers wish to find evidence for through statistical testing.

The formulation of the alternative hypothesis guides the direction of our test; it determines whether we use a one-tailed or two-tailed test. Here it's a one-tailed test since we are only interested in whether the majority (more than 50%) of Americans have checked their scores.

Numerically, the alternative hypothesis for this context is framed as:
  • \(H_1: p > 0.5\)
This suggests a proportion greater than 0.5 signifies a majority. The alternative hypothesis guides the statistical process and conclusion drawn from the data.
Test Statistic
The test statistic is a standardized value that helps determine the significance of our observed results, compared to the null hypothesis. It translates our sample data into a single number that can be compared against a statistical distribution. For this example, a z-test statistic is used because we're dealing with proportions.

Here's how it's computed:
  • Determine the sample proportion \(\hat{p} = 0.5784\)
  • Identify the null hypothesis proportion \(p_0 = 0.5\)
  • Calculate the z-score using the formula:
    \[ z = \frac{\hat{p} - p_0}{\sqrt{ \frac{p_0(1-p_0)}{n}}} \]
    \[ z = \frac{0.5784 - 0.5}{\sqrt{ \frac{0.5(1-0.5)}{1668}}} \approx 5.831 \]
This z-value indicates how many standard deviations our observed proportion is from the null hypothesis proportion. A higher absolute value of the test statistic suggests that the sample proportion is significantly different from \(p_0\), leading to potential rejection of the null hypothesis.

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Most popular questions from this chapter

In the report "Healthy People 2020 Objectives for the Nation," The Centers for Disease Control and Prevention (CDC) set a goal of 0.341 for the proportion of mothers who will still be breastfeeding their babies one year after birth (www.cdc.gov/breastfeeding/policy /hp2020.htm, April 11, 2016, retrieved November 28, 2016). The CDC also estimated the proportion who were still being breastfed one year after birth to be 0.307 for babies born in 2013 (www.cdc.gov/breastfeeding /pdf/2016breastfeedingreportcard.pdf, retrieved November 28,2016) . This estimate was based on a survey of women who had given birth in 2013 . Suppose that the survey used a random sample of 1000 mothers and that you want to use the survey data to decide if there is evidence that the goal is not being met. Let \(p\) denote the population proportion of all mothers of babies born in 2013 who were still breast-feeding at 12 months. (Hint: See Example \(10.10 .)\) a. Describe the shape, center, and variability of the sampling distribution of \(\hat{p}\) for random samples of size 1000 if the null hypothesis \(H_{0}: p=0.341\) is true. b. Would you be surprised to observe a sample proportion as small as \(\hat{p}=0.333\) for a sample of size 1000 if the null hypothesis \(H_{0}: p=0.341\) were true? Explain why or why not. c. Would you be surprised to observe a sample proportion as small as \(\hat{p}=0.310\) for a sample of size 1000 if the null hypothesis \(H_{0}: p=0.341\) were true? Explain why or why not. d. The actual sample proportion observed in the study was \(p=0.307\). Based on this sample proportion, is there convincing evidence that the goal is not being met, or is the observed sample proportion consistent with what you would expect to see when the null hypothesis is true? Support your answer with a probability calculation.

Public Policy Polling conducts an annual poll on sportsrelated issues. In 2015 , they found that in a sample of 1222 adult Americans, 794 said that they thought the designated hitter rule in professional baseball should be eliminated and that pitchers should be required to bat (www.publicpolicypolling.com/pdf/2015/PPP_Release_National_51216.pdf, retrieved December 1,2016 ). Suppose that this sample is representative of adult Americans. Based on the given information, is there convincing evidence that a majority of adult Americans think that the designated hitter rule should be eliminated and that pitchers should be required to bat?

Let \(p\) denote the proportion of students at a large university who plan to purchase a campus meal plan in the next academic year. For a large-sample \(z\) test of \(H_{0}: p=0.20\) versus \(H_{i}: p<0.20,\) find the \(P\) -value associated with each of the following values of the \(z\) test statistic. (Hint: See page 496.) a. -0.55 b. -0.92 c. -1.99 d. -2.24 e. 1.40

Refer to the instructions prior to Exercise \(10.90 .\) The paper "I Smoke but I Am Not a Smoker" ( Journal of American College Health [2010]: \(117-125\) ) describes a survey of 899 college students who were asked about their smoking behavior. Of the students surveyed, 268 classified themselves as nonsmokers, but said "yes" when asked later in the survey if they smoked. These students were classified as "phantom smokers" meaning that they did not view themselves as smokers even though they do smoke at times. The authors were interested in using these data to determine if there is convincing evidence that more than \(25 \%\) of college students fall into the phantom smoker category.

The article "Public Acceptability in the UK and the USA of Nudging to Reduce Obesity: The Example of Reducing Sugar-Sweetened Beverages" (PLOS One, June 8,2016 ) describes a survey in which each person in a representative sample of 1082 adult Americans was asked about whether they would find different types of interventions acceptable in an effort to reduce consumption of sugary beverages. When asked about a tax on sugary beverages, 459 of the people in the sample said they thought that this would be an acceptable intervention. These data were used to test \(H_{0}: p=0.5\) versus \(H_{a^{*}}: p<0.5\) and the null hypothesis was rejected. a. Based on the hypothesis test, what can you conclude about the proportion of adult Americans who think that taxing sugary beverages is an acceptable intervention in an effort to reduce consumption of sugary beverages? b. Is it reasonable to say that the data provide strong support for the alternative hypothesis? c. Is it reasonable to say that the data provide strong evidence against the null hypothesis?

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